Help needed!!! SQL Query..

[xiaoxin22]

i did this..

 

$query2 = sprintf("SELECT *  FROM (c_details INNER JOIN c_color ON c_details.cc_id = c_color.cc_id)  WHERE c_color.cc_id = %s",

mysql_real_escape_string($colour));

$result2 = mysql_query($query2);

$print2 = mysql_fetch_assoc($result2);

 

it is working but it keep showing error message..

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/sip/public_html/test/calender.php on line 90

 

can someone tell me what's wrong with it?

[Ruzzas]

WHERE %s?

 

I don't think PHP Supports that.

[xiaoxin22]

it's working.. cause i use this for all my php.. but never encounter this problem before..

[aneesme]

Are you sure you selected a database?

 

Regards,

Anees

http://ask.amoeba.co.in

[xiaoxin22]

yea.. everything is fine.. it is appearing.. what bothers me is that error message...

[trq]

That error message indicates that whatever you are passing to mysql_fetch_assoc on line 90 is not a result resource.

[xiaoxin22]

what does it mean by not result resource? can elaborate more?

[robert_gsfame]

Single quotes missing

 

%s  should be '%s'

[xiaoxin22]

OH!! AMAZING!!! it works!!! thank you very much.. i will continue to work on it..

[trq]

The problem is you where passing $result2 to mysql_fetch_assoc without first checking to make sure your query succeeded. mysql_fetch_assoc expects a result resource, which is what mysql_query returns if it succeeds, but when it fails it returns false.

 

While your at it, you should likely check your query returns some records as well.

 

$query2 = sprintf("SELECT *  FROM (c_details INNER JOIN c_color ON c_details.cc_id = c_color.cc_id)  WHERE c_color.cc_id = %s", mysql_real_escape_string($colour));
if ($result2 = mysql_query($query2)) {
  if (mysql_num_rows($result2)) {
    $print2 = mysql_fetch_assoc($result2);
  } else {
    // no records found.
  }
} else {
  // query failed
}