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Found 261 results

  1. darbok

    connecting to msqli

    Looking at how form items are put into a mysql table, the part that seems daunting is if ones form has like 50 items... having to list all those IDs with their variables makes me wonder if there is a better way or if its just " it up buttercup.".
  2. So my question is: how can I insert variables into my mysql queries in php? This is my code now, I would like to use the $me and the $friend variables to work in the sql query: function newChat($me, $friend) { global $conn; $sql = "SELECT id FROM chats WHERE (person1 = $me AND person2 = $friend) OR (person1 = $friend AND person2 = $me);"; mysqli_query($conn, $sql) or die('Error querying database.'); echo "check <br />"; $result = mysqli_query($conn, $sql); $row = mysqli_fetch_array($result); while ($row = mysqli_fetch_array($result)) { echo $row['id'] . ' - ' . $row['person1'] . ' - ' . $row['person2'] . ' - ' . $row['date'] .'<br />'; } } newChat('John', 'Marie'); Also if I set $sql to "SELECT * FROM chats" it only gives me one row of my two test rowes in my database. (it only gives the second row, and just ignores the first one? --> a screenshot of the database is at the end of this post) Can anyone please help me with this? Thanks already!
  3. Balderick

    updating column with increment in mysql

    In phpmyadmin I did the following query and had it output in php like this: $sql = "UPDATE `table` SET count = count + 1 WHERE unique_nr = 5175781"; this all worked well. though when I tried to implement it in my script I had an error: script: <?php // code to establish dbase connection: var_dump($unique_nr); // $stmt = $conn->prepare('UPDATE table SET count = ? WHERE unique_nr = ? '); $stmt = $conn->prepare('UPDATE table SET count = count + 1 WHERE unique_nr = ? '); $stmt->bind_param("is", $count, $unique_nr); $stmt->execute(); ?> resulting in an error message. I also tried this way of writing for bind_param: $stmt->bind_param("s", $unique_nr); but that didnt update the column, though there wasnt an error message anymore
  4. smus

    MySQL query is not working

    MySQL queries are not working only inside this block: ... if (isset($_POST['send'])) { } ... The variable itself is set and transferred from the form. What could be the reasons of this and how is better to check?
  5. j.silver

    InnoDB for Image Gallery Tables

    Dear all, For photo gallery tables (named: gallery_album, gallery_image, gallery_relation), I have seen a recommendation to use InnoDB engine in preference to MyISAM. I thought MyISAM would be the choice for all tables but tables for transactions. Do you agree with the recommendation to using InnoDB for photo gallery tables and why?
  6. j.silver

    Index vs Key

    Dear all, I am trying to dig into the exact difference between key and index in MySQL tables. Some say they are synonymous, others give some differences. I still don't feel confident enough to decide when to use each on a table. I would appreciate if someone well-versed in their difference explain such difference and when to use each. Thanks.
  7. j.silver

    Different Image Size

    Hi all, I have uploaded an image to a folder/directory and stored its size using the $_FILES['image']['size'] superglobal. While the image source size was reading 153kb, the stored size is 157317. Why there is a difference in size?
  8. j.silver

    Why Do We Store tmp_name of a File?

    Dear all. For an image uploaded to a folder/directory and stored its name, type, and size in a DB table, what could be the merits of also storing its temporary name, noting that such temporary name is deleted by the server upon moving the image to the desired folder/directory?
  9. j.silver

    Unknown Printed Charector

    Hi all, In fetching a result from a table (below code), 1 is also printed after the fetched record (fetched record below). There is no 1 in the table, nor have I accidentally included it in my script. What is it and why does it appear? <?php require 'db/connect.php'; if($result = $db->query("SELECT * FROM user")) { if ($count = $result->num_rows) { $rows = $result->fetch_assoc(); echo '<pre>', print_r($rows), '</pre>'; } } ?> Array ( [id] => 1 [first_name] => Bill [last_name] => John [bio] => I'm a web developer [created] => 2016-11-21 12:50:12 ) 1
  10. j.silver

