How to make more than one thing display from a table?? Please help

[kangbroke]

SQL Problem - Simple fix
All I want to do is show the top 10 viewed videos in my admin of a porn site. I know the names of my tables. I know which content I want to show. I am learning PHP and have not learned SQL. I can make it show the top 10 Viewed videos in descending order but the titles do not display so I can know which video has been viewed the most.

This -
$result = mysql_query(' SELECT *
FROM `content_views`


ORDER BY views DESC
LIMIT 0 , 10 ');


Makes it show my top 10 views but it does not display the title of the video
It displays Just like this

Top 10 Videos
118 views
75 views
71 views
70 views
66 views
65 views
65 views
64 views
61 views
56 views

[php_daemon]

What you need is a join on your videos table. Something like this...

SELECT content_views.views
     , videos.title
FROM content_views
JOIN videos 
  ON videos.id = content_views.id
ORDER BY views DESC 
LIMIT 0 , 10

[kangbroke]

I did that and it disapeared all together.
I even tried to change to the table names like below and it didnt work either.


$result = mysql_query('SELECT content_views.views
, content.title
FROM content_views
JOIN content
ON content.id = content_views.id
ORDER BY views DESC
LIMIT 0 , 10');

If it helps here is my SQL database

[php_daemon]

This is likely wrong:

ON content.id = content_views.id

How does the structure of 'content' table look like?

[kangbroke]

that table looks like this

title
filename
orig_filename
thumbnail
description
paysite
keywords
pornstars
scheduled_date
encoded_date
date_added
views
rating
length
submitter
ip
approved
hotlinked
record_num

[php_daemon]

Is record_num the unique identifier? Does it correspond to 'content' column in content_views? If so, then it should be:

SELECT content_views.views
     , content.title
FROM content_views
JOIN content
  ON content.record_num = content_views.content
ORDER BY views DESC
LIMIT 0 , 10

[kangbroke]

That did the trick thank you very much

[kangbroke]

Should I make another thread for another problem on the same script

[php_daemon]

Yeah, unless it's about the same query...