Jump to content

Recommended Posts

It's the strangest error. When ever I use the code:

 

	$query = "SELECT $fld_email FROM $tbl_name WHERE $fld_email = '$email'";
	$result = mysql_query($query, $conn);
	return (mysql_num_rows($result) > 0);

 

I always get a syntax error:

 

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/~/public_html/~/~/FILE.php on line 100

 

However, if i choose to substitute the variables for their direct values, it seems to work:

 

	$query = "SELECT email FROM clients WHERE email = '$email'";
	$result = mysql_query($query, $conn);
	return (mysql_num_rows($result) > 0);

 

The variable's look like:

 

$tbl_name = "clients";
$fld_email = "email";

 

I've tried single quoting the variable names (in the sql query), as well as changing the names of the variables. It's not really a problem, however, I am curious as to why. So, any ideas?

Link to comment
https://forums.phpfreaks.com/topic/84726-solved-small-sql-syntax-problem/
Share on other sites

This thread is more than a year old. Please don't revive it unless you have something important to add.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Restore formatting

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.