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Found 66 results

  1. First off, thanks for letting me into the forums. Excuse a newbie but I have a question about fetching data from a database with PHP and AJAX. Let's say I wanted to fetch data from the database like in this example from W3Schools, but I had modified the dropdown to for example fetch all entries with last name Griffin, Swanson or Quagmire. I made the q variable take a string instead of an integer. $q = strval($_GET['q']); Now I want to expand the SQL query to also take another condition. "SELECT * FROM user WHERE LastName = '".$q."' AND Hometown = '".$q2."'"; I understand how to build the query and that I need to declare another variable (q2), but how do I modify the AJAX request? xmlhttp.open("GET","getuser.php?q="+str,true); Also, would I need to have a submit button now that there are two conditions? Thanks!
  2. I was wondering if there is any way to set unique values to each cookie. I want to use these cookies on my website to see if the visitor has already visited my website once, but without asking any information from them. This way I want to store this unique value in my database each time someone opens my website, and I will be able to see if there have been - for example - 20 different visitors in the past month, or if the same person has visited my website 20 times. All my code is already written and working, I only need a way to set a different unique value to each cookie I set.
  3. Hi all, I'm trying to create a query which selects fields from 2 tables. I have a table 'stock' to store cars for sale and a table called 'custstock' which has the customer username and the vehID. How do I show all the cars that relate the that user who is logged in? My current query that doesn't work; $username = "-1"; if(isset($_SESSION['username'])){ $username = $_SESSION['username']; } // executeable query $sql = ("SELECT * FROM [stock] JOIN custstock ON [stock].custstockvehID = custstock.vehID WHERE username = '$username'"); Thanks, Jack
  4. Desperately needed: a way to retrieve all records from a database table while using prepared statements. The script now works but only for 1 record. <?php $_POST['id']= '2'; // set connection variables // Create connection $conn = new mysqli($servername, $username, $password, $dbname); // Check connection if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error); } $stmt = $conn->prepare("SELECT * FROM table WHERE id LIKE ? ;"); if ( !$stmt ) { yourErrorHandler($conn->error); // or $mysqli->error_list } else if ( !$stmt->bind_param('i', $id ) ) { yourErrorHandler($stmt->error); // or $stmt->error_list } else if ( !$stmt->execute() ) { yourErrorHandler($stmt->error); // or $stmt->error_list } else { $result = $stmt->get_result(); while ($row = mysqli_fetch_assoc($result)) { var_dump($row); $data[]=$row; var_dump($id); var_dump($data); $array= implode("|" , $row) ; var_dump($array); foreach( $data as $row ) { $id = $row['id']; $name = $row['name']; var_dump($id); var_dump($name); } } } ?> I hope someone can give a query that executes what is desired, but maybe there is also a possibilty to put the query in a loop? Please help.
  5. 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.
  6. <?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>
  7. 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?
  8. What is BMC remedy database? How to fetch data from this DB to display as XML files. Please provide any links ?
  9. Hi, I'm currently experiencing difficulties updating the database. I'm currently creating a 'Forgotten Password' feature. I've finished most of it, however when I attempt to update the database with a new password, nothing happens and the old password remains in the database. I've followed W3Schools example but I'm stuck. --------- The new password is generated by numbers 4 in total, ie $new_pass = (rand(1000,9999); and from the form I've used $user_email= $_POST["retrieve_email"]; then I use the line sql = "UPDATE registration SET Password=$new_pass WHERE Email=$user_email"; I can't get it to work, and I don't know why. Could anyone please write the code I need to make it work, or let me know how to proceed? Thank You. ps. I thought about deleting the entry and re-inserting it. Would that work (out of interest)?
  10. I have a piece of code where I use MYSQL to add records into a database. Here I use INSERT, but I want to work with UPDATE. The user selects records which have to be updated. The records have a unique number. I hope people can help me how to convert MYSQL code into PHP. The MYSQL code used in the console is this: UPDATE table_name SET input2='my input here ' WHERE rec_number = 2; The part of the script I use looks like this: <?php $name1 = $conn->real_escape_string($input1); $name2 = $conn->real_escape_string($input2); // $sql = "INSERT INTO $table (input1, input2 ) VALUES (' $name1 ', ' $name2 ')" ; // olde code // $sql = "UPDATE $table SET input2, input1 WHERE rec_number = $rec_number_selected VALUES (' $name1 ', ' $name2 ')" ; $sql = "UPDATE $table (input1, input2) VALUE S (' $name1 ', ' $name2 ') WHERE rec_number = $rec_number_selected " ; ?> Anyone any idea what the right code is in PHP?
