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PHP - Mysql Update Query Not Working
Hi guys need some help.I created a simple Update Customer Details page Where you Enter the Customer ID ,The Customer Details Get Displayed inside a Form and you make changes within the Form.But the update query I wrote is not working as it should be.It Executes the Query ,when I hardcode the Customer ID inside the Update query but fails when I need to Enter the Customer ID directly from the form using post.In This Code I tried to Update only the Customer's Firstname.Thanks
Here Is The Code Code: [Select] <?php $db=mysql_connect("localhost","root","") or die('Unable to Connect To Database.Please check the Database Parameters'); mysql_select_db('ecommerce') or die(mysql_error()); if(isset($_POST['enterid'])) { $_POST['inputid']; } $query="SELECT * FROM customer WHERE Cid='$_POST[inputid]'"; $result=mysql_query($query) or die(mysql_error()); $row=mysql_fetch_array($result); $new_cid=$row['Cid']; $new_firstname=$row['Cfname']; $new_lastname=$row['Clname']; $new_email=$row['Email_id']; $new_address=$row['Address']; $new_pincode=$row['Pincode']; $new_payment=$row['Mode_of_payment']; $new_city=$row['City']; $new_state=$row['State']; $new_phone=$row['Phone']; $html1=<<<HTML1 <form action="updatecustomer.php" method="post"> <p class="inputidentifier">Please Enter The Customer's ID</p><input type="text" style="width:375px;height:40px;font-size:30px;" name="inputid" size="20"><br><br> <input type="submit" name="enterid" style="height:40px"> HTML1; print($new_cid); if(isset($_POST['update'])) { $query1="UPDATE customer SET Cfname='$_POST[firstname]' WHERE Cid='$_POST[inputid]'"; mysql_query($query1) or die(mysql_error()); if(mysql_affected_rows()==1) print("Query sucessful"); else print("Something went wrong"); } ?> <html> <head> <style type="text/css"> .inputtext {width:300px; height:40px;font-size:30px;} .inputidentifier{font-size:25px;font-family:"Arial"} .h1type{font-family:"Arial"} </style> <title>Test</title> </head> <body> <?php print($html1); ?> <h1 align="center" class="h1type">Update Customer Details</h1> <form action="updatecustomer.php" method="POST"> <table align="center" cellspacing="10" cellpadding="10" border="0" width="60%"> <tr> <td align="right" class="inputidentifier">First Name</td> <td align="left"><input type="text" class="inputtext" name="firstname" placeholder="eg:Kevin" value="<?php print($new_firstname) ?>"></td> </tr> <tr> <td align="right" class="inputidentifier">Last Name</td> <td><input type="text" class="inputtext" name="lastname" placeholder="eg:Aloysius" value="<?php print($new_lastname) ?>"></td> </tr> <tr> <td align="right" class="inputidentifier">E-mail</td> <td align="left"><input type="email" style="width:500" class="inputtext" name="email" placeholder="yourname@email.com" value="<?php print($new_email) ?>"> </tr> <tr> <td align="right" class="inputidentifier">Phone Number</td> <td align="left"><input type="text" class="inputtext" name="phone" placeholder="How Do We Call You?" value="<?php print($new_phone) ?>"></td> </tr> <tr> <td align="right" class="inputidentifier">Address</td> <td><textarea style="width:500;height:150" wrap="virtual" class="inputtext" name="address" placeholder="Where is your Crib?"><?php print($new_address) ?></textarea></td> </tr> <tr> <td align="right" class="inputidentifier">State</td> <td align="left"><input type="text" style="width:500" class="inputtext" name="state" placeholder="State" value="<?php print($new_state) ?>"> </tr> <tr> <td align="right" class="inputidentifier">City</td> <td align="left"><input type="text" class="inputtext" name="city" placeholder="City" value="<?php print($new_city) ?>"> </tr> <tr> <td align="right" class="inputidentifier">Pin Code</td> <td><input type="text" class="inputtext" name="pincode" placeholder="Mulund 400080" maxlength="6" value="<?php print($new_pincode) ?>"></td> </tr> <tr> <td align="right" class="inputidentifier">How do Pay for your Bling?</td> <td align="left"> <input type="text" class="inputtext" name="payment" value="<?php print($new_payment)?