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rich_web_development

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  1. rich_web_development
    Thanks for the help
  2. rich_web_development
    <script> var buttons = document.getElementsByClassName("add-panel"); for(var i = 0; i < buttons.length; i++) { buttons[i].addEventListener("click", addPanel, false); } var panels = []; function addPanel(e) { // Get a reference to the button that was clicked var button = e.currentTarget; var value = button.getAttribute("data-value"); if(panels.indexOf(value) == -1) { // Add the button's value to the array panels.push(value); // Display the contents of the array on the page by generating <li> elements. var demo = document.getElementById("demo"); demo.innerHTML = ""; var li; for(var i = 0; i < panels.length; i++) { li = document.createElement("li"); li.innerHTML = panels[i]; demo.appendChild(li); } } } </script> Thanks. Got it. "This method returns -1 if the value to search for never occurs." It works You're amazing
  3. rich_web_development
    is this any good? <script> var buttons = document.getElementsByClassName("add-panel"); for(var i = 0; i < buttons.length; i++) { buttons[i].addEventListener("click", addPanel, false); } var panels = []; function addPanel(e) { // Get a reference to the button that was clicked var button = e.currentTarget; //----------------------------------------------- //Code to stop items being applied twice to array //----------------------------------------------- for(var i = 0; i < panels.length; i++){ if(panels[i] != button.getAttribute("data-value")){ console.log(panels[i]); }else{ return; } } //----------------------------------------------- //Code to stop items being applied twice to array //----------------------------------------------- // Add the button's value to the array panels.push(button.getAttribute("data-value")); // Display the contents of the array on the page by generating <li> elements. var demo = document.getElementById("demo"); demo.innerHTML = ""; var li; for(var i = 0; i < panels.length; i++) { li = document.createElement("li"); li.innerHTML = panels[i]; demo.appendChild(li); } } </script>
  4. rich_web_development
    Ok. I'll try and understand what you've given me first then I'm sure I'll be able to work the rest out. Thanks for the quick reply
  5. rich_web_development
    WOW! Thanks. I was trying all sorts of stuff for the past two days and just couldn't get my head around it. Thank you very, very, very much! Just tried it and it works I don't suppose, if it's not to much of a cheek, you can tell me how I can adjust it so that it can only add the specific product once? If I click on a button for a product twice it adds the product twice. I plan to have a quantity input underneath that I will try and implement. Again, thanks
  6. rich_web_development
    I thought you'd just be able to do something like: <!DOCTYPE html> <html> <body> <button onclick="myFunction1"> Click to add 4 Panel Smooth</button> <button onclick="myFunction2"> Click to add 4 Panel Grained</button> <p id="demo"></p> </body> </html> <script> var products = []; var products, text, fLen, i; fLen = products.length; text = "<ul>"; for (i = 0; i < fLen; i++) { text += "<li>" + products[i] + "</li>"; } text += "</ul>"; document.getElementById("demo").innerHTML = text; var btn = document.querySelector("button"); var btnClicked = btn.onclick; function myFunction1(){ if(btnClicked == true){ products.push("4 Panel Smooth"); } } function myFunction2(){ if(btnClicked == true){ products.push("4 Panel Grained"); } } </script> <!DOCTYPE html> <html> <head> <title>What I want it to do</title> </head> <body> <h1>What I want to achieve</h1> <button type="button" id="button1" onclick="myFunction()">Add 4 Panel Smooth</button><br><br> <button type="button" id="button2" onclick="myFunction1()">Add 4 Panel Grained</button><br><br> <button type="button" id="button3" onclick="myFunction2()">Add 4 Panel Grained 2 Glazed</button><br><br> <p id="demo"></p> <script> var products = []; function myFunction1(){ //if button 1 clicked add "4 Panel Smooth" to products array } function myFunction2(){ //if button 2 clicked add "4 Panel Grained" to products array } function myfunction3(){ //if button 3 clicked add "4 Panel Grained Glazed" to products array } var products, text, fLen, i; fLen = products.length; text = "<ul>"; for (i = 0; i < fLen; i++) { text += "<li>" + products[i] + "</li>"; } text += "</ul>"; document.getElementById("demo").innerHTML = text; //Itterate through any items that get applied to products array </script> </body> </html> I want to be able to add products to an array, when a button is clicked for that specific product, then display the products that have been clicked on by iterating through the products. Say, just for example, I have three products and when the button is clicked on product 1 and 3 then they will be displayed by being iterated over(same kind of thing you do with PHP and MySQL, select products from database, display them by iterating over them). I have been trying all types of things to try and understand how to do it but I just don't understand JavaScript well enough. I come across a problem of the array needing to be updated from multiple functions. I thought you would just be able to have one single array then click a button to call a function that would add the product to the array but the variables in the function can't be accessed outside the function. So if, I press the button linked to a specific product that calls a function to push an item to the products array, because push would be in a variable inside a function I can't get it to access the array, the array has to be outside the function so that other functions can add products to the array. I would very much appreciate if someone could help me sort this out.
  7. rich_web_development
    Thanks dude!
