I included the or die function on the end of my query statement. When I tested this on my web page I got the following error Query2 Failed: You have an error in your SQL syntax near '(curdate())' at line 1 When this code executed: $query2 = mysql_query("SELECT dayNum FROM Days WHERE dayNum = day(curdate())") ; I can't see where the error is. Can anyone see the error? The table name is correct because I have another drop down box that gets all dates in the table and outputs it with no error. I also tested the above SQL statement on a local copy of mysql (version 5.0.18-nt) and it worked with out any errors. And this code works on a different web server. Mysql client version 3.23.49. However it does not work on mysql client version 3.23.54. Also both php versions are the same. Thanks Paul On 4/3/06, Jon Drukman <jsd@xxxxxxxxxxxxx> wrote: > Paul Goepfert wrote: > > > function determineDay () > > { > > $return = ""; > > $query1 = mysql_query("SELECT months FROM Month WHERE m_id = > > month(curdate())"); > > $query2 = mysql_query("SELECT dayNum FROM Days WHERE dayNum = > > day(curdate())"); > > $query3 = mysql_query("SELECT year FROM Year WHERE year = year(curdate())"); > > always Always ALWAYS check the error return!!! > > $query1 = mysql_query("SELECT months FROM Month WHERE m_id = > month(curdate())") or die("query1 failed: " . mysql_error()); > > do this religiously on every single mysql_query. in fact, write a > wrapper function to do it for you - that's what i do. > > > If anyone can find my error please let me know. I have looked at this > > for about an hour and I can't figure it out. > > you probably could have saved an hour by checking the error code. > > -jsd- > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php