uaeustudent Posted May 29, 2007 Share Posted May 29, 2007 I have this code for dislpay information (by the id of the user) was stored in the database but it give an error and I think it from this line $id = $_GET['id']; $query = "SELECT * FROM $table WHERE id='$id'LIMIT 1"; And this my code: ]<html> <head> <title>Title here!</title> </head> <body> <?php $host = "localhost"; $user = ""; $password = ""; $db = "Job seeker"; $table = "personalinformation"; mysql_connect($host,$user,$password) or die("Failed to connect"); mysql_select_db($db) or die("Falied to select database"); $id = $_GET['id']; $query = "SELECT * FROM $table WHERE id='$id'LIMIT 1"; //$query = "SELECT * FROM $table WHERE id='$id' LIMIT 1"; $result = mysql_query($query) or die("problem with query: ".$query); $row = mysql_fetch_array($result); if ($row){ // echo "<td><td><tr>$myrow[id]</tr><tr>$myrow[Address]</tr><tr>$myrow[City]</tr><tr>$myrow[Fax]</tr><tr>$myrow[specialization]</tr><tr>$myrow[TargetSalary]</tr><tr>$myrow[JobType]</tr><tr>$myrow[WillingToRelocate]</tr></td></td>" ; } else { echo "mysql Error: " . mysql_error(); } ?> <h3> Personal Information form</h3> <table border=1> <tr> <td>id</td> <td><?=$row['id'];?></td> </tr> <tr> <td>Address</td> <td><?=$row['Address'];?></td> </tr> <tr> <td>City</td> <td><?=$row['City'];?></td> </tr> <tr> <td>Fax</td> <td><?=$row['Fax'];?> </td> </tr> <tr> <td>Specialization</td> <td><?=$row['Specialization'];?></td> </tr> <tr> <td>TargetSalary</td> <td><?=$row['TargetSalary'];?></td> </tr> <tr> <td>JobType</td> <td><?=$row['JobType'];?></td> </tr> <tr> <td>WillingToRelocate</td> <td><?=$row['WillingToRelocate'];?> </td> </tr> </table> </body> </html> Link to comment https://forums.phpfreaks.com/topic/53442-mysql-error/ Share on other sites More sharing options...
eric1235711 Posted May 29, 2007 Share Posted May 29, 2007 what the error message tells? Link to comment https://forums.phpfreaks.com/topic/53442-mysql-error/#findComment-264075 Share on other sites More sharing options...
uaeustudent Posted May 29, 2007 Author Share Posted May 29, 2007 It display mysql Error: Link to comment https://forums.phpfreaks.com/topic/53442-mysql-error/#findComment-264126 Share on other sites More sharing options...
wildteen88 Posted May 29, 2007 Share Posted May 29, 2007 Chnage this: query = "SELECT * FROM $table WHERE id='$id'LIMIT 1"; //$query = "SELECT * FROM $table WHERE id='$id' LIMIT 1"; $result = mysql_query($query) or die("problem with query: ".$query); $row = mysql_fetch_array($result); if ($row){ // echo "<td><td><tr>$myrow[id]</tr><tr>$myrow[Address]</tr><tr>$myrow[City]</tr><tr>$myrow[Fax]</tr><tr>$myrow[specialization]</tr><tr>$myrow[TargetSalary]</tr><tr>$myrow[JobType]</tr><tr>$myrow[WillingToRelocate]</tr></td></td>" ; } else { echo "mysql Error: " . mysql_error(); } to this: $query = "SELECT * FROM $table WHERE id='$id'LIMIT 1"; //$query = "SELECT * FROM $table WHERE id='$id' LIMIT 1"; $result = mysql_query($query) or die('problem with query: ' . $query . '<br />' . mysql_error()); if(mysql_num_rows($result) == 1) { $row = mysql_fetch_array($result); echo "\n<tr> <td>{$myrow['id']}</td> <td>{$myrow['Address']}</td> <td>{$myrow['City']}</td> <td>{$myrow['Fax']}</td> <td>{$myrow['Specialization']}</td> <td>{$myrow['TargetSalary']}</td> <td>{$myrow['JobType']}</td> <td>{$myrow['WillingToRelocate']}</td> </tr>\n"; } else { echo 'No results returned'; } Link to comment https://forums.phpfreaks.com/topic/53442-mysql-error/#findComment-264225 Share on other sites More sharing options...
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