    Lack of Error Reporting

    Hi all, <?php require 'db/connect.php'; $result = $db->query("SELECT * FROM people"); ?> I obtained a blank page running the above script, which indicates that everything is fine (path and table name are correct). I then removed the * from the query to see if it would report an error, but it did not. I added print_r($result); and run it. Nothing has changed. I included the * and run, it reported the number of rows and columns in the table. Why there has a been a lack of error reporting in this instance, noting that the ini file is configured to report all sorts of errors, even the warnings and notices (and it is reporting all that in other scripts)?
  11. Hi all, In one website, I read the following note: "You should also think about storing files locations on disk. Using MySQL for storing images is thought to be Bad Idea™. Handling SQL table with big data like images can be problematic." Do you agree? If not, why? What is the best practice in storing and displaying on a website a big number of images?
  12. I'm trying to select data from a MySQL database that is hosted on a webserver. I want to be able to retrieve the data from a table within the database and then illustrate it within a HTML table. There's an example on W3Schools that I've been following, but I'm unable to retrieve the data successfully. http://www.w3schools.com/php/php_ajax_database.asp Below is the source code: (HTML) <html> <head> //Javascript code <script> function showUser(str) { if (str == "") { document.getElementById("txtHint").innerHTML = ""; return; } else { if (window.XMLHttpRequest) { // code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp = new XMLHttpRequest(); } else { // code for IE6, IE5 xmlhttp = new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange = function() { if (this.readyState == 4 && this.status == 200) { document.getElementById("txtHint").innerHTML = this.responseText; } }; xmlhttp.open("GET","getuser.php?q="+str,true); xmlhttp.send(); } } </script> </head> <body> <form> <select name="users" onchange="showUser(this.value)"> <option value="">Select a person:</option> <option value="1">Peter Griffin</option> <option value="2">Lois Griffin</option> <option value="3">Joseph Swanson</option> <option value="4">Glenn Quagmire</option> </select> PHP File: (getuser.phd) <!DOCTYPE html> <html> <head> <style> table { width: 100%; border-collapse: collapse; } table, td, th { border: 1px solid black; padding: 5px; } th {text-align: left;} </style> </head> <body> <?php $q = intval($_GET['q']); $con = mysqli_connect('www.example.com','user_Admin','12345-678','my_DB'); if (!$con) { die('Could not connect: ' . mysqli_error($con)); } mysqli_select_db($con,"ajax_demo"); $sql="SELECT * FROM user WHERE id = '".$q."'"; $result = mysqli_query($con,$sql); echo "<table> <tr> <th>Firstname</th> <th>Lastname</th> <th>Age</th> <th>Hometown</th> <th>Job</th> </tr>"; while($row = mysqli_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['FirstName'] . "</td>"; echo "<td>" . $row['LastName'] . "</td>"; echo "<td>" . $row['Age'] . "</td>"; echo "<td>" . $row['Hometown'] . "</td>"; echo "<td>" . $row['Job'] . "</td>"; echo "</tr>"; } echo "</table>"; mysqli_close($con); ?> </body> </html> *MySQL table is attached I think the issue might exist from mysqli_select_db($con,"ajax_demo"); onwards inside the PHP file. Should I be referring to the table that contains the data inside the database? I have the PHP File hosted on my webserver, so I'm not sure why it won't retrieve that data when a person is selected from the list of options on the HTML page. Any help would be much appreciated.
  13. j.silver