  11. So recently I got my hands on one of youtube videos, it's quit old, and the guy explaining how to create a form for uploading image and all data that is connected with it, to MySQL and then showing on your web site. So everything is ok, but webpage does not display any images, it just show me white picture frame with picture icon on the top. This is my connection file with database: <?php $hostname_phpimage = "***"; $username_phpimage = "***"; $password_phpimage = "***"; $database_phpimage = "***"; // Create connection $inkedmen_marko = mysql_pconnect($hostname_phpimage, $username_phpimage, $password_phpimage); // Check connection if ($phpimage->connect_error) { die("Connection failed: " . $phpimage->connect_error); } ?> Then the file that uploads the images: <?php require_once('php/connect.php'); if($_POST['submit']) { $name=basename($_FILES['file_upload']['name']); $t_name=$_FILES['file_upload']['tmp_name']; $dir='/home4/inkedmen/public_html/images'; $cat=$_POST['cat']; if(move_uploaded_file($t_name,$dir."/".$name)) { mysql_select_db($database_phpimage, $inkedmen_marko); $qur="insert into anglija (mid, cid, name, path) values ('','$cat','$name','/home4/inkedmen/public_html/images/$name')"; $res=mysql_query($qur,$inkedmen_marko); echo'file upl success'; } else { echo 'not uploaded'; } } ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <meta name="viewport" content="width=device-width, initial-scale=1"> </head> <body> <form action="upload.php" method="post" enctype="multipart/form-data"> <input type="file" name="file_upload"/> cat_id <input type="text" name="cat" /><br/> <input type="submit" name="submit" value="upload"/> </form> </body> </html> Then this one shows me what countries I have in one of the data base tables, so there is two england and slovenia, they both have a separate id that I have to type when uploading an image: <?php require_once('php/connect.php'); mysql_select_db($database_phpimage, $inkedmen_marko); $qur="select * from cat"; $res=mysql_query($qur,$inkedmen_marko); ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <meta name="viewport" content="width=device-width, initial-scale=1"> </head> <body> <?php while($row= mysql_fetch_array($res)) { ?> <h1><a href="image.php?cid=<?php echo $row['id']?>"><?php echo $row['name'] ?> </a></h1><br/> <?php } ?> </body> </html> And finally, when I push on each country it directs me to the pictures that are connected with that countries according to the id given: <?php require_once('php/connect.php'); mysql_select_db($database_phpimage, $inkedmen_marko); $id=$_GET['cid']; $qur="select * from anglija where cid='$id'"; $res=mysql_query($qur,$inkedmen_marko); ?> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <meta name="viewport" content="width=device-width, initial-scale=1"> </head> <body> <?php while ($row=mysql_fetch_array($res)){ ?> <img src="<?php echo $row['path'] ?>" width="300px" height="200px"> <br/> <?php } ?> </body> </html> So in the end it directs me to the pictures, but I can't see the pictures, it's just white windows with icon on the top. The pictures uploads to the image folder, the id, names of the file, path also uploads good, but somehow I can't show the images. I know it should be something small, but I can't figure it out. Maybe someone have any ideas????
