ourloop

The below code displays correctly, I need an "IF" statement that prevents displaying rows with 0% in the percent column and I've tried several conditions. None are working. I only want to display rows that has at least 1% in the percent column. Here is the code: $result = mysql_query("SELECT …

I have a MySQL database with a table that has a field/column labled "user8" (which is the field for users' state). I have this PHP call and its working fine if I call all states individually and I only want numbers of rows. <?PHP mysql_select_db("freeskat_promail") or die(mysql_error()); $result = mysql_query("SELECT …

Hi, This code is from an older thread. I'm want to use the code, but I'm having some difficulty with the MySQL table creation. Here's the code: [CODE]<? session_start(); require("connection.php"); if(isset($_POST['submit_btn_name']) && $_POST['submit_btn_name']!='') { $title = mysql_real_escape_string($_POST['title']); $date = mysql_real_escape_string($_POST['date']); $author = mysql_real_escape_string($_POST['author']); $type = mysql_real_escape_string($_POST['type']); $description = mysql_real_escape_string($_POST['description']); $vidName …