Displaying Most Viewed Image

[RAH]

Hi,

I'm new here! 🙂 Looks like a great forum - hope to learn lots here.

I'm currently trying to get a media script to display the most viewed image.

I have worked out a query to do this:

'SELECT * FROM files WHERE category_id = 5 ORDER BY files . views DESC LIMIT 0, 1 ';

However I'm not sure how to get the appropriate values I need into variables. The above query returns 1 entire row and I have to grab several values from that row (eg. filename).

When I get the necessary data into variables I can do things like the following:

<p>Top Viewed Image</p>
<p>
<img src="http://mydomain.com/' . $top_viewed_image . 'jpg '" width="276" height="110" /></p>

I'm just sort of lost a bit, would appreciate any help.

Thanks!

[dagon]

you need something like this

$result = mysql_query("SELECT * FROM files WHERE category_id = 5 ORDER BY files . views DESC LIMIT 0, 1");

$row = mysql_fetch_array($result);

<p>Top Viewed Image</p>
<p>
<img src="http://mydomain.com/' . $row[image_name] . 'jpg '" width="276" height="110" /></p>

[RAH]

That is giving:

Parse error: syntax error, unexpected '<' in /home/nexvids/public_html/test2.php on line 10

[dagon]

post the full code

[RAH]

I just used exactly what you posted here with my domain:

dagon wrote:

you need something like this

$result = mysql_query("SELECT * FROM files WHERE category_id = 5 ORDER BY files . views DESC LIMIT 0, 1");

$row = mysql_fetch_array($result);

<p>Top Viewed Image</p>
<p>
<img src="http://mydomain.com/' . $row[image_name] . 'jpg '" width="276" height="110" /></p>

[dagon]

well I did say something like that, i assumed you knew a little of what your were doing; that cant be the whole thing as there's no db connection or starting\closing php tags and you have to either echo out the html or call the image name var in php