PHP error

Could you please post the lines above this as well? Up to the query?

Where did that variable $query come from ?

It should come from a mysql_query eg:

$query = mysql_query("SELECT... blah blah blah ");

(for the SQL purists out there: blah is not valid SQL syntax... i just used it as placeholder for a valid SQL query )

I think the error is quite clear. $query is not valid. Come on, if you learn 1+1 you surely should know what 1+1+1 would be.

--
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Rick Wong
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As aforementioned, $query is invalid. No need to post the code preceding this line.

In order to debug your code easier try using something like this right after the query is made:

if (!$query)
{
die(''Query failed: ''.mysql_error());
}

If the query has an error, then $query will return false. This if statement determines if there is a nerror and reports it, then terminates the script if there is an error. I hope this helps.

Actually its not the query first off. The actual function is mysql_num_rows. However its not to wise to put the actual query in a function like that.

Try This:

$connection = mysql_connect( "$dbhost", "$dbuser", "$dbpass");
$query = mysql_query("QUERY GOES HERE");
$results = mysql_query($query, $connection) OR DIE ("Error in query: $query. " . mysql_error());

Then use:

$num = mysql_num_rows ( $results );

That works. Hope it helps ya.

Chip Lambert
President/Lead Programmer LambertSoft

[edited by - Chip on June 7, 2004 10:09:43 AM]

quote:

---

Original post by Chip
Try This:

$connection = mysql_connect( "$dbhost", "$dbuser", "$dbpass");
$query = mysql_query("QUERY GOES HERE");
$results = mysql_query($query, $connection) OR DIE ("Error in query: $query. " . mysql_error());

---

That''s quite bizarre and wrong. You wouldn''t pass the result of mysql_query to another mysql_query.



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Ooops my bad, I meant

$query = "QUERY GOES HERE";
$results = mysql_query($query, $connection) OR DIE ("Error in query: $query. " . mysql_error());

I apologize about that.

thanks for the help, this is what finally worked...

mysql_connect($server,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$result=mysql_query("SELECT * FROM accounts WHERE username=''$loginid''");
$num=mysql_num_rows($result);