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1064 Error when trying to DELETE FROM table

Posted on 2008-06-17
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Medium Priority
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482 Views
Last Modified: 2013-12-12
Assistance needed for the following error:

Your request has been sent.
There was an error in your request. Please contact support and reference: 1064

Code is snipped.

The following script is deleting two records from two tables that are relationshiped by their id and key.

<?php
session_start();
if(!isset($_SESSION["sessioname"])){
header("location: http://www.site.com/beta/login2.html");
}else{
////////////////////////////////////////////////////////////////////////////////////////////////////MY PROTECTED CONTENT
include 'header.php';
require_once "dbconnection.php";
 
$id = ($_GET['id']);
 
$sql = "DELETE FROM vendor WHERE id= '$id'";
$sql2 = "DELETE FROM scope WHERE key= '$id'";
 
$res = mysql_query($sql,$mysql);
$res2 = mysql_query($sql2,$mysql);
$link="<a href=\"http://www.site.com/beta/home.php\">Return</a>";
 
 
//deleting vendor from database
if ($res === TRUE) {
            echo "<br/>";
			echo "Your request has been sent.<br/>";
			
			} else {
            printf("There was an error in your request. Please contact support and reference: %s\n", mysql_errno($mysql));
      }
	  
//deleting scope from database
if ($res2 === TRUE) {
            echo "<br/>";
			echo "Your request has been sent.<br/>";
			
			} else {
            printf("There was an error in your request. Please contact support and reference: %s\n", mysql_errno($mysql));
      }
     
}?>

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Question by:digarati
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7 Comments
 
LVL 1

Expert Comment

by:oku86
ID: 21804327
Regarding

if ($res === TRUE) {

it should be if ($res == TRUE) (

It should only be two equal signs
0
 
LVL 1

Author Comment

by:digarati
ID: 21804350
oku86 thanks for that update. was working. but this i guess is more formal.
i still have the 1064 error, anyone?
0
 
LVL 4

Expert Comment

by:DrBrainiac
ID: 21805537
Try this:

$sql = "DELETE FROM `vendor` WHERE `id`= '$id'";
$sql2 = "DELETE FROM `scope` WHERE `key`= '$id'";

You get this error when you use reserved words from MySQL. I suspect it would be the "key" word in this case. To make sure you never get this problem again use the backtick character as in the code above.

You can find more informatiom about this on: http://dev.mysql.com/doc/refman/5.0/en/function-resolution.html
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LVL 1

Author Comment

by:digarati
ID: 21805565
I now get the following error

Parse error: parse error, unexpected '=' in /home/content/site/html/beta/delete_vendor.php on line 12
0
 
LVL 1

Accepted Solution

by:
oku86 earned 2000 total points
ID: 21807999
$sql = "DELETE FROM vendor WHERE id=$id"
$sq2l = "DELETE FROM scope WHERE key=$id"

 - This should work fine - i use the same in my scripts and it deletes the record without a problem
0
 
LVL 1

Author Closing Comment

by:digarati
ID: 31467989
$sql = "DELETE FROM vendor WHERE id=$id";
$sql2 = "DELETE FROM scope WHERE `key`=$id";
what worked for me......
0
 
LVL 1

Expert Comment

by:oku86
ID: 21813656
Im glad we managed to figure it out in the end :)
0

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