starvinmarvin14

Nevermind, I know what's wrong. The site is blocked from using iframes.

I have a page with an iframe in it like so. <iframe src="page.php" width="100" height="100"></iframe> The page.php file, I have a meta refresh in it. <meta http-equiv="refresh" content="2; url=http://www.blah.com"> Now when I load the website, it doesn't refresh the page.php iframe. It just stays loaded without refreshing. How can I reload an iframe without the user clicking on a link inside the iframe?? I also tried a header redirect but that didn't work. Thanks.

Sorry for being a newb . How exactly do I do that?

I have this... And I am getting this error...

I have two tables. One is a table with information about a certain id number. The other table is a list of votes (either up or down) with the id from the first table. I want to be able to list the id from the first table descending in the order of the highest number of upvotes from the second table. I want to do this with one mysql query. Or any way I can get the $statement variable out of the while loop. This is what the two tables look like... Here is what I currently have which I would like to accomplish in only one mysql_query instead of two... How can I do this with one query? I would need to call both tables in the same query. Let me know if you need more clarification. Please help. Thanks!

Sorry... Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/user/public_html/mysite/toprated.php on line 52 The query is failing.

How would I call this query? I am using mysql_fetch_assoc and am getting an error. Is there anything wrong with this code? $statement1 = "SELECT picid FROM ratings GROUP by picid ORDER by AVG(rating)"; $row2 = mysql_fetch_assoc($statement1);

I have a table which holds each individual rating as a row. An example of some data would be: picid | rating 12 2 43 3 12 4 36 7 43 9 I also have another table which holds image info and has a column with the id that matches picid on the first table. How would I find the average rating of each 'picid' to be displayed on a top rated page? So it should output picid | rating 36 7 43 6 12 3 So far I have... $statement1 = "SELECT picid, SUM(rating) AS total FROM ratings GROUP BY picid ORDER by total"; Let me know if you need further clarification.

Hey, I want to be able to have a 100x100 thumbnail of pictures I am uploading while cropping the image to maintain the ratio. Right now I am stretching the image and it looks bad but it works. Here is my code... How could I change this to make the thumbnails maintain the aspect ratio but still be 100x100?

Thanks this worked!

When i echo the row I get "Resource id #2" every time.

Tried this... $query = mysql_query("SELECT id FROM image WHERE active='yes' ORDER BY RAND() LIMIT 1"); if(mysql_num_rows($query)){ $query_result = mysql_result($query,0); echo <<<OPT <meta http-equiv="REFRESH" content="0;url=http://www.mysite.com/view.php?id={$query_result}"> OPT; }else{ echo 'no records found'; } Now I am getting this error... Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/.../public_html/.../random.php on line 9 no records found I only have to rows at the moment with the id 18 and 19. How would i use mysqli_error()?

$query=mysql_query("SELECT id FROM image WHERE active='yes' ORDER BY RAND() LIMIT 1"); $query_result = mysql_result($query,0); echo <<<OPT <meta http-equiv="REFRESH" content="0;url=http://www.mysite.com/view.php?id={$query_result}"> OPT; Tried that and got this... "Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/.../public_html/.../random.php on line 9"

I'm trying to select a random id from a table called 'image'. I want the id number to be at the end of a link such as view.php?id=123. Here is my code: $query=mysql_query("SELECT id FROM image WHERE active='yes' ORDER BY RAND() LIMIT 1"); echo <<<OPT <meta http-equiv="REFRESH" content="0;url=http://www.mysite.com/view.php?id={$query}"> OPT; It does not display the number in the link. What have i done wrong?

I have a table in mysql that has the path's of images stored. The image path is stored under the column "photo". The table looks like this: 1 id 2 name 3 age 4 category 5 photo 6 rating I want to create a page such as view.php?id=1 where the image is displayed according to the id. It would have to call upon the path to display it. How would I do this?