counting the results of only one cell

SureFire

This is close to what I want to do, but I only want to show the results of one of the pages visited.

I want to replace the * with "page1.php" but it gives me an array error.

Any Ideas?

<?


mysql_connect(localhost,$username,$password); 
@mysql_select_db($database) or die( "Unable to select database"); 
$query = "SELECT page_visited, COUNT(*) as numArticles FROM site_stats GROUP BY page_visited ORDER BY page_visited ASC";  

$results=mysql_query($query); 

mysql_close();



$i=0;
while ($row = mysql_fetch_array($results)) {

echo "




$row[numArticles]<br> ";


++$i;
}

?>

barand

Add

... HAVING page_visited = 'page1.php'

to the end of your query

hth

SureFire

now my querey says the following

 
$query = "SELECT page_visited, COUNT(*) as numArticles FROM site_stats GROUP BY page_visited ORDER BY page_visited ASC HAVING page_visited = 'page1.php'";

but I get this error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/internet/www/www/test/stattest.php on line 16

Thanks though!

barand

Don't close connection until you have finished using it!

SureFire

I tried moving the close statement to the end of my code and the error message remains the same.

Thanks again!

SureFire

ok, I added a WHERE command to my mess of code and it worked Thanks for all the input barand!