>"> </tr> <tr> <td></td> <td><input type="submit" name="update" value="Update!" style="width:100px;height:60px;"></td> </tr> </table> </form> </body> </html> Similar TutorialsHi there, I'm having a problem with updating a record with an UPDATE mysql query and then following that query with a SELECT query to get those values just updated. This is what I'm trying to do...I'd like a member to be able to complete a recommended task and upon doing so, go to a page in their back office where they can check off that task as "Completed". This completed task would be recorded in their member record in our database so that when they return to this list, it will remain as "Completed". I'm providing the member with a submit button that will call the same page and then update depending on which task is clicked as complete. Here is my code: Code: [Select] $memberid = $_SESSION['member']; // Check if form has been submitted if(isset($_POST['task_done']) && $_POST['task_submit'] == 'submitted') { $taskvalue = $_POST['task_value']; $query = "UPDATE membertable SET $taskvalue = 'done' WHERE id = $memberid"; $result = mysqli_query($dbc, $query); } $query ="SELECT task1, task2, task3 FROM membertable WHERE id = $memberid"; $result = mysqli_query($dbc, $query); $row = mysqli_fetch_array($result, MYSQLI_ASSOC); $_SESSION['task1'] = $row['task1']; $_SESSION['task2'] = $row['task2']; $_SESSION['task3'] = $row['task3']; ?> <h4>Task List</h4> <table> <form action="" method="post"> <tr> <td>Task 1</td> <td><?php if($_SESSION['task1'] == 'done') {echo '<div>Completed</div>';} else{ echo '<input type="submit" name="task_done" value="Mark As Completed" />';} ?></td> </tr> <input type="hidden" name="task_value" value="task1" /> <input type="hidden" name="task_submit" value="submitted" /> </form> <form action="" method="post"> <tr> <td>Task 1</td> <td><?php if($_SESSION['task1'] == 'done') {echo '<div>Completed</div>';} else{ echo '<input type="submit" name="task_done" value="Mark As Completed" />';} ?></td> </tr> <input type="hidden" name="task_value" value="task2" /> <input type="hidden" name="task_submit" value="submitted" /> </form> <form action="" method="post"> <tr> <td>Task 1</td> <td><?php if($_SESSION['task1'] == 'done') {echo '<div>Completed</div>';} else{ echo '<input type="submit" name="task_done" value="Mark As Completed" />';} ?></td> </tr> <input type="hidden" name="task_value" value="task3" /> <input type="hidden" name="task_submit" value="submitted" /> </form> </table> The problem that I am having is that the database is not updated with the value "done" but after submission, the screen displays "Completed" instead of "Mark As Completed". So the value is being picked up as "done", but that is why I have the SELECT after the UPDATES, so that there is always a current value for whether a task is done or not. Then I refresh and the screen returns the button to Mark As Complete. Also, when I try marking all three tasks as, sometimes all three are updated, sometimes only one or two and again, I leave the page or refresh and the "Marked As Completed" buttons come back. Bizarre. If anyone can tell me where my logic is going wrong, I would appreciate it. Hi, i am currently designing a new website and am testing a few different compnents of it. One part gathers two variables of the url using the GET method and selects the relevant data from the database. It then performs an UPDATE of a field called balance before INSERTING a new record. Everything works and gets inserted except for the UPDATE. Here is the code Code: [Select] <?php include ("connect.php"); $id = ""; $offerid = ""; $id = $_GET['user_id']; $offerid = $_GET['offer_id']; $sql_user = mysql_query("SELECT * FROM user_info WHERE id = $id"); $sql_offer = mysql_query("SELECT * FROM offers WHERE offerid = $offerid"); $balance += $payout; mysql_query("UPDATE user_info SET balance = $balance WHERE id = $id"); mysql_query ("INSERT INTO offer_credited (id, offerid, date_credited, payout_to_user) VALUES ('$id', '$offerid', CURDATE(), '$payout')") or die(mysql_error()); echo "Your Offer has been recorded and Updated" ?