  12. Hi I'm looking for php/mysql code to find identical input in a mysql database column.
  13. I uploaded website files to the web root and when i tried to access it via www.mydomain.com it gave following error A Database Error OccurredError Number: 1146 Table 'webpk_16901224_app.sesion' doesn't exist INSERT INTO `sesion` (`session_id`, `ip_address`, `user_agent`, `last_activity`, `user_data`) VALUES ('1912dd6c2dd5c411cde798d9496c9a9a', '119.157.163.66', 'Mozilla/5.0 (Windows NT 5.1; rv:42.0) Gecko/20100101 Firefox/42.0', 1448544788, '') Filename: libraries/Session.php Line Number: 328 I tried to create a table like following through phpmyadmin, but i get an error "key id does not exist". My database name is webpk_16901224_app and it is empty. There are no tables. create table tablename( id INT NOT NULL, name VARCHAR (20) NOT NULL, age INT NOT NULL, PRIMARY KEY (id) ); The website script has 3 sql files, but i don't know how to use them. SQL files are attached. Desktop.zip
  14. Hello. I'm currently developing an app in the Intel XDK for my local high school. One of the features I would like to add is something where school administrators (such as the athletic director) can login and then submit data into an HTML5 form, which is then submitted to a server and stored in a database. Then, after it is stored, I would like the information to be able to be viewed by regular users of the app, in something such as an HTML page. THe purpose of this would be for the school admins to submit scores from the games such as football, etc. My question is, how do I setup the server (im assuming would run PHP) to be able to store the submitted data, and then would edit an HTML page to include the new text? Any help is appreciated! Thanks!
  15. I am trying to make a result page so I can get some information from the database by using php coding. I need it to be able to find a patient name and date of service. Which will be if I type in (Search:) Joe Doe and add (from:) 02/02/2015 (to:) 02/14/2015. It will find the patient name and all date of service. Here is my code ... I know mysql is not good but it will work for now. <?phpmysql_connect('localhost', 'root', 'no_password');mysql_select_db('timeclock');$results = array();if(isset($_GET['search'])) { $search = $_GET['search_box']; $from = $_GET['from']; // or $from = $_get['from']; $to = $_GET['to']; // or $to = $_get['to']; $sql = mysql_query("SELECT * FROM info WHERE patient_name = '".$search."' AND (date_of_service BETWEEN '".$from."' AND '".$to."')"); while($row = mysql_fetch_assoc($sql)) { $results[] = $row; }}?><!-- form --><form name="search_form" method="GET" action="">Search: <input type="text" name="search_box" value="" />Dates From : <input type="text" name="from" value=""/> To : <input type="text" name="to" value=""/><input type="submit" name="search" value="Look up Patient ..."></form><!-- end --><table width="70%" cellpadding="5" cellspace="5"><tr> <td><strong>Care Provider</strong></td> <td><strong>Patient Name</strong></td> <td><strong>Date of Time</strong></td> <td><strong>Time In</strong></td> <td><strong>Time Out</strong></td> <td><strong>Remarks</strong></td></tr><?php if(!empty($results)): ?><?php foreach($results as $row){ ?><tr> <td><?php echo $row['care_provider']; ?></td> <td><?php echo $row['patient_name']; ?></td> <td><?php echo $row['date_of_service']; ?></td> <td><?php echo $row['time_in']; ?></td> <td><?php echo $row['time_out']; ?></td> <td><?php echo $row['remarks']; ?></td></tr><?php } ?><?php endif; ?></table>
  16. When I add a new column in MYSQL in phpmyadmin a new row is created (i guess its called a row) for example: if i first did this query with a php script. $sql = "INSERT INTO MyGuests (firstname, lastname, email)VALUES ('John', 'Doe', 'john@example.com')"; And later I go to phpmyadmin and add in MyGuests (telephonenumber) and VALUE this with 12345678 then this is done in a new row. I would rather have this all in one row. Can you maximize the number of rows and still add columns.
  17. I am trying to start this online ordering system for a business. Basically identical to how the pizza-hut and dominos online ordering! How do I go about that? Do I learn how to write code? Buy software for my business? And if I do pursue software from another company, if the business becomes bigger, will that limit the rights that's I have for it to be mine? Please help. I'm new to this and really want to learn the ropes here!
