[RESOLVED] Probelm with if/then statement

sleepyw

I have a drop-down list on a page that has either "NEW" as a value, or an existing item from the db. When that form is submitted, it goes to a page where I have another form. I want an if/then statement that says if the value from the previous form = "NEW", then don't do anything. Otherwise, run a db query to get the info on the item the user DID select to populate the form on the new page.

Here's what I have:

//posted value from previous page form
$Course_Code = $_POST['Course_Code'];


if ( $Course_Code == NEW ){
die();
}
else {
$sql = "SELECT TOP 1 * FROM table_name WHERE Course_Code=$Course_Code ORDER BY Date";
$result=mysql_query($sql);
$rows=mysql_fetch_array($result);
}

However I get this error:

Parse error: parse error, unexpected ')', expecting T_STRING or T_VARIABLE or '$' in /......... on line 24

Line 24 is where the if statement begins. What am I doing wrong?

coldwerturkey

'NEW' or "NEW" not NEW

google 'php strings' to find out why

also, if $Course_Code isn't a integer, you'll get a mysql phase error.. it needs single quotes around it (I'm just bringing that up because most course codes i've seen in my days have characters) ..oh also you should escape it with mysql_real_escape_string() since you're using it in a query

sleepyw

Man, I'm such a n00b. Thanks.

While that fixed the problem, I'm getting another error with the way I'm trying to populate to new form. :o

I get a 'supplied argument is not valid' error for this line:

$rows=mysql_fetch_array($result);

Oh, and $Course_Code is alphanumeric.

Never mind. The error was caused by the "Top 1" in the select statement. Apparently that doesn't work. I'll have to find a workaround for that.

Thanks again for your help.

bradgrafelman

sleepyw wrote:
I'll have to find a workaround for that.

Namely MySQL's LIMIT syntax.

Don't forget to mark this thread resolved (if you haven't already). Feel free to also post an updated version of your code that works as well.