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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