> The balance field is a float that i want to increase by the amount of the payout variable which is from the offers table. However, the balance never gets updated and remains at 0. Any ideas what it could be, i am thinking it is something to do with, Code: [Select] $balance += $payout; But i can't think of what else to do. Hopefully this isn't too confusing. Thanks for the help I would like to edit a record that I already have in my database but my query is not working for some reason. Hope someone can help. The errors i get are Notice: Undefined variable: first_name in C:\chemicaluser\www\editinfo.php on line 40 Notice: Undefined variable: last_name in C:\chemicaluser\www\editinfo.php on line 40 Notice: Undefined variable: log_number in C:\chemicaluser\www\editinfo.php on line 40 .... line 40 = my query $query = "UPDATE table1 SET first_name = '$first_name', last_name = '$last_name' WHERE log_number = '$log_number' "; Very simple database, 1 table w/ 2 fields table name = table1 fields = first_name, last_name my php file Code: [Select] <?php require ('/menu.php'); require_once('sqlconnect.php'); if (isset($_GET['log_number']) && isset($_GET['first_name']) && isset($_GET['last_name']) ) { $log_number = $_GET['log_number']; $first_name = $_GET['first_name']; $last_name = $_GET['last_name']; } else if (isset($_POST['log_number']) && isset($_POST['first_name']) && isset($_POST['last_name']) ) { $log_number = $_POST['log_number']; $first_name = $_POST['first_name']; $last_name = $_POST['last_name']; } if (isset($_POST['submit'])) { $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); $query = "UPDATE table1 SET first_name = '$first_name', last_name = '$last_name' WHERE log_number = '$log_number' "; $result = mysqli_query($dbc, $query) or die ('Error querying database'); mysqli_close($dbc); } ?> my form Code: [Select] <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <label for="first_name">First name</label> <input type="text" id="first_name" name="first_name" value="<?php if (!empty($first_name)) echo $first_name; ?>"/> <label for="last_name">Last Name</label> <input type="text" id="last_name" name="last_name" value="<?php if (!empty($last_name)) echo $last_name; ?>" /> <input type="submit" value="Edit Information" name="submit" /> </form> I have a value that in parts of my code i need to update. Its a number value, stored as TEXT. which i need to add numbers to it, at the moment I am using this to get the current value: Code: [Select] $sql2="SELECT * FROM weekly LIMIT 1"; $result2=mysql_query($sql2); $row2 = mysql_fetch_array( $result2 ); $rollingjackpot = $row2['rollingjackpot']; then i am adding the value to the varible, then using UPDATE to update it again, cant this be done in a simple query? something like: Code: [Select] $query = mysql_query("UPDATE weekly SET `rollingjackpot` + $amount WHERE `id` = '1'") or die(mysql_error()); $db->query( "UPDATE user_resources, user_buildings, user_units_nonbattle SET user_resources.uGold=user_resources.uGold+(user_buildings.uTownCenterLevel*(user_units_nonbattle.uMiner*10))" ); When the cron runs, it updates everyone the same regardless. Any idea what i'm doing wrong? This isn't the entire code just enough to see what I'm trying to do. Everything was working until I added the mysql update query in the if statement. Is this possible or am I doing something wrong? When I run the script it just echos "No results found" twice as $num_results = 2. Code: [Select] <?php include("../includes/connect.php"); $query = "SELECT ........ "; $result = $db->query($query); $num_results = $result->num_rows; if ($num_results == 0) { exit; } else { $i=0; while ($i < $num_results) { $row = $result->fetch_assoc(); $id = $row['id']; if ($expiration_date > $today) { ### EMAIL CODE HERE ### $update = "UPDATE model SET reminder_sent = '1' WHERE id = '$id' "; $result_2 = $db->query($update); $i++; } else { echo "No results found."; $i++; } } } ?> I made a small editing system for my news page, and I need to update three columns within my table "announcements" in the database. I tried a method of updating all of them with one MySQL query instead of using three as it just isn't neat. I've searched several methods via google and I've tried all of them, but just can't seem to get it to work. Is this MySQL query correct? mysql_query("UPDATE announcements SET title = {$title} WHERE id = '$id', content = {$content} WHERE id = '$id', lastmodified = ". date('M-d-Y') ." WHERE id = '$id'"); Ok, I'm going start off simple. If I have to provide more code I will. I am doing an update on a table called countries. Yet my query just will not update the db. Is there anything wrong with this query? mysql_query("UPDATE countries SET country_id = '{$_POST['update_value']}' WHERE country_id = '{$_POST['original_html']}'") or die(mysql_error()); I have the following PHP script to update two time/date fields in the database. When i run this the fields are not updated. Can anyone see where i m going wrong. <?php $con = mysql_connect("localhost","dbname","dbpassword"); if (!