Hi,
I tried you 'as' example and got this:
Warning: mysql_fetch_array(): supplied argument is not
a valid MySQL result resource in message.php
My code is this:
$qr = mysql_query("SELECT IF( emailerror = ('y' ),
'error yes', 'do nothing' ) AS answer FROM tipping
WHERE sid =
'$sid'");
$result = mysql_query($qr);
$row = mysql_fetch_array($result);
echo $row['answer'];
Any ideas?
Jerry
--- "John W. Holmes" <holmes072000@charter.net>
wrote: > > I am using a php page to grab information
using
> the if
> > statement. Now the mysql query works fine but I
> can't
> > get the output to output into php. I have tried
> many
> > things and have failed to get it to display.
> >
> > This is the code:
> >
> > <?php
> > mysql_connect ("localhost", "$usr", "$pass") or
> die
> > ('I cannot connect to the database.');
> > mysql_select_db ("$db");
> >
> > $qr = mysql_query("SELECT IF( emailerror = ('y' ),
> > 'error yes', 'do nothing' ) FROM tipping WHERE sid
> =
> > '$sid'");
> >
> > ?>
> >
> > If I do the above query in mysql it returns
> correctly.
> > If Y is in the table it dispalys error yes. But
> how
> > do I get it to output to a php page with the above
> > function?
>
> Run the query through mysql_query(), fetch the
> result row with
> mysql_fetch_row() and display $row[0].
>
> $result = mysql_query($qr);
> $row = mysql_fetch_row($result);
> echo $row[0];
>
> OR
>
> $result = mysql_query($qr);
> echo mysql_result($result,0);
>
> OR
>
> $qr = mysql_query("SELECT IF( emailerror = ('y' ),
> 'error yes', 'do nothing' ) AS answer FROM tipping
> WHERE sid =
> '$sid'");
> $result = mysql_query($qr);
> $row = mysql_fetch_array($result);
> echo $row['answer'];
>
> Notice the "AS answer" added to the query in the
> last example.
>
> ---John W. Holmes...
>
> PHP Architect - A monthly magazine for PHP
> Professionals. Get your copy
> today. http://www.phparch.com/
>
>
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