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  #1  
Old 04-16-2008, 04:04 AM
noonespecial noonespecial is offline
 
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Default Driving Me Crazy (MYQL Help Please)

I have a table (links_favorites) with two columns (linkid and userid).

I want to write something that counts the number of matches (in linkid) between two different userids.

So it something that grabs all the linkids from userid A - and then all the linkids from userid B - and then counts all the places where this intersects.

However, I simply can't figure out the best/easiest way to do this.

Could someone help please?
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  #2  
Old 04-16-2008, 04:41 AM
nighthalk nighthalk is offline
 
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Default

I don't have your database to test this. So here you can test it.

Code:
<?php
   $sql="SELECT linkid, userid, FROM links_favorites WHERE MATCH (linkid,userid)";
   $result = mysql_query($sql);
   $resultsnumber = mysql_numrows($result);

echo "$resultsnumber";
?>
Let me know if this works.
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  #3  
Old 04-16-2008, 07:35 AM
noonespecial noonespecial is offline
 
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Default

Problem with this is that I need to be able to specify the userids ...

IE: there are 4 matches in LINKID between userid-A and userid-B

Where would I specify those in this example?

--------------- Added [DATE]1208335849[/DATE] at [TIME]1208335849[/TIME] ---------------

Code:
   $sql="SELECT linkid, userid FROM links_favorites WHERE userid=1083";
   $result = mysql_query($sql);  
   $number1 = mysql_fetch_array($result);
   
   $sql2="SELECT linkid, userid FROM links_favorites WHERE userid=1";
   $result2 = mysql_query($sql2);  
   $number2 = mysql_fetch_array($result2);
   
   $result3 = array_intersect($number1, $number2);
   
   $joined = count($result3);
   
   echo $joined;
Also doesn't work.

--------------- Added [DATE]1208335899[/DATE] at [TIME]1208335899[/TIME] ---------------

Code:
     $sql="SELECT linkid FROM links_favorites WHERE userid=1083";
   $result = mysql_query($sql);  
   $number1 = mysql_fetch_array($result);
   
   $myids=implode(",", $number1);
   
   $sql2="SELECT COUNT(linkid) as count FROM links_favorites WHERE userid=1 AND linkid IN (".$myids.")";
   $result2 = mysql_query($sql2);  
   $number2 = mysql_fetch_array($result2);
Also doesn't work.

--------------- Added [DATE]1208337252[/DATE] at [TIME]1208337252[/TIME] ---------------

Code:
$fandom = $vbulletin->db->query_read("SELECT *
FROM links_favorites AS links_favorites
LEFT JOIN links_favorites AS favs2 ON (links_favorites.linkid=favs2.linkid)
WHERE links_favorites.userid=".$vbulletin->userinfo['userid']."
AND favs2.userid=".$friend[userid]."
");

$joined= $db->num_rows($fandom);
I think this works ..... ugly.
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  #4  
Old 04-16-2008, 07:37 PM
MoT3rror MoT3rror is offline
 
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Default

PHP Code:
$linkcount $db->query_first("
     SELECT COUNT(linkid) AS linkcount 
     FROM links_favorites 
     WHERE userid IN (1, 2)
"
); 
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  #5  
Old 04-16-2008, 07:53 PM
Farcaster Farcaster is offline
 
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Default

If I am understanding your question correctly and you want to know how many links User 1 and User 2 share, then you could use this query:

[SQL]SELECT
COUNT(DISTINCT t1.linkid) AS shared_links
FROM
link_favorites t1
INNER JOIN link_favorites t2 ON
t1.linkid = t2.linkid
WHERE
t1.userid = 1
AND t2.userid = 2[/SQL]

Your inclination to use a self-join in your example is pretty close to the mark, except that you could get duplicates and you might as well let MySQL handle the counting. Also, a left join would return all of user 1's links, whether they matched or not.
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  #6  
Old 04-17-2008, 05:38 AM
Eikinskjaldi's Avatar
Eikinskjaldi Eikinskjaldi is offline
 
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Default

or you could use a nested query:

[sql]
select count(distinct linkid) from link_favourites
where userid=1
and linkid in
(select distinct linkid from link_favourites where userid=2)
[/sql]
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  #7  
Old 04-17-2008, 05:41 AM
noonespecial noonespecial is offline
 
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Default

Thank you.
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