RE: [PHP-DB] mysql if statement to php

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/ > > http://mobile.yahoo.com.au - Yahoo! Mobile - Check & compose your email via SMS on your Telstra or Vodafone mobile.