A weird PHP problem

[cemlouis]

Hi,

I have a code like below:

$sql = mysql_query("SELECT id FROM data WHERE link LIKE '%$search%' LIMIT 0,30") or die (mysql_error()); 
if($sql > 0){
   echo "Yes we have the record in here!";
}

When I run the code the runner enters in the if statement because $sql has a value of 4 but it mustn't because when I run the query in phpmyadmin sql query tab there were no results. So what I am missing?

Thank you,
Cem

[blackhorse]

$sql has the value 4. the 4 must be the resouce id.

$sql = mysql_query("SELECT id FROM data WHERE link LIKE '%$search%' LIMIT 0,30") or die (mysql_error());
$sql_row=mysql_fetch_object($sql );
if($sql_row > 0){
echo "Yes we have the record in here!";
}

[cemlouis]

Hi blackhorse, thank you this is what I was looking for...

[blackhorse]

NP, usually you will use while loop with the mysql_fetch_object

$sql = mysql_query("SELECT id FROM data WHERE link LIKE '%$search%' LIMIT 0,30") or die (mysql_error());
while ($sql_row=mysql_fetch_object($sql ))
{

//handle the data one record row a time here.

}