Yet another syntax issue.

[dlyles]

The below coding is supposed to look through the db and if the record doesn't exist, insert the record into the db.

[code]
$groupname=mysql_real_escape_string($_POST['groupname']);

$sqll="select * from groups where GroupName = '". $groupname . "'";
$nresult=mysql_query($sqll);
$nrow=$row['groupname'];
$count=mysql_num_rows($nresult);
if($count==1){

$sql = "insert into groups (GroupName) VALUES ('" . $groupname . "')";
$result = mysql_query($sql) or die("Error 2: query: $sql<br>" . mysql_error());
echo $sql;
echo "1 record added";

}
else
{
echo "nah";
}

?>

[/code]
I've put data in the field that I know is already in there and some I know aren't and I get the same result.  It's assuming the value is there and not updating the db.

[Psycho]

[quote]It's assuming the value is there and not updating the db.[/quote]

The logic in your code is bacwards. Basically you are checking the database to see if a particular groupname already exists. If it DOES exist then you add it to the database again. Shouldn't that add the group name if it does NOT already exist? Also, unless you have set the groupname field int he database to be a unique column you shouldn't use if($count==1){, instead use if($count>0){

Also this line [code]$nrow=$row['groupname'];[/code] has no purpose and wouldn't equal anything anyway since you did not retrieve the data from the query using something liek mysql_fetch_assoc().

[kenrbnsn]

If there is no record there, then $count would be zero, not 1, so the "if" statement should be
[code]<?php
if($count == 0){
?>[/code]

Also, you have this statement
[code]<?php
$nrow=$row['groupname'];
?>[/code]
which won't give you anything meaningful, since you haven't fetched any rows at that time.

Ken

[dlyles]

Thanks guys!