  8. rich_web_development
    Hi, Sorry, in advance, if this is a simple question. I want to be able to use CSS combinators to select two child elements of a parent element. I know it can be done various other ways. Could someone please let me know if I can use combinators to achieve this effect. So for example: <!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8"> <meta name="viewport" content="width=device-width, initial-scale=1.0"> <meta http-equiv="X-UA-Compatible" content="ie=edge"> <title>Document</title> <style> /* I want to achieve this effect WITH COMBINATORS! but header h1 and p selected in one go instead of having to do it twice */ header > h1{ color:red; } header > p{ color:red; } main > h1{ color:green; } main > p{ color: green; } /* texts stays black/* /*header h1 > p{ color: red; }*/ /* text stays black */ /*header + h1 p{ color:red; }*/ /* Text stays black/* /*header + h1 + p{ color:red; }*/ /*Only the p is red header > h1 + p{ color:red; }*/ /*header > h1 p{ color:red; }*/ /*Doesn't work header > h1 > p{ color:red; }*/ /*Doesn't work header h1 p{ color:red; }*/ </style> </head> <body> <header> <h1>Some text</h1> <p>Some more text</p> </header> <main> <h1>Some text</h1> <p>Some more text</p> </main> </body> </html>
  9. rich_web_development
    I didn't mean that you showed me only static. I rushed my last post and I'm still learning about all this web dev stuff. Apologies for my post being unclear. Thanks for all your help and advice.
  10. rich_web_development
    Thanks for that. I just commented out the w3custom-list > p b + span {display: inline-block; width: 68%;} bit and now the static text and the info line up side by side. I have another problem now. Now that I have the photos and info coming from a database so I have no need for the static information for example: <div class="w3-third w3-container w3-margin-bottom"> <img src="parkhome8.jpg" alt="Norway" style="width:100%" class="w3-hover-opacity"> <div class="w3-container w3-white w3custom-list"> <p><b>Name</b> <span>Holiday home</span></p> <p><b>Bedrooms</b> <span>3</span></p> <p><b>Width</b> <span>20</span></p> </div> So I've deleted the above and changed it to the following: <?php // edited from here $stmt = $conn->prepare("SELECT name, bedrooms, length, width, mainpic, serialno FROM holidayhomes"); $stmt->execute(); $stmt->bind_result($name, $bedrooms, $length, $width, $mainpic, $serialno); $result = $stmt->get_result(); $rows = $result->num_rows; for ($j = 0 ; $j < $rows ; ++$j) { $result->data_seek($j); $row = $result->fetch_array(MYSQLI_NUM); echo <<<_END <!-- First Photo Grid--> <div class="w3-row-padding"> <div class="w3-third w3-container w3-margin-bottom"> <img src="/uploads/$row[4]" alt="holiday home" style="width:100%;max-height:174px;" class="w3-hover-opacity"> <div class="w3-container w3-white w3custom-list"> <p><b>Name</b> <span>$row[0]</span></p> <p><b>Bedrooms</b> <span>$row[1]</span></p> <p><b>Width</b> <span>$row[2]</span></p> <p><b>Length</b> <span>$row[3]</span></p> <p><b>SerialNo</b> <span>$row[5]</span></p> </div> </div> _END; } $result->close(); $conn->close(); ?> So now that I've deleted the static part all of the grey background collapses underneath and everything is moved up underneath the photo and info of what is displayed from my database. So when the page was displayed with static content the photos and info were above a pagination which all of this was on a grey background with a footer underneath all of the above. Now I'm pulling the photos and the info from a database everything that was below and behind the photos and info have collapsed up underneath the photos and info. Actually, I think I have it. I just put some margin-top on the pagination. Although when the browser is smaller it leaves a big gap so will have to change the margin-top on the pagination when the browser is smaller.
  11. rich_web_development
    That is amazing. Thank you very much. When the browser is small the static text and info line up side by side which is exactly what I wanted to achieve. When the browser is full screen the info appears underneath the static text. Is there a way for the info to be displayed side by side with the static text when the browser is in full screen? Is there something wrong with the w3schools website? When I visit a page it jumps to the bottom of the page and starts playing some video with music. Just went to a page and it started playing some ad for Thomson Cruises.
