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 <[EMAIL PROTECTED]> 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-
>

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