基於前一種情況的SQL DB2計數

Question

有人知道基於先前案例陳述的計數解決方案嗎？

這是我在查詢中的意思， 無法正常使用

SELECT                              
    CASE 
        WHEN field1 IN ('a','b','c') THEN 'MATCH1'
        WHEN field2 IN ('d','e','f') THEN 'MATCH2'
        ELSE 'NOMATCH' END AS MATCHFIELD,
    (SELECT COUNT(*) FROM TABLE2 WHERE field3=MATCHFIELD) AS SUBCOUNT
FROM TABLE1     

我在db2 v9.X上。 我已經tryed HAVING，但無法找到一個可行的解決方案

問候

Answer 1

將SELECT與case表達式包裝在派生表中：

select MATCHFIELD,
    (SELECT COUNT(*) FROM TABLE2 WHERE field3=MATCHFIELD) AS SUBCOUNTSELECT                              
from
(
    select
          CASE 
               WHEN field1 IN ('a','b','c') THEN 'MATCH1'
               WHEN field2 IN ('d','e','f') THEN 'MATCH2'
               ELSE 'NOMATCH'
          END AS MATCHFIELD
    FROM TABLE1
) dt

Answer 2

我會用JOIN和GROUP BY

SELECT MATCHFIELD, COUNT(*)
FROM TABLE2
JOIN
(
    SELECT                              
        CASE 
            WHEN field1 IN ('a','b','c') THEN 'MATCH1'
            WHEN field2 IN ('d','e','f') THEN 'MATCH2'
            ELSE 'NOMATCH' END AS MATCHFIELD
    FROM TABLE1  
) t ON field3=t.MATCHFIELD
GROUP BY MATCHFIELD

Answer 3

嘗試這個 ：

with tmp as (
SELECT  CASE 
        WHEN field1 IN ('a','b','c') THEN 'MATCH1'
        WHEN field2 IN ('d','e','f') THEN 'MATCH2'
        ELSE 'NOMATCH' END AS MATCHFIELD
FROM TABLE1
)

select tmp.*,  (SELECT COUNT(*) FROM TABLE2 WHERE field3=MATCHFIELD) AS SUBCOUNT
from tmp