[SOLVED] MySql Search Problem !

[nita]

 

HI

I have following code for search my recipe database. Once I input search word (existing in database)

is displaying 0 results.

and error:

 

 

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in .... on line 217

 

 

My code:

$PHP_SELF = $_SERVER['HTTP_REFERER'];
$search=$_POST["search"];

$minchar = 2;
if (strlen($search) < $minchar)
{
	echo "Search must be longer than 2 characters.
} 
else 
{
    $result1 = mysql_query("SELECT * FROM recipe WHERE name LIKE '%$search%' 
    OR desc LIKE '%$search%'");

    $numrows1=mysql_num_rows($result1); // line 217
    if ($numrows1 == 0)
  	    {
		echo "<p>Sorry, your search returned 0 results</p>
  	    }
    else 
    { 
        while($row=mysql_fetch_array($result1))
        {	
       	        include "display_recipe.php";
        }
    }
}

 

I'm using similar code in another application and it works as it should be.

 

 

If some one can help me to find solution for this problem... please

 

Thanks a lot in advance...

Nita

 

[MadTechie]

change

<?php
$result1 = mysql_query("SELECT * FROM recipe WHERE name LIKE '%$search%' 
    OR desc LIKE '%$search%'");
?>

 

to

 

<?php
$result1 = mysql_query("SELECT * FROM recipe WHERE name LIKE '%$search%' 
    OR desc LIKE '%$search%'") or die(mysql_error());
?>

 

any errors ?

[nita]

it was about that "desc"  is one of the reserved words in php, i renamed column to "description" and works perfectly.

[MadTechie]

yeah noticed that

 

but error check is always worth adding