whats wrong with this? (mysql results/array)

abda53

<?php 
if (isset($search))
{
	$search = $_POST['search'];
	$sql = "SELECT * FROM `logs` WHERE station LIKE \"%$search%\" OR url LIKE \"%$search%\" ORDER BY id DESC";
	$result = mysql_query($sql) or die ("Couldn't execute query 1.");
	$found = mysql_num_rows($result);
	//$row = mysql_fetch_array($result);
	//$url = $row['url'];

$url_times = array();
foreach($found as $url) 
{
	if( !in_array($result, $search_results_array) ) 
    {
		//do yo shit 
        $tmp = array($result); 
        $search_results_array = array_merge($search_results_array,$tmp);
		$found_num++;
    }
	if( in_array($result, $search_results_array) )
	{
		if( !array_key_exists($result, $url_times) )
		{
			$tmp = array($result => 1);
			$url_times = array_merge($url_times,$tmp);
		}
		else
		{
			$url_times[$result] = $url_times[$result] + 1; 
		}
	}
}

echo "Found <b>".$found_num."</b> urls containing the word <b>".$search."</b><br>";

foreach($url_times as $site => $times)
{
	echo "<i>".$site."</i> was found a total of <i>".$times."</i> times";
}
echo "</i> was found a total of <i>".$times."</i> times";

}

?>

basically it will search the database for a url, show the url only once, and then show how many times that url was found.

Piranha

What does it show now? Do you get any error messages?

abda53

sorry.

it shows this

Warning: Invalid argument supplied for foreach() in D:\www\webroot\search.php on line 46

Notice: Undefined variable: found_num in D:\www\webroot\search.php on line 69
Found urls containing the word my

Notice: Undefined variable: times in D:\www\webroot\search.php on line 75
was found a total of times

line 46 - foreach($found as $url)
line 69 - echo "Found ".$found_num." urls containing the word ".$search." ";
line 75 - echo " was found a total of ".$times." times";

thanks in advance

Piranha

$found that is used i row 46 is an integer, and it should be an array to be used in a foreach loop. It is assigned in the row $found = mysql_num_rows($result);

The rest of the errors are not worth looking at before you have corrected this one. It is pretty common that one error leads to another, that's why you should only look at the first one.

NogDog

As currently coded, $found is an integer (the result of mysql_num_rows()), not an array. The typical method for looping through the results of mysql_query() is a while loop:

$result = mysql_query($sql) or die('error msg');
while($row = mysql_fetch_array($result))
{
   // do whatever you need to do with array $row for this result row
}