sqlite 查询对子查询的结果进行简单的计算答案

【问题标题】：sqlite query do simple calculations on result of subquerysqlite 查询对子查询的结果进行简单的计算
【发布时间】：2013-09-02 12:29:30
【问题描述】：

我有这个查询需要在 sqlite (iOS) 上进行。

SELECT k.ZKUNDENNR, k.ZNAME1,
(SELECT SUM(vk.ZNETTO) FROM ZPBSROW r LEFT JOIN ZWARENGRUPPEVK vk ON vk.ZPBSROW=r.Z_PK WHERE r.ZKUNDE=k.Z_PK AND ZJAHR=2013 AND ZMONAT>=1 AND ZMONAT<=6) AS summeJ,
(SELECT SUM(vk.ZNETTO) FROM ZPBSROW r LEFT JOIN ZWARENGRUPPEVK vk ON vk.ZPBSROW=r.Z_PK WHERE r.ZKUNDE=k.Z_PK AND ZJAHR=2012 AND ZMONAT>=1 AND ZMONAT<=6) as summeVJ,
(SELECT SUM(vk.ZDB_BASIS) FROM ZPBSROW r LEFT JOIN ZWARENGRUPPEVK vk ON vk.ZPBSROW=r.Z_PK WHERE r.ZKUNDE=k.Z_PK AND ZJAHR=2012 AND ZMONAT>=1 AND ZMONAT<=6) as summeDBVJ,
(SELECT SUM(vk.ZDB_BASIS) FROM ZPBSROW r LEFT JOIN ZWARENGRUPPEVK vk ON vk.ZPBSROW=r.Z_PK WHERE r.ZKUNDE=k.Z_PK AND ZJAHR=2013 AND ZMONAT>=1 AND ZMONAT<=6) as summeDBJ, 
((summeJ-summeVJ)/summeVJ*100) AS variance FROM ZKUNDE k WHERE summeJ>0 ORDER BY summeJ DESC LIMIT 0,10

我需要做的是计算summeJ 和summeVJ 之间的差异，但我总是收到这个错误：“没有这样的列：summeJ”。任何人都可以帮助我吗？提前致谢

【问题讨论】：

标签： sql sqlite subquery

【解决方案1】：

您需要使用子查询，因为列别名不能在select的同一级别重复使用：

SELECT t.*, ((summeJ-summeVJ)/summeVJ*100) AS variance
FROM (SELECT k.ZKUNDENNR, k.ZNAME1,
             (SELECT SUM(vk.ZNETTO) FROM ZPBSROW r LEFT JOIN ZWARENGRUPPEVK vk ON vk.ZPBSROW=r.Z_PK WHERE r.ZKUNDE=k.Z_PK AND ZJAHR=2013 AND ZMONAT>=1 AND ZMONAT<=6) AS summeJ,
             (SELECT SUM(vk.ZNETTO) FROM ZPBSROW r LEFT JOIN ZWARENGRUPPEVK vk ON vk.ZPBSROW=r.Z_PK WHERE r.ZKUNDE=k.Z_PK AND ZJAHR=2012 AND ZMONAT>=1 AND ZMONAT<=6) as summeVJ,
             (SELECT SUM(vk.ZDB_BASIS) FROM ZPBSROW r LEFT JOIN ZWARENGRUPPEVK vk ON vk.ZPBSROW=r.Z_PK WHERE r.ZKUNDE=k.Z_PK AND ZJAHR=2012 AND ZMONAT>=1 AND ZMONAT<=6) as summeDBVJ,
             (SELECT SUM(vk.ZDB_BASIS) FROM ZPBSROW r LEFT JOIN ZWARENGRUPPEVK vk ON vk.ZPBSROW=r.Z_PK WHERE r.ZKUNDE=k.Z_PK AND ZJAHR=2013 AND ZMONAT>=1 AND ZMONAT<=6) as summeDBJ
      FROM ZKUNDE k
     ) t
WHERE summeJ>0
ORDER BY summeJ DESC
LIMIT 0,10

【讨论】：