$con) { die('Could not connect: ' . mysql_error()); } echo 'Connected successfully'; mysql_select_db("my_db", $con); mysql_query("UPDATE msm_content SET created = '2011-01-02 00:00:00', modified = '2011-01-01 00:00:00'"); echo 'Query Updated successfully'; mysql_close($con); ?> Your guidance is much appreciated. When I echo the POST, it echoes the correct value. The MySQL portion seems to just ignore it all together. I've tried changing the dropdown option to just a text field, same thing occurred. I have text fields right above this particular one that update just fine with the SAME exactly scripting. if POST, update query, done. Works. This one for some reason will not. MySQL portion: if ($_POST['bUpdate']){ mysql_query("UPDATE `Patients` SET `b` = '$_POST[bUpdate]' WHERE `id` = '".$_GET['id']."'"); } echo $_POST['bUpdate']; Form Portion: Code: [Select] <tr onmouseover="color(this, '#baecff');" onmouseout="uncolor(this);"> <td width="310" colspan="2" align="center"><span class="fontoptions">Postcard Status </span><br /> <? if ($data['b'] == 1){ echo '<select name="bUpdate"><option value="1" selected>Yes</option><option value="0">No</option></select>'; } else { echo '<select name="bUpdate"><option value="1">Yes</option><option value="0" selected>No</option></select>'; } ?> </td> </tr> Hello, I am trying to pick up php again and just exercising my skills. So I have it so that it fills my form with the values of what I want to edit, and when I click the edit button, it doesn't edit any of the information. When I echo out $result, I get a MYSQL query string that has the same values as the table, so its not getting the new values that are edited. <?php @mysql_connect('localhost', 'root', '') or die("Could not connect to Mysql Server. " . mysql_error()); @mysql_select_db('tutorials') or die("Could not connect to Database. " . mysql_error()); if(isset($_GET['edit'])) { $id = $_GET['edit']; $query = "SELECT `username`, `password` FROM `users` WHERE `id` = '$id'"; $result = mysql_query($query); $row = mysql_fetch_array($result); $name = $row['username']; $password = $row['password']; } if(isset($_POST['edit'])) { $id = $_GET['edit']; $query = "UPDATE `users` SET `username` = '$name', `password` = '$password' WHERE `id` = '$id'"; $result = mysql_query($query); echo $query; if(!$result) { echo mysql_error(); }else{ echo 'updated post'; } } ?> <form method="POST" action="" > <input type="text" name="name" value="<?php echo $name; ?>" /> First name <br /> <input type="text" name="password" value="<?php echo $password; ?>" /> Last name <br /> <input type="submit" name="edit" value="edit" /> </form> I believe it has something to do with the values of $name and $password in the form conflicting with the first if isset and the second if isset. Thanks for any help possible Hi all, I have the following MySQL insert query: Code: [Select] $insert= mysql_query ("INSERT INTO tablename (column1,`".$EXPfields."`) VALUES ('$something','".$EXPvalues."')"); where $EXPfields is an array of table-field-names and $EXPvalues is an array of table-field-values. Now I want to write an equivalent query, but using UPDATE instead of INSERT INTO, but I don't want to write out all the field names/values separately, but again want to use $EXPfields and $EXPvalues. So something like this: Code: [Select] $update = mysql_query ("UPDATE tablename SET (column1,`".$EXPfields."`) = ('$something','".$EXPvalues."') WHERE .... "); Is this possible? If so, what is the proper syntax? Thanks! I've messed around with this for three hours now, and it's driving me batty. I'm running the following Query: mysql_select_db("myDB", $con); $query = "UPDATE mytable SET Name = '$Name', Address = '$Address', City = '$City', State = '$State', Zip = '$Zip', Phone = '$Phone', Website = '$Website', Type = '$Type' WHERE id = '$ID'"; The problem is that it's not updating my database. I've tried everything. I've forced the page to Echo out the variables so that I know they're right. I've run (successful) Delete queries to make sure I'm connecting to the right table. I've forced it to spit out the variables at the end of the Update. Everything I've tried works beautifully... except the Update itself. It doesn't error out or anything- it acts like it's Updating, but the data never changes. I can add to the database and delete from it, but it refuses to let me Update anything. I've looked all