  18. hello there!!i m building an app for the elections..in this app i ve build a rest service that takes data(votes,etc) from a .js file of another website,and presents them to some fields.the problem is that.on that js file,the creators provide only "party_id" and not "party_name" which is contained in another .js file.i copied the id's and the names to my database.and now i want that:when the first js file gives the "Party_id",to search the database and provide me the "party_name"..what i want to do is to compare the values given from the rest service and the database collection and to return the correct name. i tried this code: var objectId = value; //Debug information that you can see in browser console.console.log("objectId = " + objectId);var whereObject = {"kwdikos": objectId};return JSON.stringify(whereObject); as shown here,it returns all the database array!! http://imageshack.com/a/img673/9877/EkpHcb.png but as it seems,it returns it without comparing..i mean,i want it when the id sent is 2,to return ND,and when id_sent is 1,to return PASOK.it just displays all the database array without criteria match!i tried to do it by using IF statement,but it didnt let me do it!! the correct results should be these:http://imagizer.imageshack.us/a/img537/7291/d2mTx9.png thanks for your time guys!!!
  19. How would you go about making a hyperlink that can only be visited once before changing to another one? (Like tracking the amount of times the link was visited, and when it reaches a certain number it switches.)
  20. i have a form in which i have a field date of birth , and a textbox to fill date of birth , when a person fills his date of birth like 01/01/1990 , then it will be saved into database , and when i retrieve this date form database then this comes 01/01/1990 12:00 AM , i want only date part from it. This is my form to fillup date of birth form.aspx <tr> <td width="30%"> Date of Birth <br /> </td> <td style="width: 1px"> :<asp:TextBox ID="txtDateofBirth" runat="server" placeholder="dd/mm/yyyy"></asp:TextBox> </td> <td class="td2"> </td> </tr> this is code for form.aspx.cs SqlCommand cmd = new SqlCommand("insert into ApplicantForm(NameofApplicant, dateofbirth)" + "values( @NameofApplicant, convert(Datetime, @dateofBirth, 103)", con); cmd.Parameters.AddWithValue("@dateofBirth",txtDateofBirth.Text); this is code for print of this date of birth print.aspx <asp:Label ID="txtDOB" runat="server" ></asp:Label> this is code print.aspx.cs SqlCommand cmd = new SqlCommand("Select * from ApplicantForm Where ApplicantId=" + ((Request.QueryString["ApplicantId"])) + "", con); dr = cmd.ExecuteReader(); txtDOB.Text = dr["dob"].ToString();
  21. http://stackoverflow.com/q/26809213/4080419 GOAL: To have multiple people edit a Google Spreadsheet and have all edits display on a webpage (not Google site) where my css takes over the formatting and display of the content coming from the Spreadsheet. PROBLEM: I am brand new to PHP, as in I've been doing tutorials all week and this is my first time working with it. Likewise, I'm brand new to Google API's and don't fully understand how they work. I don't know what code to use or really how to modify it. WHAT I'VE TRIED: Google API obviously has the code for how to fetch the data to put into a page, so I've copied this, though I don't know if I put it in correctly/Dreamweaver says "There is a syntax error on line 32. Code hinting may not work until you fix this error". Additionally, I've read that perhaps Google Fusion Tables might be better because I can download it as a CSV version and then have that be what used in my actual script (though I don't fully understand how/if that would continually update itelf?): <html xmlns="http://www.w3.org/1999/xhtml"><head><title>Untitled Document</title><!--#include file="php-5.6.2.tar.bz2" --><?php include("php-5.6.2.tar.bz2"); ?