Maybe try NOW() or CURRENT_DATE()?

Just a guess and sorry if it's already been mentioned

Dan

 
-------------------
http://chrome.me.uk
 

-----Original Message-----
From: Joe Wollard [mailto:[EMAIL PROTECTED] 
Sent: 04 April 2006 00:38
To: Paul Goepfert
Cc: Jon Drukman; php-general@lists.php.net
Subject: Re: [PHP] Re: mysql_fecth_array() and function call as parameter

Paul,

Try TO_DAYS(curdate()) instead - if not you can't say we didn't try! ;-)

If that doesn't work you may want to try a MySQL list.

- Joe



On 4/3/06, Paul Goepfert <[EMAIL PROTECTED]> wrote:
>
> 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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