over the net, and tried every variation I can find. Nothing works. I also tried this: $query = "UPDATE `testbed` SET `Name`=[$Name],`Address`=[$Address],`City`=[$City],`State`=[$State],`Zip`=[$Zip],`Type`=[$Type],`Phone`=[&Phone],`website`=[$Website] WHERE `ID`=[$ID]"; And $result = mysql_query("UPDATE mytable SET Name = '$Name', Address = '$Address', City = '$City', State = '$State', Zip = '$Zip', Phone = '$Phone', Website = '$Website', Type = '$Type' WHERE id = '$ID'"); None of them do anything at all. Hi everyone, I have been trying to get this code working but no matter what I try it will not work, I am trying to allow a user to update their personal details when they login to my website. I have created a form and submit button, the code is shwn below: <?php echo $_SESSION['myusername']; ?> <br><br> <form action="updated.php" method="post"> Firstname: <input type="text" name="firstname" /><br><br> Surname : <input type="text" name="surname" /><br><br> Date Birth: <input type="text" name="dob" /><br><br> Total Wins: <input type="text" name="wins" /> Total Loses: <input type="text" name="loses" /><br><br> Email Add: <input type="text" name="email" /><br><br> Country : <input type="text" name="born" /><br><br> Other Info: <input type="text" name="other" /><br><br> <input type="submit" name="Submit" value="Update" align="right"></td> </form> The updated.php file <?php mysql_connect ("localhost","root","") or die("Cannot connect to Database"); mysql_select_db ("test"); $sql=mysql_query("UPDATE memberdetails SET firstname='{$_POST['firstname']}', surname='{$_POST['surname']}', dob='{$_POST['dob']}', totalwins='{$_POST['wins']}', totalloses='{$_POST['loses']}', email='{$_POST['email']}', country='{$_POST['born']}', info='{$_POST['other']} WHERE username=$_SESSION['myusername']); if (!mysql_query($sql)) { die('Error: ' . mysql_error()); } echo "Details Updated"; ?> This is the error i recieve: Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in C:\wamp\www\website\updated.php on line 45 Line 45 is $sql=mysql_query line If anybody can help it will be very much appreciated! hi - i'm trying to update mysql databse but not working... any ideas??? Code: [Select] [php] <?php ob_start(); require_once('../xconfig.php'); if(isset ($_GET['quote'])&& $_GET['quote']!=""){ $quote_id=$_GET['quote']; mysql_select_db($database, $makeconnection); $sql_find_quote = "SELECT * FROM tbl_quote WHERE quote_id = $quote_id"; $find_quote = mysql_query($sql_find_quote, $makeconnection) or die(mysql_error()); $row = mysql_fetch_assoc($find_quote); $totalRows = mysql_num_rows($find_quote); $quote_author = $_POST['quote_author']; $quote_desc = $_POST['quote_desc']; if (isset($_POST['submitted'])&&($_POST['submitted'] == "yes")) { $register_query = "SELECT quote_desc FROM tbl_quote WHERE quote_desc='$quote_desc'"; mysql_select_db($database, $makeconnection); $register_check=mysql_query($register_query, $makeconnection); $register_found_quote = mysql_num_rows($register_check); if($register_found_quote>1){ header ("Location: quote_modify.php?error=quoteexists"e=$quote_id"); }else{ $sql_modify = "UPDATE tbl_quote SET quote_desc = '$quote_desc' quote_author = '$quote_author' WHERE quote_id = '$quote_id'"; } mysql_select_db($database, $makeconnection); $Result1 = mysql_query($sql_modify, $makeconnection) or die(mysql_error()); header ("Location: quote.php"); } } ob_flush(); ?> [/php] here's my HTML form Code: [Select] <form action="" method="post" enctype="multipart/form-data" name="form1" id="form1"> <p> <input name="quote_author" type="text" class="active-field" id="quote_author" value="<? echo $row['quote_author'];?>"/> <br /> <input name="quote_desc" type="text" class="active-field" id="quote_desc" value="<? echo $row['quote_desc'];?>" /> <br /> <input name="submitted" type="hidden" id="submitted" value="yes" /> <input name="submit" type="submit" id="submit" class="submit-button" value="modiy"/> <br /> </p> </form> view_leader.php
