> 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/
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
