RE: [PHP] Passing by conditional IF statement...why?

2004-03-22 Thread Pablo Gosse 
Ryan A wrote:
 Hi,
 I have this simple code in my php script:
 
 * * * * *
 $res = mysql_query(SELECT product_id, now()-1 FROM .$tc._prods
 where cno=$cno AND product_id='$product_id' LIMIT 1); 
 
 if($res)
  {
  $r = mysql_fetch_row($res);
  $product_id2   = $r[0];
  $th_pres= $r[1];
 echo debug echo;
  }else {echo No results, sorry;}
 * * * * *
 
 its working great when the data actually exists but when there are no
 matches it still executes the if($res) part instead of displaying
 No results, sorry. Why is that? or am I using the syntax wrong?  
 
 Thanks,
 -Ryan

You need to test the number of records being returned.

By just using if ($res) you're simply ensuring that it is returning a
valid resource.

It is, but it's just that the resource has zero rows.

Change the test such that it check the number of results to be = 1 and
it should work.

Cheers,
Pablo

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RE: [PHP] Passing by conditional IF statement...why?

2004-03-22 Thread Martin Towell 
Ryan

Because you are actually getting a result, a result with no records...
Might be better to use
http://au2.php.net/manual/en/function.mysql-num-rows.php instead

Martin

 -Original Message-
 From: Ryan A [mailto:[EMAIL PROTECTED]
 Sent: Tuesday, 23 March 2004 11:37 AM
 To: [EMAIL PROTECTED]
 Subject: [PHP] Passing by conditional IF statement...why?
 
 
 Hi,
 I have this simple code in my php script:
 
 * * * * *
 $res = mysql_query(SELECT product_id, now()-1 FROM 
 .$tc._prods where
 cno=$cno AND product_id='$product_id' LIMIT 1);
 
 if($res)
  {
  $r = mysql_fetch_row($res);
  $product_id2   = $r[0];
  $th_pres= $r[1];
 echo debug echo;
  }else {echo No results, sorry;}
 * * * * *
 
 its working great when the data actually exists but when there are no
 matches it still executes the if($res) part instead of
 displaying No results, sorry.
 Why is that? or am I using the syntax wrong?
 
 Thanks,
 -Ryan

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To unsubscribe, visit: http://www.php.net/unsub.php