如何在一个查询中进行多个计数/存在？

Question

I'm struggling to figure out how I can check for two things from one table without making two calls to mysql.

I have a Members table. I'd like to test whether a certain value exists in the MemberID column, and whether a certain value exists in the PhoneNumber column. MemberID and PhoneNumber are both indexed.

But there's something wrong with the syntax I'm trying. For example, each of

SELECT COUNT(1) AS IDExists FROM Members WHERE MemberID = '999999999999' LIMIT 1

and

SELECT COUNT(1) AS PhoneExists FROM Members WHERE PhoneNumber = '5555555555' LIMIT 1

works. Why can't they be combined, somehow, like

SELECT (COUNT(1) AS IDExists FROM Members WHERE MemberID = '999999999999' LIMIT 1), (COUNT(1) AS PhoneExists FROM Members WHERE PhoneNumber = '5555555555' LIMIT 1)

?

Or perhaps, since I only care whether the value occurs zero times, or at all, something like

SELECT EXISTS (SELECT 1 FROM Members WHERE MemberID = '999999999999')

?

But, unfortunately, there's something wrong with that syntax even for the case of one of my inquiries.

Any suggestions on how to accomplish this?

Answer 1

You want a conditional SUM

SQL FIDDLE

SELECT 
    SUM( CASE 
             WHEN PhoneNumber = '5555555555' THEN 1
             ELSE 0
         END) PhoneExists,
    SUM( CASE 
             WHEN MemberID = '999999999999' THEN 1
             ELSE 0
         END) IDExists
FROM Members

Answer 2

You can combine two working query like this:

select a.IDExists, b.PhoneExists
from (SELECT COUNT(1) AS IDExists FROM Members WHERE MemberID = '999999999999' LIMIT 1) a,
     (SELECT COUNT(1) AS PhoneExists FROM Members WHERE PhoneNumber = '5555555555' LIMIT 1) b

Though the query join two results with cartesian product, it is guaranteed that each result has only one row, it would not be a problem.

I'm not sure how big is the table and what kinds of index it has. If the table is big, then perhaps you have indexes on column MemberID and PhoneNumber respectively, then the above query will return the result quickly.