  12. rich_web_development
    Hi, I'm trying to edit the W3.CSS portfolio template. The part I want to edit looks as follows: <!-- First Photo Grid--> <div class="w3-row-padding"> <div class="w3-third w3-container w3-margin-bottom"> <img src="parkhome8.jpg" alt="Norway" style="width:100%" class="w3-hover-opacity"> <div class="w3-container w3-white"> <p><b>Lorem Ipsum</b></p> <p>Praesent tincidunt sed tellus ut rutrum. Sed vitae justo condimentum, porta lectus vitae, ultricies congue gravida diam non fringilla.</p> </div> </div> <div class="w3-third w3-container w3-margin-bottom"> <img src="parkhome8.jpg" alt="Norway" style="width:100%" class="w3-hover-opacity"> <div class="w3-container w3-white"> <p><b>Lorem Ipsum</b></p> <p>Praesent tincidunt sed tellus ut rutrum. Sed vitae justo condimentum, porta lectus vitae, ultricies congue gravida diam non fringilla.</p> </div> </div> <div class="w3-third w3-container"> <img src="parkhome8.jpg" alt="Norway" style="width:100%" class="w3-hover-opacity"> <div class="w3-container w3-white"> <p><b>Lorem Ipsum</b></p> <p>Praesent tincidunt sed tellus ut rutrum. Sed vitae justo condimentum, porta lectus vitae, ultricies congue gravida diam non fringilla.</p> </div> </div> </div> I want to edit it so that instead of just some text I want to achieve something like as below the following: Name: Holiday home Bedrooms: 3 Width: 20 So that 'Name: Bedrooms: Width:' will be static text and the info next to it will be pulled from a database. The problem is getting the info to line up nice and neat and getting it to act responsive when the size of the screen/browser is different/changed. I've tried it with a list but I just cannot work out how to get the static part 'Name: Bedrooms: Width' and then the info part 'Holiday home, 3, 20' to line up nice an neat (like it is above) and get it responsive. If someone could suggest the best way to get this to work I would be very grateful.
  13. rich_web_development
    Thank you very much. I think I have it now. Well, it works anyway. Although where should I place the '$result->close(); $conn->close();' within the below code? I have changed the code as follows: <?php require_once 'login.php'; $conn = new mysqli($hn, $un, $pw, $db); if ($conn->connect_error) die($conn->connect_error); if (isset($_POST['delete'])&& isset($_POST['serialno'])) { $serialno = mysql_entities_fix_string($conn, 'serialno'); $query = "DELETE FROM holidayhomes WHERE serialno='$serialno'"; $result = $conn->query($query); if (!$result) echo "DELETE failed: $query<br>" . $conn->error . "<br><br>"; } if (isset ($_POST['name']) && isset ($_POST['bedrooms']) && isset ($_POST['length']) && isset ($_POST['width']) && isset ($_POST['serialno'])) { $name = mysql_entities_fix_string ($conn, 'name'); $bedrooms = mysql_entities_fix_string ($conn, 'bedrooms'); $length = mysql_entities_fix_string ($conn, 'length'); $width = mysql_entities_fix_string ($conn, 'width'); $serialno = mysql_entities_fix_string ($conn, 'serialno'); // $query = "INSERT INTO holidayhomes (name, bedrooms, length, width, serialno) VALUES ('$name','$bedrooms', '$length', '$width', '$serialno')"; $stmt = $conn->prepare("INSERT INTO holidayhomes (name, bedrooms, length, width, serialno)VALUES(?,?,?,?,?)"); $stmt->bind_param("sssss", $name, $bedrooms, $length, $width, $serialno); $stmt->execute(); $result = $stmt->get_result(); //$result = $conn->query($query); //if (!$result) echo "INSERT failed: $query<br>" . $conn->error . "<br><br>"; } if(isset($_POST["submit"])){ $target_dir = "uploads/"; $target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]); $uploadOk = 1; $imageFileType = pathinfo($target_file, PATHINFO_EXTENSION); } if(isset($_POST["submit"])){ $check = getimagesize($_FILES["fileToUpload"]["tmp_name"]); if($check !== false){ // echo "FIle is an image - " . $check["mime"] . "." ; $uploadOk = 1; } else { // echo "File is not an image."; $uploadOk = 0; } } // if ($uploadOk == 0){ // // echo "Sorry, your file was not uploaded."; // } else { if (isset($_POST["submit"])){ (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)); } // echo "The file " . basename( $_FILES["fileToUpload"]["name"]) . " has been uploaded."; // else { // echo "Hello!"; // } // } if(isset($_POST["submit"])){ $mainpic = basename( $_FILES["fileToUpload"]["name"]); $serialno = $_POST['serialno']; $query = "UPDATE holidayhomes SET mainpic='$mainpic' WHERE serialno='$serialno'"; $result = $conn->query($query); } // $query = "SELECT * FROM holidayhomes"; // $result = $conn->query($query); // if (!$result) die ("Database access failed: " . $conn->error); // $rows = $result->num_rows; // for ($j = 0 ; $j < $rows ; ++$j) // { // $result->data_seek($j); // $row = $result->fetch_array(MYSQLI_NUM); // } // $result->close(); // $conn->close(); // function get_post($conn, $var) // { // return $conn->real_escape_string($_POST[$var]); // } function mysql_entities_fix_string($conn, $var){ return htmlentities(mysql_fix_string($conn, $var)); } function mysql_fix_string($conn, $var){ if (get_magic_quotes_gpc()) $var = stripslashes($var); return $conn->real_escape_string($_POST[$var]); } ?