Hello, I have this simple php form that updates the table, and also shows what the information in it is. Here it is: <strong>Update multiple rows in mysql</strong><br> <?php $host="localhost"; // Host name $username=""; // Mysql username $password=""; // Mysql password $db_name="test"; // Database name $tbl_name="test_mysql"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $sql="SELECT * FROM $tbl_name"; $result=mysql_query($sql); // Count table rows $count=mysql_num_rows($result); ?> <table width="500" border="0" cellspacing="1" cellpadding="0"> <form name="form1" method="post" action=""> <tr> <td> <table width="500" border="0" cellspacing="1" cellpadding="0"> <tr> <td align="center"><strong>Id</strong></td> <td align="center"><strong>Month</strong> <strong>Date</strong> </td> <td align="center"><strong>Lastname</strong></td> <td align="center"><strong>Email</strong></td> </tr> <?php while($rows=mysql_fetch_array($result)){ ?> <tr> <td align="center"><?php $id[]=$rows['id']; ?><?php echo $rows['id']; ?></td> <td align="center"><input name="month[]" MAXLENGTH="3" size="3" type="text" id="month" value="<?php echo $rows['month']; ?>">- <input name="date[]" MAXLENGTH="2" size="2" type="text" id="date" value="<?php echo $rows['date']; ?>"> </td> <td align="center"><input name="lastname[]" type="text" id="lastname" value="<?php echo $rows['lastname']; ?>"></td> <td align="center"><input name="email[]" type="text" id="email" value="<?php echo $rows['email']; ?>"></td> </tr> <?php } ?> <tr> <td colspan="4" align="center"><input type="submit" name="Submit" value="Submit"></td> </tr> </table> </td> </tr> </form> </table> <?php // Check if button name "Submit" is active, do this if (isset($_POST['Submit'])) { for($i=0;$i<$count;$i++){ $month = $_POST['month']; $date = $_POST['date']; $lastname = $_POST['lastname']; $email = $_POST['email']; $name = $month."<br>".$date; $sql1="UPDATE $tbl_name SET name='$name[$i]', month='$month[$i]', date='$date[$i]', lastname='$lastname[$i]', email='$email[$i]' WHERE id='$id[$i]'"; $result1=mysql_query($sql1); } } if($result1){ header("location:update2.php"); } ?> <?php $result = mysql_query("SELECT * FROM test_mysql") or die(mysql_error()); echo "<table border='1'>"; echo "<tr> <th>Name</th> <th>Age</th> </tr>"; // keeps getting the next row until there are no more to get while($row = mysql_fetch_array( $result )) { // Print out the contents of each row into a table echo "<tr><td>"; echo $row['name']; echo "</td><td>"; echo $row['lastname']; echo "</td></tr>"; } echo "</table>"; ?> What it does is this, it gets the row from the database, echos it and I am able to edit it and then save. The thing is, it works with the updating but im trying to combine 2 variables into 1 and update that one also. Im trying to combine month and date textboxes into one variable in the database that is name. It for some reason updates the column name to A instead of what I had in the 2 fields month and date. Here is where I combine both fields into one. $name = $month."<br>".$date; And here is where $month and $date is set. $month = $_POST['month']; $date = $_POST['date']; Here is a picture. Hi there, I have this query, which outputs the records properly in PhpMyAdmin. But when using this query inside my PHP code, the result is incomplete. Instead of retrieving the expected 15 results, I only get the first one twice. Also, I don't get any errors. Code: (php) [Select] <?php $get_files = "SELECT `id` FROM `files` WHERE `category` = 'W';"; include_once('connection.inc.php'); $con = mysql_connect(MYSQL_SERVER, MYSQL_USERNAME, MYSQL_PASSWORD) or die ("Connection failed: " . mysql_error()); $dbname = "myDB"; $select_db = mysql_select_db($dbname, $con) or die ("Can't open database: " . mysql_error()); $result_files = mysql_query($get_files) or die ("Query failed: " . mysql_error()); $files = mysql_fetch_array($result_files); $W_files = "'".implode("', '", $files)."'"; // echo $W_files; outputs the first number twice, instead of the expected 15 different ones $get_checklist = "SELECT * FROM `checklist` WHERE `id` IN ($W_files) ORDER BY `id` ASC;"; $result_checklist = mysql_query($get_checklist) or die ("Query failed: " . mysql_error()); // .... code for displaying ?> Hiya peeps! I have this code; $q3 = "SELECT `data` FROM `firm_text_strings` WHERE `id` = '100' AND `languages_id` = 'EN'"; $q4 = "SELECT `data_2` FROM `firm_text_strings` WHERE `id` = '100' AND `languages_id` = 'EN'"; $r3 = mysql_query($q3); $r4 = mysql_query($q4); if($r3 && $r4) { while($w3 = mysql_fetch_assoc($r3) && $w4 = mysql_fetch_assoc($r4)) { $resultStr = $w3['data'] . '<br />' . $w4['data_2']; } } else { $resultStr = 'error'; } everything is working up to the while function, it just wont do the loop for some reason. Anyone have any ideas. Many thanks, James. |
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