mysql_num_row() error

Question

This is the error that I am getting while using mysql_num_rows() function

Error: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource

Here is my source code:

$title = "SELECT * FROM item WHERE item.title LIKE % " .implode("% OR item.title LIKE % ", $data);

$title_result = mysql_query($title, $this->dbh);

echo mysql_num_rows($title_result);

What I am doing is creating a search for a website. Currently I am doing 4 Select statements and then I would like to total everything found in my database from all 4 of those sql statements. If I am thinking correctly the mysql_num_rows() returns the number of rows that was found from a Select statement. So in theory I was thinking that I could add them together to get my total found in a search, am I correct in this thinking if not how could I add 4 different sql select statements together?

Also my test search is in the database.

Thanks in advance for the help.

Answer 1

Looks like your query failed to execute. You need to check the return value of mysql_query before you try mysql_num_rows.

$title_result = mysql_query($title, $this->dbh);

if($title_result)
{
echo mysql_num_rows($title_result);
}
else
{
die(mysql_error());
}

What actually is happening in your case is mysql_query is returning false, which is not a valid MySQL result resource and when you pass it to mysql_num_rows it returns the following error:

Error: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource

Answer 2

Your MySQL statement did not execute correctly. This is just a hunch, but try again after replacing your $title with the following:

$title = 'SELECT * FROM item WHERE item.title LIKE "%'.implode('%" OR item.title LIKE "%', $data).'%"';

Then later, execute it as follows:

$result = mysql_query ($title, $this->dbh) or die (mysql_error($this->dbh));

Note: this is assuming you trust $data. If it comes from external sources (such as user-specified search), you really ought to sanitise via mysql_real_escape_string or prepared statements

Answer 3

Thanks a lot that fixed all of my problems. I can't tell you how thankful I am cause I looked at this for about 2 hours trying to figure it out and I was beginning to get very frustrated. :)