> <!DOCTYPE html> <head> </head> <html> <body> <form action="PTestPreStmnt05.php" method="POST"><pre> Name <input type="text" name="name"> Bedrooms <input type="text" name="bedrooms"> Length <input type="text" name="length"> Width <input type="text" name="width"> Serial Number <input type="text" name="serialno"> <input type="submit" value="ADD RECORD"> </pre> </form> <br><br><br> <?php // $query = "SELECT * FROM holidayhomes"; // $result = $conn->query($query); // if (!$result) die ("Database access failed: " . $conn->error); $stmt = $conn->prepare("SELECT name, bedrooms, length, width, mainpic, serialno FROM holidayhomes"); $stmt->execute(); $stmt->bind_result($name, $bedrooms, $length, $width, $mainpic, $serialno); $result = $stmt->get_result(); $rows = $result->num_rows; for ($j = 0 ; $j < $rows ; ++$j) { $result->data_seek($j); $row = $result->fetch_array(MYSQLI_NUM); echo <<<_END <pre> Name: $row[0] Bedrooms: $row[1] Length: $row[2] Width: $row[3] Serial No: $row[5] MainPic: $row[4] <img src="uploads/$row[4]" width=200 height=200> </pre> _END; } $result->close(); $conn->close(); ?> <pre> <form action="PTestPreStmnt05.php" method="POST"> <input type="hidden" name="delete" value="yes"> <input type="hidden" name="serialno" value="<?php echo $row[5]?>"> <input type="submit" value="DELETE RECORD"> </form> </pre> <pre> <form action="PTestPreStmnt05.php" method="post" enctype="multipart/form-data"> Select main photo: <input type="hidden" name="serialno" value="<?php echo $row[5]?>"> <input type="file" name="fileToUpload" id="fileToUpload"> <input type="submit" value="Upload Image" name="submit"> </form> </pre> <br><br> </body> </html>
  14. rich_web_development
    I still get the same error if I remove the parens $stmt = $conn->prepare("SELECT name, bedrooms, length, width, mainpic, serialno FROM holidayhomes VALUES (?,?,?,?,?,?)"); $stmt->bind_param("ssssss", $name, $bedrooms, $length, $width, $mainpic, $serialno); $stmt->execute(); $result = $stmt->get_result();
  15. rich_web_development
    I have tried lots of different combinations to try and use prepared statements with the 'SELECT * FROM' part on line 95. I just cannot work out how to use prepared statements on this 'SELECT * FROM' part. I have tried replacing from line 95 to 97 with the following: $stmt = $conn->prepare("SELECT (name, bedrooms, length, width, mainpic, serialno) FROM holidayhomes VALUES (?,?,?,?,?,?)"); $stmt->bind_param("ssssss", $name, $bedrooms, $length, $width, $mainpic, $serialno); $stmt->execute(); $result = $stmt->get_result(); When I replace line 95 to 97 with the above code I get the following error: Fatal error: Call to a member function bind_param() on boolean in C:\xampp\htdocs\PTestPreStmnt02.php on line *the line that the bind_param is on If you could show me how to get the prepared statement working when using SELECT from the database I would be very grateful.
  16. rich_web_development
    Hi, I've been following the php prepared statements on the w3schools website http://www.w3schools.com/php/php_mysql_prepared_statements.asp I have made amendments to my code (well, not really mine as most of it I have had help with from this forum ) - on line 47 - to use prepared statements when INSERTING info from a user to MySQL, as follows: <!DOCTYPE html> <head> </head> <html> <body> <form action="PTestPreStmnt.php" method="POST"><pre> Name <input type="text" name="name"> Bedrooms <input type="text" name="bedrooms"> Length <input type="text" name="length"> Width <input type="text" name="width"> Serial Number <input type="text" name="serialno"> <input type="submit" value="ADD RECORD"> </pre> </form> <br><br><br> <?php require_once 'login.php'; $conn = new mysqli($hn, $un, $pw, $db); if ($conn->connect_error) die($conn->connect_error); if (isset($_POST['delete'])&& isset($_POST['serialno'])) { $serialno = get_post($conn, 'serialno'); $query = "DELETE FROM holidayhomes WHERE serialno='$serialno'"; $result = $conn->query($query); if (!$result) echo "DELETE failed: $query<br>" . $conn->error . "<br><br>"; } if (isset ($_POST['name']) && isset ($_POST['bedrooms']) && isset ($_POST['length']) && isset ($_POST['width']) && isset ($_POST['serialno'])) { $name = get_post ($conn, 'name'); $bedrooms = get_post ($conn, 'bedrooms'); $length = get_post ($conn, 'length'); $width = get_post ($conn, 'width'); $serialno = get_post ($conn, 'serialno'); // $query = "INSERT INTO holidayhomes (name, bedrooms, length, width, serialno) VALUES ('$name','$bedrooms', '$length', '$width', '$serialno')"; //PREPARED STATEMENT $stmt = $conn->prepare("INSERT INTO holidayhomes (name, bedrooms, length, width, serialno)VALUES(?,?,?,?,?)"); $stmt->bind_param("sssss", $name, $bedrooms, $length, $width, $serialno); $stmt->execute(); $result = $stmt->get_result(); //$result = $conn->query($query); //if (!