[SOLVED] PHP log warning for mysql_num_rows

[oskom]

Hello all,

In trying to clean up some PHP errors, I frequently get a "PHP Warning"(mysql_num_rows(): supplied argument is not a valid MySQL result resource in /usr/local/ftp/path/to/file/content.php on line 30). Here's the code in question:

$result = mysql_query('query here');
if(mysql_num_rows($result) == 0) { //this is the suspect line

I'm pretty sure the fix is:

$result = mysql_query('query here');
if(mysql_num_rows($result) == false) { //0 changed to false

In reading PHP.net, http://us2.php.net/mysql_num_rows, it states for returned values "The number of rows in a result set on success, or FALSE on failure."

 

Am I on the right track? I'm asking only because success with bug-checking using the error log tends to be spotty at best.

[teng84]

your query maybe failed thats why your getting that error post here your actual sql statement

[oskom]

The code for the query is:

$result = mysql_query("SELECT * FROM content WHERE contentID = ".$_REQUEST['contentID']."");

[teng84]

$result = mysql_query("SELECT * FROM content WHERE contentID = ".$_REQUEST['contentID']."") or die (mysql_error());

 

try and print here the result

[cooldude832]

you shouldn't use the raw data from any form like REQUEST GET POST etc

[rab]

Filter that variable as an integer and quote it in the SQL command.

[PFMaBiSmAd]

I am surprised there is not a sticky post concerning that error. It has been asked about four times in the past couple of hours.

[oskom]

Point well taken, cooldude832, about sticking raw data into a query...note to self. On the error, I should say that I'm getting all the desired results from this query. Essentially, the if statement is checking if the ID passed in a URL string exists as an ID for a row in the database. If it returns nothing, an error message displays. It all works fine, if tenuously, as the PHP error suggests.

 

Let me ask, however, if doing this would be just as well...

$result = mysql_query("SELECT * FROM content WHERE contentID = ".$_REQUEST['contentID']);
if(!$result) {

 

Or is it 6-or-one-half-dozen?

[rab]

<?php
$_REQUEST['contentID'] = (int)$_REQUEST['contentID'] < 0 ? 0 : (int)$_REQUEST['contentID'];

if( $_REQUEST['contentID'] > 0 ) {
   $result = mysql_query("SELECT * FROM content WHERE contentID = '{$_REQUEST['contentID']}'");
   if(!$result) {
      // ERROR
   }
} else {
   // Need an ID
}
?>

[oskom]

Thanks, rab. By the way(stinking novice question to follow), the first part of this code I get...

$_REQUEST['contentID'] = (int)$_REQUEST['contentID']

It's this part that tweaks my gray matter...

 < 0 ? 0 : (int)$_REQUEST['contentID'];

I've seen this syntax before but never quite understood it's function. Is that a short-form of an if/else statement?