从 SQL 数据库中选择存在具有用户 ID 的行，然后获取具有相似值的所有其他行

Question

I'm not sure why but I'm hitting an absolute wall trying to come up with this select statement. Maybe there is a PHP/MYSQL function that I'm not familiar with that would help. The idea is simple for this user management software: there are managers, and managers can (but do not have to) share clients. Amongst the manager and shared client relationship, one of the managers can be assigned as a lead. So here's how the basic example of what the database looks like for 1 client that is shared between 2 managers and assigned, and another client that is also shared but NOT assigned (represented by zero).

DROP TABLE IF EXISTS clients;

CREATE TABLE clients
(client_id SERIAL PRIMARY KEY
,client_name VARCHAR(12) UNIQUE
);

INSERT INTO clients VALUES
(555,'Jimmy'), 
(789,'Tyler'); 

DROP TABLE IF EXISTS managers;

CREATE TABLE managers
(manager_id SERIAL PRIMARY KEY
,manager_name VARCHAR(12)UNIQUE
);

INSERT INTO managers VALUES
(123,'Michael'),
(456,'David');

DROP TABLE IF EXISTS relationships;

CREATE TABLE relationships
(client_id INT NOT NULL
,manager_id INT NOT NULL
,assigned INT NOT NULL
,PRIMARY KEY(client_id,manager_id)
);

INSERT INTO relationships VALUES
(555, 123, 0),
(555, 456, 1),
(789, 123, 0),
(789, 456, 0);

To get to the point: the statement I'm trying to make is for a manager to be shown all the clients that he has a relationship with, but are NOT assigned to him or anyone else on his team, ie select all of my clients where no one is assigned as the lead.

Expected input: Show all clients that manager 123 has a relationship with, but have yet to be assigned to any manager Expected result: client 789

Happy to clarify as I can see this being overtly confusing as described.

Answer 1

So you will have to start with finding all clientid's from the manager in RELATIONS, but remove all that have a manager assigned already.

Depending on the size of the table you might want to rewrite this, but here is one approach:

1) Get all clientids that have no manager:

SELECT R1.client_id, SUM(R1.assigned) as sumassigned FROM relationships AS R1 GROUP BY R1.client_id HAVING ( SUM(R1.assigned) = 0)

Now it is easier, you just join, eg:

SELECT R2.client_id, R2.manager_id FROM relationships AS R2
INNER JOIN 
(SELECT R1.client_id, SUM(R1.assigned) as sumassigned FROM relationships AS R1 GROUP BY R1.client_id HAVING ( SUM(R1.assigned) = 0) ) AS DRVNOMANAGER
ON (R2.client_id = DRVNOMANAGER.client_id)
WHERE (R2.manager_id = 123)

Not tested. (meaning you might have to fix it)

The idea is to create a derived (temp) table containing all clients without a manager, then do an inner join on your original question ("what clients does this manager 123 know that do not have another assigned)

Does that solve your problem?

PS: Such things are much easier solved in PHP, but if your dataset is huge, that is not feasible.

OP asked for clientnames, so just add that:

        SELECT R2.client_id, R2.manager_id, C.client_name  FROM relationships AS R2
INNER JOIN 
        (SELECT R1.client_id, SUM(R1.assigned) as sumassigned FROM relationships
 AS R1 GROUP BY R1.client_id HAVING ( SUM(R1.assigned) = 0) ) AS DRVNOMANAGER
        ON (R2.client_id = DRVNOMANAGER.client_id)
    INNER JOIN clients AS C ON (C.client_id = R2.client_id)
        WHERE (R2.manager_id = 123)

And last request: removed deleted relations:

Simply add a WHERE clause to the inner DRVNOMANAGER with your restriction on the rows that are used in the GROUP BY. eg:

SELECT R2.client_id, R2.manager_id, C.client_name  FROM relationships AS R2
INNER JOIN 
        (SELECT R1.client_id, SUM(R1.assigned) as sumassigned FROM relationships
 AS R1 WHERE (NOT(R1.deleted = 1) ) GROUP BY R1.client_id HAVING ( SUM(R1.assigned) = 0) ) AS DRVNOMANAGER
        ON (R2.client_id = DRVNOMANAGER.client_id)
    INNER JOIN clients AS C ON (C.client_id = R2.client_id)
        WHERE (R2.manager_id = 123)

==============================================

THIS WAS MY OLD ANSWER. NOT RELEVANT ANYMORE.

"select all of my clients where no one is assigned as the lead."

If I read you well that means: Get all clientid from RELATIONS where some managerid is given, AND assigned=0. (assigned=0 meaning "no one is assigned as the lead.")

Is that correct?

Then you end up with something like this (for managerid 123):

SELECT R.clientid, C.clientname FROM RELATIONSHIPS AS R WHERE ( (R.managerid = 123) AND (R.assigned=0))
INNER JOIN CLIENTS AS C ON (C.clientid = R.clientid)

I removed the spaces in the columnnames because I hate spaces in columnnames.

Answer 2

SELECT c.* 
  FROM managers m 
  JOIN relationships r 
    ON r.manager_id = m.manager_id 
  JOIN clients c 
    ON c.client_id = r.client_id 
  LEFT 
  JOIN relationships x 
    ON x.client_id = c.client_id 
   AND x.assigned = 1 
 WHERE m.manager_id = 123 
   AND r.assigned = 0 
   AND x.client_id IS NULL;
+-----------+-------------+
| client_id | client_name |
+-----------+-------------+
|       789 | Tyler       |
+-----------+-------------+