Substition in a list for Connections

6 years 5 months ago #298312 by chanteur94
Replied by chanteur94 on topic Substition in a list for Connections
I think i dont explain correctly myself.

Here is a simple select in my CB list :

(SELECT COUNT(*) FROM ems0_comprofiler_members AS m WHERE m.referenceid = 554 AND m.accepted = 1 )

With PhpMyAdmin, count return 2
In CB List , count return 15

Why ?

With debug mode, i can see that the Select return a long sentence that i dont write :

SELECT COUNT( DISTINCT u.id )
FROM ems0_users u
JOIN ems0_user_usergroup_map g
ON g.`user_id` = u.`id`
JOIN ems0_comprofiler ue
ON ue.`id` = u.`id`

WHERE u.block = 0

AND ue.approved = 1

AND ue.confirmed = 1

AND ue.banned = 0

AND g.group_id IN (2)
AND ( (SELECT COUNT(*)
FROM ems0_comprofiler_members AS m
WHERE m.referenceid = 554
AND m.accepted = 1 ) )

And this Select, return 15 as well in PhpMyAdmin

So, what's wrong ?

Thank you.

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6 years 5 months ago #298327 by krileon
Replied by krileon on topic Substition in a list for Connections
You're just adding a select query to a WHERE statement. That won't do anything. You need to actually condition it. Example as follows.

( SELECT COUNT( * ) FROM `#__comprofiler_members` AS m WHERE m.`referenceid` = u.`id` AND m.`accepted` = 1 ) > 0

Your query doesn't make any sense either as it has no relation to the tables being queried like my above example so it won't filter anything out. You'd need something like the below for example to filter the userlist to users that 554 is connected to.

( SELECT COUNT( * ) FROM `#__comprofiler_members` AS m WHERE m.`referenceid` = 554 AND m.`memberid` = u.`id` AND m.`accepted` = 1 ) > 0


Kyle (Krileon)
Community Builder Team Member
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The following user(s) said Thank You: nant, chanteur94

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