digarati
asked on
1064 Error when trying to DELETE FROM table
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.
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));
}
}?>
ASKER
oku86 thanks for that update. was working. but this i guess is more formal.
i still have the 1064 error, anyone?
i still have the 1064 error, anyone?
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
$sql = "DELETE FROM `vendor` WHERE `id`= '$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
ASKER
I now get the following error
Parse error: parse error, unexpected '=' in /home/content/site/html/be ta/delete_ vendor.php on line 12
Parse error: parse error, unexpected '=' in /home/content/site/html/be
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ASKER
$sql = "DELETE FROM vendor WHERE id=$id";
$sql2 = "DELETE FROM scope WHERE `key`=$id";
what worked for me......
$sql2 = "DELETE FROM scope WHERE `key`=$id";
what worked for me......
Im glad we managed to figure it out in the end :)
if ($res === TRUE) {
it should be if ($res == TRUE) (
It should only be two equal signs