表who
wid--name-------father---mother
1----Daisy------David----Liza
2----Jenny------Joe------Judy
3----Meggy------Mike-----Manuela
4----Sarah------Joe------Judy
5----Chelsea----Bill-----Hillary
6----Cindy------David----Liza
7----Kelly------Joe------Judy
表ages
aid---whoid---age
1-----1--------0
2-----2--------0
3-----3-------14
4-----4-------30
5-----5-------22
6-----6-------17
7-----1-------18
我希望该列表成为结果:
id---name------age
1----Meggy-----14
2----Cindy-----17
3----Daisy-----18 (Selected data that bigger than 0)
4----Chelsea---22
5----Sarah-----30
6----Jenny-----30 (Her age is 0 on ages table and Sarah's age with same father and mother)
7----Kelly-----30 (No data on ages table and Sarah's age with same father and mother)
我尝试了这个查询:
SELECT
*,
(CASE age
WHEN '0' THEN (
SELECT age
FROM ages a
LEFT JOIN who w
ON w.wid = a.whoid
WHERE
w.father = father
AND
w.mother = mother
ORDER BY a.age DESC LIMIT 1
)
ELSE age
END
) AS newage
FROM who
LEFT JOIN ages
ON wid = whoid
ORDER BY newage
这有什么问题?
答案 0 :(得分:21)
CASE … WHEN NULL
永远不会匹配任何内容,CASE NULL
将始终匹配ELSE
子句(在您的情况下返回age
,即NULL
)。< / p>
使用此:
CASE COALESCE(age, 0) WHEN 0 THEN … ELSE age END
<强>更新强>
您还需要对表进行别名并使用字段说明中的别名:
SELECT *,
CASE COALESCE(age, 0)
WHEN '0' THEN
(
SELECT MAX(age)
FROM who wi
JOIN ages ai
ON ai.whoid = wi.wid
WHERE wi.father = w.father
AND wi.mother = w.mother
)
ELSE
age
END AS newage
FROM who w
LEFT JOIN
ages a
ON a.whoid = w.wid
ORDER BY
newage
答案 1 :(得分:0)
您的子选择中的父亲和母亲可能在您的外部查询中与父亲和母亲混淆。或者你在问为什么它不是最优的?