SQL syntax error

[NZ_Kiwis]

I have a code which lists all my results from results_group then I edit a number the "group1" box and I want to set this number as order in the update code.... when clicking save it does this

<?php 
if ($action == "save") 
{ 
$c = 1;
$result5 = mysql_query ("SELECT * From results_group where comp_run_id = $comp_run_id");
while($row5 = mysql_fetch_array($result5)) {
$group_id = $row5[group_id];

echo $group_id;

$result = mysql_query ("update results_group set order = '$group1' where group_id = $group_id") or die("INSERT error0: ".mysql_error()); 


}
?>
 

This is my error
38INSERT error0: You have an error in your SQL syntax near 'order = '38' where group_id = 38' at line 1

Any ideas how to fix it..

[planetsim]

<?php 
if ($action == "save") 
{ 
$c = 1;
$result5 = mysql_query ("SELECT * From results_group where comp_run_id = $comp_run_id");
while($row5 = mysql_fetch_array($result5)) {
$group_id = $row5[group_id];

echo $group_id;

$result = mysql_query ("UPDATE results_group set
                       `order` = '$group1' 
                       where group_id = $group_id
                       ") or die("INSERT error: ".mysql_error()); 


}
?>

I modified your SQL format so you can tell where exactly the error is however your using a field name order which is a reserved word for SQL for ORDER BY, you will need to use backticks if you wish to use reserved words.