    utf8 VS utf8mb4

    Dear all, I came across the following statement on one website: "Note: Since MySQL 5.5.3 you should use utf8mb4 rather than utf8. They both refer to the UTF-8 encoding, but the older utf8 had a MySQL-specific limitation preventing use of characters numbered above 0xFFFD." 1) Is there any drawbacks in using utf8mb4, or it is recommended for use from now on?
  14. Hi I have been working on building a custom post a result set to a database which thankfully I have now managed with some help, now I am working on the other end of this project the displaying the saved data back to users, From the start of this I have an online tournament hosting group that go to my form and post a list of player names with points earned per tournament. Once the form is submitted it's sent to my form_processing.php where the results are exploded and split into 2 arrays "$player_name, $points" Then on the INSERT I have an INSERT $sql = "INSERT into `bg_points` (`player_name`, `points`) values (?, ?) on duplicate key update points = points + ?"; to of course update current player points if the exist or add new records if not. Then finally saved into my DB table, now I am working on building the page to display these points back to the players so they can track their earned points and standings each month. which is easy enough I just call an MySQL SELECT * FROM statement on the DB table that's done, Now I need to sort these results in lowest to highest order by points so if player1 has 120 points player2 has 50 points player3 has 75 points player4 has 5 points then I need to be displaying these back as player4 player2 player3 player1 I have had a look online for some possible examples to help guide me into this and I either get HTTP 500 or just no effect at all. here is the very basic code I have put together so far which I can and will update and add into my site framework once I have it in order and working... <HTML> <head></head> <body bgcolor="#0000FF"> <?PHP $servername = "localhost"; $username = "***********"; $password = "*******"; $dbname = "***********"; // Create connection $conn = new mysqli($servername, $username, $password, $dbname); // Check connection if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error); } //output the saved data $sql = "SELECT * FROM `bg_points` "; $result = $conn->query($sql); if ($result->num_rows > 0) { // output data of each row while($row = $result->fetch_assoc()) { echo "id: " . $row["id"]. " $nbsp $nbsp Playerv Name: " . $row["player_name"]. " $nbsp $nbsp Points: " . $row["points"]. "<br>"; } } else { echo "0 results"; } $conn->close(); ?> </body> </html> thanks for any help or ideas in advanced
  15. Hi I am having trouble inserting data into my MySQL database. This is my PHP code ini_set('display_errors', 1); ini_set('display_startup_errors', 1); error_reporting(E_ALL); static $connection; if(!isset($connection)) { $connection = new mysqli("localhost","username","password"); } if ($connection->connect_error) { die("Connection failed: " . $connection->connect_error); } $stmt = $connection->prepare("INSERT into Product(product_title,product_price,product_availability,productImage_1,productImage_2,productImage_3,productImage_4,product_description,product_shipping,product_pickup) VALUES(?,?,?,?,?,?,?,?,?,?)"); $product_title = $_POST['title']; $product_price = $_POST['price']; $product_availability = $_POST['stock']; $null = NULL; $product_description = $_POST['description']; $product_shipping = $_POST['postage']; $product_pickup = $_POST['pickup']; $stmt->bind_param('sisbbbbsis',$product_title,$product_price,$product_availability,$null,$null,$null,$null,$product_description,$product_shipping,$product_pickup); $stmt->send_long_data(3, file_get_contents($_FILES['img1']['tmp_name'])); $stmt->send_long_data(4, file_get_contents($_FILES['img2']['tmp_name'])); $stmt->send_long_data(5, file_get_contents($_FILES['img3']['tmp_name'])); $stmt->send_long_data(6, file_get_contents($_FILES['img4']['tmp_name'])); $stmt->execute(); $stmt->close(); $connection->close(); echo "Product inserted successfully"; I am also getting these errors. Notice: Undefined index: title in /Applications/MAMP/htdocs/Bourke/insertproduct.php on line 61 Notice: Undefined index: price in /Applications/MAMP/htdocs/Bourke/insertproduct.php on line 62 Notice: Undefined index: stock in /Applications/MAMP/htdocs/Bourke/insertproduct.php on line 63 Notice: Undefined index: description in /Applications/MAMP/htdocs/Bourke/insertproduct.php on line 67 Notice: Undefined index: postage in /Applications/MAMP/htdocs/Bourke/insertproduct.php on line 68 Notice: Undefined index: pickup in /Applications/MAMP/htdocs/Bourke/insertproduct.php on line 69 Fatal error: Call to a member function bind_param() on boolean in /Applications/MAMP/htdocs/Bourke/insertproduct.php on line 71 Thanks in advance.
  16. <?php $host = 'localhost'; $username = 'root'; $password = ''; $datadase = 'registerfinal'; $connect = mysqli_connect($host, $username, $password) or die ('error to connect to datadase'.mysqli_error()); if ($connect) { echo 'mysqli connect succsessfull'; } echo '<br /><br />'; $selectdb = mysqli_select_db($connect, $datadase) or die ('unable to select datadase'.mysqli_error()); if($selectdb) { echo 'database selected succsessfully'; } if(isset($_POST['savedetails'])) { $firstname = $_POST['firstname']; $lastname = $_POST['lastname']; $username = $_POST['username']; $password = $_POST['password']; $repeat_password = $_POST['repeat_password']; $gender = $_POST['gender']; $country = $_POST['country']; if(isset($_POST['food'])) { $food = $_POST['food']; $favfood = ""; foreach($food as $meal ) { $favfood = $meal.","; print_r($favfood); } } if(isset($_POST['imageUpload'])) { $imageUploadname = $_FILES['imageUpload']['name']; $imageUploadsize = $_FILES['imageUpload']['size']; $imageUploadtmp_name = $_FILES['imageUpload']['tmp_name']; $imageUploadtype = $_FILES['imageUpload']['type']; $uploadFolder = "uploadFolder/"; $destinationName = rand(1000, 10000).$imageUploadname; move_uploaded_file($imageUploadtmp_name, $uploadFolder.$destinationName); echo "$imageUploadname"; echo "$imageUploadsize"; echo "$imageUploadtmp_name"; echo "$imageUploadtype"; echo "$destinationName"; } $sqltwo = "INSERT INTO `registerfinaltable` (`id`, `firstname`, `lastname`, `username`, `password`, `repeat_password`, `gender`, `food`, `country`, `imageUploadname`, `imageUploadsize`, `imageUploadtype`) VALUES (NULL, '$firstname', '$lastname', '$username', '$password', '$repeat_password', '$gender', '$favfood', '$country', '$destinationName', '$imageUploadsize', '$imageUploadtype')"; $results = mysqli_query($connect, $sqltwo) ; if($results){ echo "inserted successfully"; } } ?> <html> <head> <title>register</title> </head> <body> <form action = "" method = "post" enctype = "multipart/form-data" > <label>first name : <input type = "text" name = "firstname" /> </label> <br /><br /> <label>last name : <input type = "text" name = "lastname" /> </label><br /><br /> <label>username : <input type = "text" name = "username" /> </label><br /><br /> <label>password : <input type = "password" name = "password" /> </label><br /><br /> <label>repeat password : <input type = "password" name = "repeat_password" /> </label><br /><br /> <label>Male : <input type = "radio" name = "gender" value = "Male" /> </label><br /><br /> <label>Female : <input type = "radio" name = "gender" value = "Female" /> </label><br /><br /> <label>pizza : <input type = "checkbox" name = "food[]" value = "pizza"/> </label><br /><br /> <label>burger : <input type = "checkbox" name = "food[]" value = "burger"/> </label><br /><br /> <label>chips : <input type = "checkbox" name = "food[]" value = "chips"/> </label><br /><br /> <label>sausage : <input type = "checkbox" name = "food[]" value = "sausage"/> </label><br /><br /> <label>sandwich : <input type = "checkbox" name = "food[]" value = "sandwich"/> </label><br /><br /> <label>Image : <input type = "file" name = "imageUpload" /> </label><br /><br /> <select name = "country"> <?php $sql = 'SELECT * FROM `countrie` '; $querry = mysqli_query($connect, $sql); while($country = mysqli_fetch_array($querry)):; ?> <option value = "<?php echo $country['country']; ?>"><?php echo $country['country']; ?></option> <?php endwhile;?> </select> <br /> <input type = "submit" name = "savedetails" /> </form> <table border = "1" bgcolor = "" width = "100%"> <tr><th>id</th><th>Firstname</th><th>Lastname</th><th>Username</th><th>Password</th><th>Password 2</th><th>Gender</th><th>Fav. Food</th> <th>Image</th> <th>Country</th><th>imageUploadname</th><th>imageUploadsize</th><th>imageUploadtype</th></tr> <?php $sqldata = "SELECT * FROM registerfinaltable"; $querysqldata = mysqli_query($connect, $sqldata); while($rows = mysqli_fetch_array($querysqldata) ):; ?> <tr> <td><?php echo $rows['id'];?></td> <td><?php echo $rows['firstname'];?></td> <td><?php echo $rows['lastname'];?></td> <td><?php echo $rows['username'];?></td> <td><?php echo $rows['password'];?></td> <td><?php echo $rows['repeat_password'];?></td> <td><?php echo $rows['lastname'];?></td> <td><?php echo $rows['gender'];?></td> <td><?php echo $rows['food'];?></td> <td><?php echo $rows['country'];?></td> <td><?php echo $rows['imageUploadname'];?></td> <td><?php echo $rows['imageUploadsize'];?></td> <td><?php echo $rows['imageUploadtype'];?></td> <?php endwhile;?> </tr> </table> </body> </html>
  17. Balderick

