mysql 中的多个连接与 3 个表

Question

i have 3 tables

branch-
  br_id,
  br_name

users 
 user_id
 user_name
 user_type (2 types 'owner' 'admin')
 user_branch (associated branch )
 user_branches (comma seperated branch ids in case of 'admin' type )

item 
  it_id
  it_amount
  it_owner
  it_admin 
  it_activated

Here each item is owned by as user type "owner"

Each item is also associated by a user type "admin"

what I want is list of branches and associated total amount and total count

**Branch**             **total**  **total amount**
some branch Name    5   500
another Branch Name 7   780

How can i do it in single query

i tried this which is showing amount of all branches

SELECT br_name,
count(it_id),
SUM(it_amount),
FROM branch
LEFT JOIN users ON FIND_IN_SET(br_id,user_branches)>0
LEFT JOIN item ON it_admin=ad_id
WHERE it_activated=1
GROUP BY br_name

but I am getting same count and amount in all branches

Answer 1

First, you should really provide bare minimums, especially when asking with bounty. Not knowing your table structures and sample content makes it harder for others to assist. I have created sample create tables and inserts, then the query. Correct me if anything is inaccurate.

CREATE TABLE branch (
  br_id int unsigned NOT NULL,
  br_name varchar(10) NOT NULL,
  PRIMARY KEY (br_id)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

CREATE TABLE users (
  user_id int unsigned NOT NULL,
  user_name varchar(10) NOT NULL,
  user_type varchar(10) NOT NULL,
  user_branch int NOT NULL,
  user_branches varchar(10) NOT NULL,
  PRIMARY KEY (user_id)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

CREATE TABLE item (
  it_id int unsigned NOT NULL,
  it_amount int NOT NULL,
  it_owner int NOT NULL,
  it_admin int NOT NULL,
  it_activated int NOT NULL,
  PRIMARY KEY (it_id)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

insert into branch ( br_id, br_name )
  values 
  ( 1, 'b1' ), 
  ( 2, 'b2'), 
  ( 3, 'b3' );

insert into users ( user_id, user_name, user_type, user_branch, user_branches )
  values 
  ( 1, 'n1', 'owner', 1, '' ),
  ( 2, 'n2', 'admin', 2, '2, 3' ),
  ( 3, 'n3', 'owner', 3, '' );

insert into item ( it_id, it_amount, it_owner, it_admin, it_activated )
  values
  ( 1, 100, 1, 2, 1 ),
  ( 2, 150, 3, 2, 1 ),
  ( 3, 125, 1, 2, 1 ),
  ( 4, 30, 3, 2, 0 );

Now, what I believe you need is a UNION query covering individually the owner vs the admin, otherwise, you will be getting double-counts across the board. Secondly, I believe not the greatest layout of table design whenever you are using a comma separated list to hold multiple IDs that someone/something can qualify under, such as your ADMIN users. But if that is what you have and cant change it, so be it.

So, now on with the query. Look at each one individually. If doing the query by the OWNER type of user, join just on that user type and tie exclusively to the user's branch, then tie to the item table based on the item's owner. You get one result.

Finally, similar join, but only based on the ADMIN type of user, and join to the item based on the item's ADMIN ID.

Again, without your sample source records and expected output, this is the best interpretation I can come up with (and probably others too) at this time. So, with the queries in a similar column result structure they can be brought together with a "UNION". I added an additional column to show the origin of the numbers as based on the Source of owner/admin respectively.

select
        b.br_name,
        'By Owner' Source,
        count(i.it_id) numOfItems,
        SUM(i.it_amount) sumOfAmount
    FROM
        branch b
            JOIN Users u
                on u.user_type = 'owner'
                AND b.br_id = u.user_branch
                JOIN Item i
                    on u.user_id = i.it_owner
                    AND i.it_activated = 1
    group by
        b.br_name
UNION        
select
        b.br_name,
        'By Admin' Source,
        count(i.it_id) numOfItems,
        SUM(i.it_amount) sumOfAmount
    FROM
        branch b
            JOIN Users u
                on u.user_type = 'admin'
                AND FIND_IN_SET(b.br_id, u.user_branches) > 0
                JOIN Item i
                    on u.user_id = i.it_admin
                    AND i.it_activated = 1
    group by
        b.br_name;
        

Answer 2

I am not sure what you want from the query, so I have taken a guess. My answer query finds the total count and totaled amounts for the items linked to each branch, first by owner and then by admins.

In your query, you have a field "ad_id":

    LEFT JOIN item ON it_admin=ad_id

I assume that is meant to be "user_id"?

I also notice that, in your query, you have not referenced the owners at all. If you are in fact looking for the results from items linked to branches just by the admins, then you can just remove/ignore the first two subqueries in the following query:

       SELECT   br_name as branchname,
            (SELECT COUNT( DISTINCT it_id) 
                FROM item i1 
                JOIN users u1 ON i1.it_owner = u1.user_id AND i1.it_activated=1
                WHERE u1.user_branch = b.br_id AND u1.user_type = "owner") as ownertotal,
            (SELECT COUNT( DISTINCT it_id) 
                FROM item i2 
                JOIN users u2 ON i2.it_admin = u2.user_id AND i2.it_activated=1
                WHERE FIND_IN_SET(b.br_id, u2.user_branches) != 0 AND u2.user_type = "admin") as admintotal ,
            (SELECT SUM(i3.it_amount) 
                FROM item i3 
                JOIN users u3 ON i3.it_owner = u3.user_id AND i3.it_activated=1
                WHERE u3.user_branch = b.br_id AND u3.user_type = "owner") as owneramount,
            (SELECT SUM(i4.it_amount) 
                FROM item i4 
                JOIN users u4 ON i4.it_admin = u4.user_id AND i4.it_activated=1
                WHERE FIND_IN_SET(b.br_id, u4.user_branches) != 0 AND u4.user_type = "admin") as adminamount
    FROM branch b;