$result) echo "INSERT failed: $query<br>" . $conn->error . "<br><br>"; } if(isset($_POST["submit"])){ $target_dir = "uploads/"; $target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]); $uploadOk = 1; $imageFileType = pathinfo($target_file, PATHINFO_EXTENSION); } if(isset($_POST["submit"])){ $check = getimagesize($_FILES["fileToUpload"]["tmp_name"]); if($check !== false){ // echo "FIle is an image - " . $check["mime"] . "." ; $uploadOk = 1; } else { // echo "File is not an image."; $uploadOk = 0; } } // if ($uploadOk == 0){ // // echo "Sorry, your file was not uploaded."; // } else { if (isset($_POST["submit"])){ (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)); } // echo "The file " . basename( $_FILES["fileToUpload"]["name"]) . " has been uploaded."; // else { // echo "Hello!"; // } // } if(isset($_POST["submit"])){ $mainpic = basename( $_FILES["fileToUpload"]["name"]); $serialno = $_POST['serialno']; $query = "UPDATE holidayhomes SET mainpic='$mainpic' WHERE serialno='$serialno'"; $result = $conn->query($query); } //SHOULD I USE A PREPARED STATEMENT HERE? $query = "SELECT * FROM holidayhomes"; $result = $conn->query($query); if (!$result) die ("Database access failed: " . $conn->error); $rows = $result->num_rows; for ($j = 0 ; $j < $rows ; ++$j) { $result->data_seek($j); $row = $result->fetch_array(MYSQLI_NUM); echo <<<_END <pre> name $row[0] Bedrooms $row[1] Length $row[2] Width $row[3] Serial No $row[10] MainPic $row[4] <img src="uploads/$row[4]" width=200 height=200> </pre> <pre> <form action="PTestPreStmnt.php" method="POST"> <input type="hidden" name="delete" value="yes"> <input type="hidden" name="serialno" value="$row[10]"> <input type="submit" value="DELETE RECORD"> </form> </pre> <pre> <form action="PTestPreStmnt.php" method="post" enctype="multipart/form-data"> Select main photo: <input type="hidden" name="serialno" value="$row[10]"> <input type="file" name="fileToUpload" id="fileToUpload"> <input type="submit" value="Upload Image" name="submit"> </form> </pre> <br><br> _END; } $result->close(); $conn->close(); function get_post($conn, $var) { return $conn->real_escape_string($_POST[$var]); } ?> </body> </html> I was just wondering if I should use a prepared statement on the part that selects all from the database to be displayed - line 94 - ? If I do need to use prepared statements when information from the database could someone be so kind as to advise me on how to go about achieving this. I have tried but just cannot work it out. I noticed on the next section on the w3schools website it has a bit about 'select data' and in that part there is a bit for select data using prepared statements but only for PDO http://www.w3schools.com/php/php_mysql_select.asp. If anyone could help I would be extremely grateful. Oh yeah, love the new templates http://www.w3schools.com/w3css/w3css_templates.asp
  17. rich_web_development
    Oh right. I think I have it now Thanks very much for the help!
  18. rich_web_development
    Thanks for the reply. Sorry about the messy code and not using the code tags. I'll use them in the future. I've taken all the echo parts out of the mainphoto01.php. Changed the form for uploading photos so that the action points back to the same file. I put the code for uploading photos in the PicTest03.php file. It all works and everything stays on the same page now. The only thing is I keep getting the following Notice: Undefined index: fileToUpload in C:\xampp\htdocs\PicTest03.php on line 54 Notice: Undefined index: fileToUpload in C:\xampp\htdocs\PicTest03.php on line 70 Notice: Undefined index: fileToUpload in C:\xampp\htdocs\PicTest03.php on line 77 Notice: Undefined variable: serialno in C:\xampp\htdocs\PicTest03.php on line 79 <!DOCTYPE html> <head> </head> <html> <body> <form action="PicTest03.php" method="POST"><pre> Name <input type="text" name="name"> Bedrooms <input type="text" name="bedrooms"> Length <input type="text" name="length"> Width <input type="text" name="width"> Serial Number <input type="text" name="serialno"> <input type="submit" value="ADD RECORD"> </pre> </form> <br><br><br> <?php require_once 'login.php'; $conn = new mysqli($hn, $un, $pw, $db); if ($conn->connect_error) die($conn->connect_error); if (isset($_POST['delete'])&& isset($_POST['serialno'])) { $serialno = get_post($conn, 'serialno'); $query = "DELETE FROM holidayhomes WHERE serialno='$serialno'"; $result = $conn->query($query); if (!$result) echo "DELETE failed: $query<br>" . $conn->error . "<br><br>"; } if (isset ($_POST['name']) && isset ($_POST['bedrooms']) && isset ($_POST['length']) && isset ($_POST['width']) && isset ($_POST['serialno'])) { $name = get_post ($conn, 'name'); $bedrooms = get_post ($conn, 'bedrooms'); $length = get_post ($conn, 'length'); $width = get_post ($conn, 'width'); $serialno = get_post ($conn, 'serialno'); $query = "INSERT INTO holidayhomes (name, bedrooms, length, width, serialno) VALUES ('$name','$bedrooms', '$length', '$width', '$serialno')"; $result = $conn->query($query); if (!$result) echo "INSERT failed: $query<br>" . $conn->error . "<br><br>"; } $target_dir = "uploads/"; $target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]); $uploadOk = 1; $imageFileType = pathinfo($target_file, PATHINFO_EXTENSION); if(isset($_POST["submit"])){ $check = getimagesize($_FILES["fileToUpload"]["tmp_name"]); if($check !