    store javascript in a database column

    My goal is to store javascript code into a database. My first idea was to use htmlspecialchars; store it in mysql in a table column and later retrieve it with htmlspecialchars_decode. All this to prevent injection / hacking. But online I read one or two warnings that it wouldnt work, which I assume is so (I didnt test it, but it seems quite obvious afterwards) . So my question is: is it possible to have a user store javascript in a database and use it in a php script for specific purposes in a secure way?
  18. Hi i am new for the PHP, i had the experienced before to do a login system page but what i want it is HTML and PHP file to separate i dont want PHP with HTML together to 1 file , cause professional people will more likely to use 2 file rather than 1 file easy to maintain and edit how to use the external PHP file coding without re-design whole page cause normally what i tested before was link or turn to other new page and result will come out but the design was totally gone so have any ways and suggestion to do that ... in the end, im sorry i not well in english and my knowledge of PHP still quite new , thanks you so much
  19. HI i trying to create a website of forum textarea editor are necessary too for recommendation which editor are best for textarea such as summernote, wysihtml5 .... and how is work to store the data/value/text with styling effect in mysql database ? Thanks all of you
  20. HI i trying to create a website of forum textarea editor are necessary too for recommendation which editor are best for textarea such as summernote, wysihtml5 .... and how is work to store the data/value/text in mysql database ? Thanks all of you
  21. HarrySeah

    textarea editor

    HI i trying to create a website of forum textarea editor are necessary too for recommendation which editor are best for textarea such as summernote, wysihtml5 .... and how is work to store the data/value/text in mysql database ? Thanks all of you
  22. Daniel On The Web

