Change your query to be:

$query_cat = "select concat(author_names, ' ', author_surnames) as 
somenewcolname from author, author-cat
where author_cat.cat_code = $code
and author.author_code = author-cat.author_code";

Change somenewcolname to whatever you want the column to be called or leave
the "as somenewcolname" part off if you don't care what the column name is.
This should help.


-------Original Message-------

From: Wilmar Perez
Date: Friday, September 06, 2002 12:16:28
Subject: RE: [PHP-DB] not a valid MySQL result

Hi guys

Thanks for your help. Peter was right I had an error in the sentence, I 
worked it out but still the same error came up. The mysql_error() function 
doesn't return any thing.

Any idea?

Thanks a lot.


>Couple of points, you probably have an SQL error,

>I use

>echo $query;
>echo '<br>'.mysql_error();

>which shows the query and the error

>you use 'author-cat' in the table name
>and 'author_cat' and 'author-cat' in the where clause which will throw an

Wilmar Pérez
Network Administrator
Library System
Tel: ++57(4)2105145
University of Antioquia
Medellín - Colombia

PHP Database Mailing List (
To unsubscribe, visit:


Reply via email to