== false){ // echo "FIle is an image - " . $check["mime"] . "." ; $uploadOk = 1; } else { // echo "File is not an image."; $uploadOk = 0; } } if ($uploadOk == 0){ // echo "Sorry, your file was not uploaded."; } else { if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)){ // echo "The file " . basename( $_FILES["fileToUpload"]["name"]) . " has been uploaded."; } else { // echo "Hello!"; } } $mainpic = basename( $_FILES["fileToUpload"]["name"]); if(isset($_POST["serialno"])) $serialno = $_POST['serialno']; $query = "UPDATE holidayhomes SET mainpic='$mainpic' WHERE serialno='$serialno'"; $result = $conn->query($query); $query = "SELECT * FROM holidayhomes"; $result = $conn->query($query); if (!$result) die ("Database access failed: " . $conn->error); $rows = $result->num_rows; for ($j = 0 ; $j < $rows ; ++$j) { $result->data_seek($j); $row = $result->fetch_array(MYSQLI_NUM); echo <<<_END <pre> name $row[0] Bedrooms $row[1] Length $row[2] Width $row[3] Serial No $row[10] MainPic $row[4] <img src="uploads/$row[4]" width=200 height=200> </pre> <pre> <form action="PicTest03.php" method="POST"> <input type="hidden" name="delete" value="yes"> <input type="hidden" name="serialno" value="$row[10]"> <input type="submit" value="DELETE RECORD"> </form> </pre> <pre> <form action="PicTest03.php" method="post" enctype="multipart/form-data"> Select main photo: <input type="hidden" name="serialno" value="$row[10]"> <input type="file" name="fileToUpload" id="fileToUpload"> <input type="submit" value="Upload Image" name="submit"> </form> </pre> <br><br> _END; } $result->close(); $conn->close(); function get_post($conn, $var) { return $conn->real_escape_string($_POST[$var]); } ?> </body> </html> Hope I used the code tags correctly
  19. rich_web_development
    Hi, I'm trying to follow the JSON example at http://www.w3schools.com/json/json_example.asp I have adjusted the code a little but the code is as follows: jsonTest.html <!DOCTYPE html> <html> <head> <style> h1 { border-bottom: 3px solid #cc9900; color: #996600; font-size: 30px; } table, th , td { border: 1px solid grey; border-collapse: collapse; padding: 5px; } table tr:nth-child(odd) { background-color: #f1f1f1; } table tr:nth-child(even) { background-color: #ffffff; } </style> </head> <body> <h1>Customers</h1> <div id="id01"></div> <script> var xmlhttp = new XMLHttpRequest(); var url = 'jsonTestPHP.php'; xmlhttp.onreadystatechange=function() { if (this.readyState == 4 && this.status == 200) { myFunction(this.responseText); } } xmlhttp.open("GET", url, true); xmlhttp.send(); function myFunction(response) { var arr = JSON.parse(response); var i; var out = "<table>"; for(i = 0; i < arr.length; i++) { out += "<tr><td>" + arr.Name + "</td><td>" + arr.Bedrooms + "</td><td>" + arr.Length + "</td></tr>"; } out += "</table>"; document.getElementById("id01").innerHTML = out; } </script> </body> </html> jsonTest.php <?php header("Access-Control-Allow-Origin: *"); header("Content-Type: application/json; charset=UTF-8"); require_once 'login.php'; $conn = new mysqli($hn, $un, $pw, $db); if ($conn->connect_error) die($conn->connect_error); $result = $conn->query("SELECT name, bedrooms, length FROM holidyhomes"); $outp = "["; while($rs = $result->fetch_array(MYSQLI_ASSOC)) { if ($outp != "[") {$outp .= ",";} $outp .= '{"Name":"' . $rs["name"] . '",'; $outp .= '"Bedrooms":"' . $rs["bedrooms"] . '",'; $outp .= '"Length":"'. $rs["length"] . '"}'; } $outp .="]"; $conn->close(); echo($outp); ?> So I have changed the var url = "http://www.w3schools.com/website/customers_mysql.php"; from the jsonTest.html to var url = 'jsonTestPHP.php'; When I run this I get "Invalid character: var arr = JSON.parse(response);" I just want to be able to grab the data from my database and display it on screen using JSON. Can anyone help me understand why I'm getting this Invalid character error or tell me if I'm going about this the totally wrong way? Thanks
  20. rich_web_development
    Hi, I'm trying to get a form where you submit information about a home to a database and below the form it displays the homes with the information and a photo. Below the home information and photo I have a button to DELETE RECORD and a button to upload a photo. All of the php and forms are in one file so submitting the home info and deleting record are just have action="samefile.php". This is so that it stays on the same page and doesn't have to jump to another page. I cannot get the upload photo to stay on the same page. The only way I can get the upload photo to work is to send it to another file. This of course takes me to another page which I don't want. I want everything to stay on the same page. The code I have (in a file called "PicTest03.php) is as follows: <!DOCTYPE html> <head> </head> <html> <body> <form action="PicTest03.php" method="POST"><pre> Name <input type="" name="name"> Bedrooms <input type="text" name="bedrooms"> Length <input type="text" name="length"> Width <input type="text" name="width"> Serial Number <input type="text" name="serialno"> <input type="submit" value="ADD RECORD"> </pre> </form> <br><br><br> <?php require_once 'login.php'; $conn = new mysqli($hn, $un, $pw, $db); if ($conn->connect_error) die($conn->connect_error); if (isset($_POST['delete'])&& isset($_POST['serialno'])) { $serialno = get_post($conn, 'serialno'); $query = "DELETE FROM holidayhomes WHERE serialno='$serialno'"; $result = $conn->query($query); if (!