><style type="text/css">#page {margin-left:auto;margin-right: auto;top: 15%;bottom: 15%;width: 1000px;}.menue {border-color:rgb(0,0,0);border-width:thick;height:300px;width: 300px;position: abosolute;margin-left: 10%;margin-right: 10%;}</style></head><body><div id="page"><div id="Skillet_Product_List_and_Price" class="menue"><?phpusing Google.GData.Client;using Google.GData.Spreadsheets;namespace MySpreadsheetIntegration{class Program{static void Main(string[] args){SpreadsheetsService service = new SpreadsheetsService("https://docs.google.com/a/emich.edu/spreadsheets/d/1OAFhMrayek9BKvO6ZI9EIPmCeY61oLL7-wfZOIg2xV8/edit#gid=724477237");// TODO: Authorize the service object for a specific user (see other sections)// Instantiate a SpreadsheetQuery object to retrieve spreadsheets.SpreadsheetQuery query = new SpreadsheetQuery();// Make a request to the API and get all spreadsheets.SpreadsheetFeed feed = service.Query(query);if (feed.Entries.Count == 0){// TODO: There were no spreadsheets, act accordingly.}// TODO: Choose a spreadsheet more intelligently based on your// app's needs.SpreadsheetEntry spreadsheet = (SpreadsheetEntry)feed.Entries[0];Console.WriteLine(spreadsheet.Title.Text);// Get the first worksheet of the first spreadsheet.// TODO: Choose a worksheet more intelligently based on your// app's needs.WorksheetFeed wsFeed = spreadsheet.Worksheets;WorksheetEntry worksheet = (WorksheetEntry)wsFeed.Entries[0];// Fetch the cell feed of the worksheet.CellQuery cellQuery = new CellQuery(worksheet.CellFeedLink);cellQuery.MinimumRow = 2;cellQuery.MinimumColumn = 1;cellQuery.MaximumColumn = 2;CellFeed cellFeed = service.Query(cellQuery);// Iterate through each cell, printing its value.foreach (CellEntry cell in cellFeed.Entries){// Print the cell's address in A1 notationConsole.WriteLine(cell.Title.Text);// Print the cell's address in R1C1 notationConsole.WriteLine(cell.Id.Uri.Content.Substring(cell.Id.Uri.Content.LastIndexOf("/") + 1));// Print the cell's formula or text valueConsole.WriteLine(cell.InputValue);// Print the cell's calculated value if the cell's value is numeric// Prints empty string if cell's value is not numericConsole.WriteLine(cell.NumericValue);// Print the cell's displayed value (useful if the cell has a formula)Console.WriteLine(cell.Value);}}}}?></div></div></body></html> But I've also tried this, which I was playing with : $file = "https://docs.google.com/a/emich.edu/spreadsheets/d/1OAFhMrayek9BKvO6ZI9EIPmCeY61oLL7-wfZOIg2xV8/edit#gid=724477237"$data = file_get_contents($file) or die('Could not read file!');echo $datafunction readRows() {var sheet = SpreadsheetApp.getActiveSheet(Eateries Menu Pricing.xlsx);var rows = sheet.getDataRange(A2:22);var numRows = rows.getNumRows(22);var values = rows.getValues();for (var i = 0; i <= numRows - 1; i++) {var row = values;Logger.log(row);}};var ss = SpreadsheetApp.getActiveSpreadsheet();var sheet = ss.getSheets()[0];// This represents ALL the datavar range = sheet.getDataRange();var values = range.getValues();//This logs the spreadsheet in CSV format with a trailing commafor (var i = 0; i < values.length; i++) {var row = "";for (var j = 0; j < values.length; j++) {if (values[j]) {row = row + values[j];}row = row + ",";}Logger.log(row);} And finally I've tried this just to try displaying a website thinking it'll work: <?php$myvar = "https://docs.google.com/a/emich.edu/spreadsheets/d/1O8-h-xmGE49K4x9sHrRL-c 97bHPXhma2eKkqrYAvog/edit#gid=1568863245";$var = fopen($myvar,"rb");echo stream_get_contents($var);?> DISCLAIMER: I understand that somewhat similar-ish questions have been asked before, but I am so new to PHP that I really need to be talked through this process. I've been looking everywhere online to try to understand this all, and I'm really trying to understand the process and how this all works
  22. hello w3school mmbrs i have a q. thats makes me confused and its how can i insert data into WEB FORM from a DATABASE Assume : i have a database called 'home' and my table called 'vals' and the table 'vals' have 3 Fields 'username','val1','val2' and i create a web form that's have 2 inputs (val1,val2) so i want to put the values FROM THE DATABASE INTO MY WEB FORM note* , i want to insert data into web form (not from webform) any idea please ?!
