Select 全部来自一个表，但如果它存在于另一个表中则删除

Question

I currently got a table called Group, it contains a bio and a name and an unique id number on the group.

Then i also got another table called user_groups which also has an unique id, a user_id and group_id which is the same as the unique number on the group is.

So currently im selecting all groups like this:

"SELECT * FROM groups WHERE 1"

Which results in all groups.

So I tried doing this to get all group from the group table except the groups the user got from the other table user_groups

$query = "SELECT * 
FROM groups 
LEFT OUTER JOIN user_groups 
ON groups.id = user_groups.group_id 
WHERE user_groups.user_id != '".$_SESSION['id']."' 
UNION 
SELECT * 
FROM groups 
RIGHT OUTER JOIN user_groups 
ON groups.id = user_groups.group_id 
WHERE user_groups.user_id != '".$_SESSION['id']."'";

When I put = to, I get the groups that exists with the user_id from user_groups table but I want all groups that is not linked to an id. How do I acheive this?

So in my case user with id 15 already "has" group 1 so I dont want to show it for them.:)

Answer 1

I suggest you use NOT EXISTS which is efficient and convenient for this style of query as it will locate just the rows in the group table where the current user is NOT a member.

SELECT *
FROM groups
WHERE NOT EXISTS (
        SELECT NULL
        FROM user_groups
        WHERE groups.id = user_groups.group_id
            AND user_groups.user_id = $_SESSION[id]
        )

Note the subquery used for NOT EXISTS does not need to return any data through its own select clause; so select NULL may be used, but some prefer to use select 1 or select * instead. Note too that because we are using NOT EXISTS we actually need to find the rows that the user IS a match so we use = $_SESSION[id] instead of <> or !=