On Saturday 25 May 2002 01:29, Steve Buehler wrote:
> I am having trouble with the following function. What it should do is to
> check one table for team_id's. Than it goes to another table and gets all
> rows with that team_id. If the team_id is in the new table, it should do
> one thing, else it should do something else. Can somebody look at this
> code and see if they can find out where my problem is besides in my
> inexperience? It always goes to the else statement inside the if
> statement.
Your biggest problem is that you're trying to nest mysql_query() but you're
only using 1 link identifier. You need to establish another connection using
another mysql_connect().
> function searchbyteamname($team_name){
> GLOBAL $PHP_SELF;
> $result=mysql_query("SELECT team_id,name FROM team WHERE name like
> '%$team_name%' AND deleted NOT LIKE '1' ORDER BY 'name'");
> $i=1;
> while (($row=mysql_fetch_object($result))){
> if(!($result1=mysql_query("SELECT * FROM team_season WHERE team_id =
> '$row->team_id' AND deleted NOT LIKE '1'"))){
The value returned from mysql_query() is an indication of whether or not the
query itself succeeded. It is not an indication of whether the query returned
any records.
--
Jason Wong -> Gremlins Associates -> www.gremlins.com.hk
Open Source Software Systems Integrators
* Web Design & Hosting * Internet & Intranet Applications Development *
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