    Database Connection Issue

    <?php $serv = "localhost:81"; $user = "root"; $password = ""; try { $con = new PDO("mysql:host=$serv;dbname=test", $user, $password); $con->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); echo "Successfully connected."; } catch (PDOException $e) { echo "Connection failed: " . $e->getMessage(); } ?> Here I'm trying to use PDO to connect to my SQL database "test". However, I receive this error each time I connect: SQLSTATE[HY000] [2002] No connection could be made because the target machine actively refused it. Am I connecting with the wrong server, user or password? Because I fail to see here the mistake.
  23. rich_web_development

    Display Everything on the same page

    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?
  24. rich_web_development

    JSON data from web server running PHP and MySQL Example

    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
  25. mysteriousmonkey29

    Interactive web comic aggregator general design?

    Hello, I am currently in the middle of creating a web comic aggregation website. So far, I have written a PHP-based RSS reader that grabs the image source links and publication dates for various web comics from their RSS feeds, and uses this to update a MySQL database. This RSS reader/database updater is configured to run once every several hours and update the database with any new comics it finds. It seems to be working fine, and now I am starting to work on the front end of the website, which needs to grab this information from the ever-growing database, and display it in a user-friendly and interactive format. I wrote the database updater in a fairly procedural way, because it didn't strike me as something that lent itself very much to object-oriented development. However, the front end of the website does strike me as something that I can save myself a lot of time on by coming up with a decent object oriented design first. Unfortunately, I don't have a lot of experience in this, so I thought I would look online for suggestions. So far, I have decided specifically what I would like the front end to do, and have come up with 2 possible approaches to designing it, but am unsure about both. Here's what I would like it to do: (To clarify, when I say the phrase "comic strip," I'm referring to a whole strip like XKCD, Dilbert, the far side, etc., and when I say the phrase "comic," it refers to an individual comic within a comic strip, i.e. the Dilbert comic for 8/30/2016). I would like my webpage, by default, to display a vertically stacked list of the latest comic images from each comic strip in the database. For customization, I want a comic strip control panel, which allows users to decide which comic strips are enabled/disabled, and in what order they are displayed. I would like each comic image to appear between text denoting the name/author of the strip and an individual set of buttons that allows the user to go forward and backward through the archived comics, and also to adjust the size/scaling of the image. Finally, I would like a similar but generalized set of buttons that allows the user to control all the comic strips in the same fashion, but at once. Based on my basic understanding of object oriented programming, and the design approach suggested in this stack overflow answer: http://stackoverflow.com/questions/1100819/how-do-you-design-object-oriented-projects, I came up with one possible design: Two classes: 1) a comic strip class that contains attributes like the comic strip's name, ID number (from my database), enabled/disabled status, and current comic to display (an object of the next class), in addition to functions to get and set each of these variables, 2) a comic class that contains attributes like the image source link, publication date, and scaling percentage, in addition to functions to get and set each of these variables Then the main webpage would contain a comic strip control panel and general comic control panel as described in the desired functionality. In addition, it would contain an array of individual comic control panels, and an array of div tags (or a similar tag), which would be displayed vertically stacked and alternating along with the comic strip names in plain text. Finally, there would be functions for updating comic and comic strip object information from the database, and functions for updating the div tag content for comic display based on the comic object information. All the functions would be called using JavaScript/Ajax from the appropriate buttons. However, I was unsure about this design, so I asked a friend, who suggested that I use a model-view-controller design pattern, which I guess is popular in web development because its potential for separating all the various languages. After reading about this online, I have come up with a possible alternative design: Model: PHP, an array of comic strips, including name and ID number, active/inactive status, and current comic image URL/publishing date View: PHP generated HTML/CSS, as described before, a comic strip control panel for enabling/disabling and ordering, a vertically stacked list of comic strip names, constituent comic images, and individual control buttons, and a general set of buttons for controlling all enabled comics at once Controller: PHP/MySQL, a function to update order and active/inactive comic strip status, a function to update a given comic strip's current comic information, and a function to update all comic strips' current comic information at once As I understand it, the controller functions would be called upon user input into the view (clicking the buttons), which would then update the information stored in the model, and then somehow that would trigger a refresh of the view based on the new model information. I am worried I may have missed something in each of these designs (especially the first one), and unsure about the potential merits of one versus another. I was also wondering if it might be possible to combine the two somehow. Does anyone have any suggestions? Help would be much appreciated
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