emilcarlo Posted August 13, 2010 Share Posted August 13, 2010 Good evening, This has been bugging me for 24 hours now. I have created 2 web applications. The first one is the dummy, and the second one is the final. Basically, this two web applications does the same, just a few modifications made on the final one. I have a process of pulling out of information from the database should the user would want to modify a field. I have a problem though, the dummy one works perfectly, but the new one don't. In fact, if I use the dummy php file together with the new files, the dummy won't be working. But if I use it together with the old files, it works perfectly fine. This is the error I am getting: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\invent-asia\edit_client.php on line 46 <?php include("dbconnection.php"); if(isset($_POST["btnSubmit"])) { $id = $_POST["id"]; $territory = $_POST["territory"]; $job_title = $_POST["job_title"]; $area_of_work = $_POST["area_of_work"]; $employer = $_POST["employer"]; $location = $_POST["location"]; $job_title = $_POST["job_title"]; $date_posted = $_POST["date_posted"]; $closing_date = $_POST["closing_date"]; $department = $_POST["department"]; $gender = $_POST["gender"]; $first_name = $_POST["first_name"]; $last_name = $_POST["last_name"]; $title = $_POST["title"]; $telephone = $_POST["telephone"]; $address_1 = $_POST["address_1"]; $address_2 = $_POST["address_2"]; $address_3 = $_POST["address_3"]; $country = $_POST["country"]; $city = $_POST["city"]; $postal_code = $_POST["postal_code"]; $website = $_POST["website"]; $email_address = $_POST["email_address"]; $comment_by_cg = $_POST["comment_by_cg"]; $date_emailed = $_POST["date_emailed"]; $mailing_comments = $_POST["mailing_comments"]; $telesales_comments = $_POST["telesales_comments"]; $query = "UPDATE contacts SET territory= '".$territory."', job_title = '".$job_title."', area_of_work = '".$area_of_work."', employer = '".$employer."', location = '".$location."', job_title = '".$job_title."', employer = '".$employer."', date_posted = '".$date_posted."', department = '".$department."', closing_date = '".$closing_date."', gender = '".$gender."', first_name = '".$first_name."', last_name = '".$last_name."', title = '".$title."', telephone = '".$telephone."', address_1 = '".$address_1."', address_2 = '".$address_2."', address_3 = '".$address_3."', country = '".$country."', city = '".$city."', postal_code = '".$postal_code."', website = '".$website."', email_address = '".$email_address."', comment_by_cg = '".$comment_by_cg."', date_emailed = '".$date_emailed."', mailing_comments = '".$mailing_comments."', telesales_comments = '".$telesales_comments."' WHERE id = '".$id."'"; mysql_query($query) or die(mysql_error()); echo "<script> alert('You have successfully updated a record'); window.location = 'view.php'; </script>"; } $query = "SELECT * FROM contacts WHERE id = '".$_GET["id"]."'"; $result = mysql_query($query, $connection); if(mysql_num_rows($result) > 0) - THIS IS THE 46th LINE IN THE CODE { $territory = mysql_result($result,0, "territory"); $job_title = mysql_result($result, 0, "job_title"); $area_of_work = mysql_result($result, 0, "area_of_work"); $employer = mysql_result($result, 0, "employer"); $status = mysql_result($result, 0, "status"); $location = mysql_result($result,0, "location"); $department = mysql_result($result,0, "department"); $date_posted = mysql_result($result,0, "date_posted"); $closing_date = mysql_result($result,0, "closing_date"); $gender = mysql_result($result,0, "gender"); $first_name = mysql_result($result,0, "first_name"); $last_name = mysql_result($result,0, "last_name"); $title = mysql_result($result,0, "title"); $telephone = mysql_result($result,0, "telephone"); $address_1 = mysql_result($result,0, "address_1"); $address_2 = mysql_result($result,0, "address_2"); $address_3 = mysql_result($result,0, "address_3"); $city = mysql_result($result,0, "city"); $country = mysql_result($result,0, "country"); $postal_code = mysql_result($result,0, "postal_code"); $website = mysql_result($result,0, "website"); $email_address = mysql_result($result,0, "email_address"); $comment_by_cg = mysql_result($result,0, "comment_by_cg"); $date_emailed = mysql_result($result,0, "date_emailed"); $mailing_comments = mysql_result($result,0, "mailing_comments"); $telesales_comments = mysql_result($result,0, "telesales_comments"); } ?> Can anyone help me solve this issue? This is the only feature in my project that is left bugged Your quick response is well appreciated. Thank you very much Quote Link to comment https://forums.phpfreaks.com/topic/210643-pulling-of-data-from-the-database/ Share on other sites More sharing options...
PFMaBiSmAd Posted August 13, 2010 Share Posted August 13, 2010 Your query failed due to some error. If you echo mysql_error(); on the next line after the line with your mysql_query() statement, it will tell you why the query failed. Quote Link to comment https://forums.phpfreaks.com/topic/210643-pulling-of-data-from-the-database/#findComment-1098854 Share on other sites More sharing options...
emilcarlo Posted August 13, 2010 Author Share Posted August 13, 2010 Hi PFMaBiSmAd, Thanks for that, problem is solved xD I am real noob here, and it's just my first time to do php programming. I was able to identify the problem I have been solving 24 hours now. The problem is the table doesn't exist on the new one. I have different tables in the dummy and the final one. Oh such an angel, thanks much much ^^ Quote Link to comment https://forums.phpfreaks.com/topic/210643-pulling-of-data-from-the-database/#findComment-1098858 Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.