将两个SQL查询合而为一

Question

How can i perform these two queries in one single query, so that it shows the eid that has more than 2 values and shows its eid as well?

select eid, count(Edited_by.eid)
from Edited_by
group by eid;


select Editor.eid  
from Editor 
where ( select count(*) 
    from Edited_by 
    where Edited_by.eid=Editor.eid ) > 2;

Answer 1

UPDATE

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A predicate in the HAVING clause can operate on aggregate expressions. I think you may be looking for something like this:

SELECT t.eid
     , COUNT(t.eid) AS cnt
  FROM Edited_by t
 GROUP BY t.eid
HAVING COUNT(t.eid) > 2

---

Whoops. I entirely missed that the two queries were referencing two different tables.

To also reference the Editor table, you could use the query above as an inline view.

SELECT c.eid                  AS edited_by_eid
     , c.cnt                  AS cnt
     , IF(c.cnt>2,e.Eid,NULL) AS editor_eid
  FROM Editor e
  JOIN ( SELECT t.eid
              , COUNT(t.eid) AS cnt
           FROM Edited_by t
          GROUP BY t.eid
       ) c
    ON c.eid = e.Eid

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FOLLOWUP

Q: "I want to get the result of BOTH queries by running one."

A: I'm not understanding exactly what you want to achieve.

To have two separate resultsets returned, you would need to run two separate statements. MySQL can return multiple resultsets from a stored procedure, for example, a procedure could execute the two queries. The client could process both resultsets (if that's supported and enabled in the client interface library.) That would be a "single statement" ( CALL my_procedure; ).

If you want to concatenate the results of two separate queries, you could use the UNION ALL set operator. Normally, I would return a discriminator column to distinguish which rows were returned by which query. To do that, the number of columns and the datatypes of each column must match between the two queries.

In the example you give, the query from the Editor table would need to return a dummy integer-type column in order to concatenate the two results.

SELECT t.eid
     , COUNT(Edited_by.eid) AS cnt
  FROM Edited_by t
 GROUP BY t.eid
 UNION ALL
 SELECT e.eid
      , 0 AS cnt
   FROM Editor e
  WHERE ( SELECT COUNT(*)
            FROM Edited_by c
           WHERE c.eid=e.eid 
        ) > 2