  23. I am making packet type system when in my **Packet Table** , parent packet ( `id` which is primary key ) and its N -Sub Packet is under (`parent_id`) is stored , below is my table structure :**Packet_table** id | packet_name | parent_id | 1 | 01 | 0 2 | 02 | 0 3 | 03 | 1 4 | 04 | 1 5 | 05 | 1 6 | 06 | 4 7 | 07 | 4 8 | 08 | 3 9 | 09 | 5 10 | 010 | 2 ........................so on and on with N packets in same table Below is what i have tried *but its not getting* `id` N sub packet detail properly: SELECT p.`packet_name` AS MAIN, s.`packet_name` AS SUB FROM packet_table s LEFT JOIN packet_table p ON s.`parent_id` = p.`id`thats as per above table : **id** ( which is primary / auto increment )id = 1 -> main packet (01) , its sub and N sub packets are :01 - > 03,04,0504 -> 06,0703 -> 0805 -> 09 **in short** 01 -> 03 -> 08 04 -> 06 , 07 05 -> 09 its not necessary above design format mysql code .. just simple N sub packet query will doabove is just few but in my case there will be N number of sub packet for each (id).So how it be can achieve .Note : it can be same as Category and N sub Category type
  24. Hello W3S! It's been a while since i've been online here... sorry about that Anyways. I have some trouble with a dynamic menu i'm trying to make with MySQLI... not sure if that is the problem anyhow... Here is the code as a start, i will explain under the code below what i'm trying to do: // File that we are on (viewing / watching)$tab = pathinfo( $_SERVER['SCRIPT_NAME'], PATHINFO_FILENAME );$menu_res = query("SELECT * FROM menu WHERE menu_file_url = ".$tab);$menu_row = mysqli_fetch_row($menu_res);if( $tab == $menu_row['menu_file_url'] || $menu_row['menu_accessible'] == "no" ) stderr("Page Error", "We are currently working on this page! Go to another page to keep browsing! Thanks for your patience! :)");if(isset($CURUSER)){ $menu_while_res = query(" SELECT * FROM menu WHERE menu_accessible = 'yes' AND menu_view = 'user' OR menu_view = 'both' ORDER BY menu_order_id ASC");}else{ $menu_while_res = query(" SELECT * FROM menu WHERE menu_accessible = 'yes' AND menu_view = 'guest' OR menu_view = 'both' ORDER BY menu_order_id ASC");}$HTMLOUT .= "<ul class='nav_first'>"; while ($menu_while_row = mysqli_fetch_array($menu_while_res, MYSQLI_ASSOC)) { // Menu Items Loaded Here $tabarray = array( $menu_while_row['menu_array_id_name'] => "<li><a href='".$menu_while_row['menu_file_url']."'>".$menu_while_row['menu_name']."</a></li>", ); // K = Key // V = Value foreach($tabarray as $k => $v) { if( $tab == $k ) $HTMLOUT .= str_replace("<li>", "<li class='nav_active'>", $v); else $HTMLOUT .= $v; } // Unset Menu For re-load again unset($tabarray); }$HTMLOUT .= "</ul>"; Currently i'm trying to make a dynamic menu with MySQLI! It's working perfectly... but when i tried to "expand" the project a bit longer and try to make a dynamic menu with errors on pages if the users are not allowed to view a specific file, then i get nothing... What i'm trying to do is to controle in the database with "Enum" as my DB setup that is "no" is has been set on one of the menu items (menu_accessible).. then the item will NOT show on the menu! AND if the user still tries to enter that specific page by URL, he will then get an error message saying that we are working on the website page... The code i'm trying to insert into this project is this little peace of code here: // File that we are on (viewing / watching)$tab = pathinfo( $_SERVER['SCRIPT_NAME'], PATHINFO_FILENAME );$menu_res = query("SELECT * FROM menu WHERE menu_file_url = ".$tab);$menu_row = mysqli_fetch_row($menu_res);if( $tab == $menu_row['menu_file_url'] || $menu_row['menu_accessible'] == "no" ) stderr("Page Error", "We are currently working on this page! Go to another page to keep browsing! Thanks for your patience! :)"); However, i get no respond on the code! Even when i have checked if the "$tab" variable is real and related to the name inside the DB (which it is!)... so if possible, can anyone help and tell me what i'm doing wrong here Oo? Thanks alot by the way! And sorry for the long goodbye hehe ...studies and all, killing me! -.-' Anyways, hope some answers or good tips... really need this one Thanks in advance! Mr rootKID