$result) echo "DELETE failed: $query<br>" . $conn->error . "<br><br>"; } if (isset ($_POST['name']) && isset ($_POST['bedrooms']) && isset ($_POST['length']) && isset ($_POST['width']) && isset ($_POST['serialno'])) { $name = get_post ($conn, 'name'); $bedrooms = get_post ($conn, 'bedrooms'); $length = get_post ($conn, 'length'); $width = get_post ($conn, 'width'); $serialno = get_post ($conn, 'serialno'); $query = "INSERT INTO holidayhomes (name, bedrooms, length, width, serialno) VALUES ('$name','$bedrooms', '$length', '$width', '$serialno')"; $result = $conn->query($query); if (!$result) echo "INSERT failed: $query<br>" . $conn->error . "<br><br>"; } $query = "SELECT * FROM holidayhomes"; $result = $conn->query($query); if (!$result) die ("Database access failed: " . $conn->error); $rows = $result->num_rows; for ($j = 0 ; $j < $rows ; ++$j) { $result->data_seek($j); $row = $result->fetch_array(MYSQLI_NUM); echo <<<_END <pre> name $row[0] Bedrooms $row[1] Length $row[2] Width $row[3] Serial No $row[10] MainPic $row[4] <img src="uploads/$row[4]" width=200 height=200> </pre> <pre> <form action="PicTest03.php" method="POST"> <input type="hidden" name="delete" value="yes"> <input type="hidden" name="serialno" value="$row[10]"> <input type="submit" value="DELETE RECORD"> </form> </pre> <pre> <form action="mainphoto01.php" method="post" enctype="multipart/form-data"> Select main photo: <input type="hidden" name="serialno" value="$row[10]"> <input type="file" name="fileToUpload" id="fileToUpload"> <input type="submit" value="Upload Image" name="submit"> </form> </pre> <br><br> _END; } $result->close(); $conn->close(); function get_post($conn, $var) { return $conn->real_escape_string($_POST[$var]); } ?> </body> </html> The code for "mainphoto01.php is: <?php require_once 'login.php'; $conn = new mysqli($hn, $un, $pw, $db); if ($conn->connect_error) die($conn->connect_error); $target_dir = "uploads/"; $target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]); $uploadOk = 1; $imageFileType = pathinfo($target_file, PATHINFO_EXTENSION); if(isset($_POST["submit"])) { $check = getimagesize($_FILES["fileToUpload"]["tmp_name"]); if($check !== false){ echo "FIle is an image - " . $check["mime"] . "." ; $uploadOk = 1; } else { echo "File is not an image."; $uploadOk = 0; } } if ($uploadOk == 0){ echo "Sorry, your file was not uploaded."; } else { if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)){ echo "The file " . basename( $_FILES["fileToUpload"]["name"]) . " has been uploaded."; } else { echo "Hello!"; } } $mainpic = basename( $_FILES["fileToUpload"]["name"]); if(isset($_POST["serialno"])) $serialno = $_POST['serialno']; $query = "UPDATE holidayhomes SET mainpic='$mainpic' WHERE serialno='$serialno'"; $result = $conn->query($query); if(!$result) echo "INSERT failed: $query<br>" . $conn->error . "<br><br>"; ?> No matter what I try I cannot get the above code to work if I keep it in the "PicTest03.php" file. Can someone please show what the code should look like so I can get everything to show on the same page? I have had a look as the PHP and AJAX and DATABASE http://www.w3schools.com/php/php_ajax_database.asp which I got the example working but it uses a drop down list. Is there a way to convert the code used for the example on http://www.w3schools.com/php/php_ajax_database.asp and instead of drop down list just have the information from the database shown under the form for submitting the home information?
  21. rich_web_development
    Oh, I think I have it now. I changed the form to include name="serialno" in details.php as follows: <form action="mainphoto.php" method="post" enctype="multipart/form-data"> Select main photo: <input type="hidden" name="serialno" value="$row[10]"> <input type="file" name="fileToUpload" id="fileToUpload"> <input type="submit" value="Upload Image" name="submit"> </form> </pre> Then included in mainphoto.php the following: $mainpic = basename( $_FILES["fileToUpload"]["name"]); if(isset($_POST["serialno"])) $serialno = $_POST['serialno']; $query = "UPDATE holidayhomes SET mainpic='$mainpic' WHERE serialno='$serialno'"; $result = $conn->query($query); if(!$result) echo "INSERT failed: $query<br>" . $conn->error . "<br><br>"; Thanks for all your help. I just need to work out how to keep it all on one page so that it doesn't keep navigating of and stays on the one page like a single page application. It would also be nice to be able to delete the photo from uploads when I click delete "DELETE RECORD". Thanks for all your help!
