Question

我正在尝试获得GROUP BY子句获得的两列总和的差异，我正在使用此查询

"SELECT " + KEY_NAME + ", SUM("
            + KEY_COL1 + ") AS s1, SUM(" + KEY_COL2 + ") AS s2, "
            + KEY_ID + " FROM " + DATABASE_TABLE + " WHERE " + COUNT
            + " BETWEEN " + start + " AND " + end + " GROUP BY "
            + KEY_NAME + " HAVING (s1 - s2) < 0 ORDER BY " + KEY_NAME
            + " ASC"

我想要满足上述条件的所有字段在我的android游标适配器中实现它。它没有给出任何错误但是它显示了结果的两面（其中s1-s2> 0＆amp; s1-s2 <0）由于我不知道的某些原因。基本上它给出与

相同的结果

"SELECT " + KEY_NAME + ", SUM("
            + KEY_COL1 + ") AS s1, SUM(" + KEY_COL2 + ") AS s2, "
            + KEY_ID + " FROM " + DATABASE_TABLE + " WHERE " + KEY_COUNT
            + " BETWEEN " + start + " AND " + end + " GROUP BY "
            + KEY_NAME + " ORDER BY " + KEY_NAME
            + " ASC"

我做错了什么？

此外，我想要满足上述条件的所有字段的s1-s2之和。我怎么能这样做？

修改

上述代码用于此目的。但是回到我的第二个问题。

我想要总和（sum（s1） - sum（s2））。基本上我想要波西米亚答案中所有“差异”的总和。

Answer 1

删除所有字符串构建代码噪声后，错误显而易见：

SELECT
    KEY_NAME,
    SUM(KEY_COL1) AS s1,
    SUM(KEY_COL2) AS s2,
    SUM(KEY_COL1) - SUM(KEY_COL2) as diff, -- ### added this to get diff
    KEY_ID
FROM DATABASE_TABLE
WHERE COUNT BETWEEN start AND end
GROUP BY KEY_NAME, KEY_ID -- ### Added KEY_ID to group by clause
HAVING (s1 - s2) < 0
ORDER BY KEY_NAME ASC

您没有按所有非聚合列进行分组：您必须将KEY_ID添加到group by子句中。

在任何其他数据库上，这会爆炸，但mysql会允许它，但会为组中列出的每个唯一非聚合列组合返回一个随机行。

我还添加了一个列，可以根据需要计算总和的差异。

Answer 2

使用

解决它

"SELECT SUM(s2) - SUM(s1) FROM ( " + "SELECT " + KEY_NAME + ", SUM("
            + KEY_COL1 + ") AS s1, SUM(" + KEY_COL2 + ") AS s2, "
            + KEY_ID + " FROM " + DATABASE_TABLE + " WHERE " + COUNT
            + " BETWEEN " + start + " AND " + end + " GROUP BY "
            + KEY_NAME + " HAVING (s1 - s2) < 0 ORDER BY " + KEY_NAME
            + " ASC)"

谢谢大家。

Mysql查询两列组和的差异

2 个答案: