SQl num_rows problem

Posted on 16th Feb 2014 07:03 pm by admin

when i try to count rows from an SQL select i get an warning

Code: [Select]$countviews = mysql_query("SELECT * FROM viewed WHERE sfw = '0' AND cookieid = $id");
$num_rows_viewed = mysql_num_rows($countviews);
Code: [Select]Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in ******.php on line 58
anybody an idea why this is?

Your Answer:

Login to answer

Other forums

How to display objects in a row
I have 7 codes that i want to display in a row one next to another.
The first is {$ads->ad

DateTimeZone::listIdentifiers headache
http://php.net/manual/en/datetimezone.listidentifiers.php


So the documentation states tha

mysql_real_escape_string
Let me preface this that I am very much a PHP noob, but I have some SQL training (not necessarily My

Strange array issue, never happened before.
mysql_fetch_array returns 1 array per call. Generally that's why it is inserted into a while statem

Custom array sort? asc then desc
Any ideas how I could sort this array? I've been trying for a while. Maybe with usort, but I have no

If statement help please.
Hello on my site I have 2 content boxes. 1 is for the main content (which is the biggest one) and th

Need help/w Page: 1,2,3,4 function!
So I have this code im working on where it's ment to display seperate pages having 20 logs each
I

Problem with variable declaration in switch statement
Hello, I am having some trouble assigning a value to a variable inside a switch statement. What I a

FILTER_CALLBACK -- Files?
Hi All,

I'm using the php filter functions to validate my form data. For custom filters, I'm

rename the file
File.txt

Code: ***DOCUMENT***
..DN:
000044255
..CB:
..SN:
..PY:
2009
..E