Count the total records containing specific values

问题:

I have a question and hope you guys can assist me.

I have a table containing two columns:

type           // contains 2 different values: "Raid" and "Hold"
authorization  // contains 2 different values: "Accepted" or "Denied"

I need to make a view that returns values like this:

TYPE:RAID     ACCEPTED:5          DENIED:7

Basically I want to know how many of the values in TYPE are "Raid" and then how many of them are "Accepted" and "Denied".

Thank you in advance!!

回答1:

SELECT
   Type
  ,sum(case Authorization when 'Accepted' then 1 else 0 end) Accepted
  ,sum(case Authorization when 'Denied' then 1 else 0 end) Denied
 from MyTable
 where Type = 'RAID'
 group by Type

回答2:

This code should work for mySQL

SELECT type, COUNT(*)
FROM table
GROUP BY type;

or

SELECT type, authorization, COUNT(*)
FROM table
GROUP BY type, authorization;

回答3:

You can use COUNT in combination with a CASE statement

SELECT COUNT(CASE authorization WHEN 'denied' THEN 1 ELSE NULL END) as denied,
  COUNT(CASE authorization WHEN 'authorized' THEN 1 ELSE NULL END) as authorized
FROM table
WHERE type = 'RAID'

SUM(CASE …) is also possible, but you'll have to return 0 in the ELSE clause instead of NULL

回答4:

select count(*) as count from tbl_name where type='Raid'

for total number of type=raid

Are you saying something like this?

回答5:

Hey this might help:-

select type as 'TYPE',sum(Denied) as 'DENIED',sum(Accepted) as 'AUTHORIZED' from
(
 SELECT type,0 as 'Denied',count(*) as 'Accepted' from t where authorization = 'Accepted'    group by type
 union all
 SELECT type,count(*) as 'Denied',0 as 'Accepted' from t where authorization = 'Denied'     group by type ) as sub_tab group by TYPE;