  22. rich_web_development
    Thanks for that Sorry for my stupidity but how do I grab the serial number from the correct row while in the mainphoto.php file?
  23. rich_web_development
    Hi, I'm trying to do a simple website where I can upload some information about homes and add photos to a MySQL database and then display the information. So I have an html form that sends the details of the home to a php file called details.php. The html form is as follows. <!DOCTYPE html> <html> <body> <form action="details.php" method="POST"> <pre> Name <input type="" name="name"> Bedrooms <input type="text" name="bedrooms"> Length <input type="text" name="length"> Width <input type="text" name="width"> Serial Number <input type="text" name="serialno"> <input type="submit" value="ADD RECORD"> </pre> </body> </html> The details.php look like the following: <?php require_once 'login.php'; $conn = new mysqli($hn, $un, $pw, $db); if ($conn->connect_error) die($conn->connect_error); if (isset($_POST['delete'])&& isset($_POST['serialno'])) { $serialno = get_post($conn, 'serialno'); $query = "DELETE FROM holidayhomes WHERE serialno='$serialno'"; $result = $conn->query($query); if (!$result) echo "DELETE failed: $query<br>" . $conn->error . "<br><br>"; } if (isset ($_POST['name']) && isset ($_POST['bedrooms']) && isset ($_POST['length']) && isset ($_POST['width']) && isset ($_POST['serialno'])) { $name = get_post ($conn, 'name'); $bedrooms = get_post ($conn, 'bedrooms'); $length = get_post ($conn, 'length'); $width = get_post ($conn, 'width'); $serialno = get_post ($conn, 'serialno'); $query = "INSERT INTO holidayhomes (name, bedrooms, length, width, serialno) VALUES ('$name','$bedrooms', '$length', '$width', '$serialno')"; $result = $conn->query($query); if (!$result) echo "INSERT failed: $query<br>" . $conn->error . "<br><br>"; } $query = "SELECT * FROM holidayhomes"; $result = $conn->query($query); if (!$result) die ("Database access failed: " . $conn->error); $rows = $result->num_rows; for ($j = 0 ; $j < $rows ; ++$j) { $result->data_seek($j); $row = $result->fetch_array(MYSQLI_NUM); echo <<<_END <pre> Name $row[0] Bedrooms $row[1] Length $row[2] Width $row[3] Serial No $row[10] MainPic <img src="$row[4]" width=auto height=auto><br><br> </pre> <pre> <form action="details.php" method="POST"> <input type="hidden" name="delete" value="yes"> <input type="hidden" name="serialno" value="$row[10]"> <input type="submit" value="DELETE RECORD"> </form> </pre> <pre> <form action="mainphoto.php" method="post" enctype="multipart/form-data"> Select main photo: <input type="file" name="fileToUpload" id="fileToUpload"> <input type="submit" value="Upload Image" name="submit"> </form> </pre> _END; } $result->close(); $conn->close(); function get_post($conn, $var) { return $conn->real_escape_string($_POST[$var]); } ?> ​So the above inserts the information (about a home entered in the html form) into the MySQL database then displays the information and an option to delete the homes information then I have a form to upload a photo for the home. For uploading the photo I have a php file called mainphoto.php which is as follows: <?php require_once 'login.php'; $conn = new mysqli($hn, $un, $pw, $db); if ($conn->connect_error) die($conn->connect_error); $target_dir = "uploads/"; $target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]); $uploadOk = 1; $imageFileType = pathinfo($target_file, PATHINFO_EXTENSION); if(isset($_POST["submit"])) { $check = getimagesize($_FILES["fileToUpload"]["tmp_name"]); if($check !== false){ echo "FIle is an image - " . $check["mime"] . "." ; $uploadOk = 1; } else { echo "File is not an image."; $uploadOk = 0; } } if ($uploadOk == 0){ echo "Sorry, your file was not uploaded."; } else { if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)){ echo "The file " . basename( $_FILES["fileToUpload"]["name"]) . " has been uploaded."; } else { echo " off!"; } } $mainpic = basename( $_FILES["fileToUpload"]["name"]); $query = "INSERT INTO holidayhomes (mainpic) VALUES ('$mainpic')"; $result = $conn->query($query); if(!$result) echo "INSERT failed: $query<br>" . $conn->error . "<br><br>"; ?> How do I get the form to submit and add the name of the photo to the same record(row) as the homes information? So then when it iterates through the record(rows) in the MySQL database I can display the photo to the relevant information about the home?
  24. rich_web_development
    Just wrote the example from PHP.net out by hand and now it works! Oh my god. All that bother because it didn't copy and paste properly. I shall have to copy the one from w3schools out by hand now. Thank you very much for your help! I was just about to chuck my laptop out of my window.
  25. rich_web_development
    I don't have any extensions.
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