Question

我的桌子像：

+-------+--------+----------+------------+-------+
| cd_hs | cd_cnt | name_cnt |   dates    | value |
+-------+--------+----------+------------+-------+
|     1 |      1 | aaa      | 2018-06-01 |    50 |
|     1 |      2 | bbb      | 2018-07-01 |   150 |
|     1 |      3 | ccc      | 2018-08-01 |    20 |
|     1 |      1 | aaa      | 2018-06-02 |    40 |
|     1 |      2 | bbb      | 2018-07-02 |    70 |
|     1 |      3 | ccc      | 2018-08-02 |    80 |
+-------+--------+----------+------------+-------+

实际上我有更多数据，但我只是显示示例，而我想做的就是

我想按cd_hs列中的name_cnt按year，dates分组并执行sum(value)，但我有2个条件。首先是显示条件为cd_cnt with 1 and 2的值，第二个条件为cd_cnt without 1 and 2的值，这意味着我除了1和2以外还有很多值，并且在一列中将别名为other

预期结果：

+-------+------+----------+-------------+
| cd_hs | year | name_cnt | total_value |
+-------+------+----------+-------------+
|     1 | 2018 | aaa      |          90 |
|     1 | 2018 | bbb      |         220 |
|     1 | 2018 | other    |         100 |
+-------+------+----------+-------------+

我该怎么做？我是查询的新手，不知道该怎么办。

Answer 1

考虑到您的规格似乎与您的要求不完全一致，您的问题有点令人困惑。

如果您提供的示例结果实际上是，那么简单的SUM和GROUP BY应该可以这里的把戏：

SELECT cd_hs, EXTRACT(YEAR from dates) as year, name_cnt, SUM(value_)
FROM foo
GROUP BY cd_hs, EXTRACT(YEAR from dates), name_cnt

结果：

| cd_hs | year | name_cnt | sum |
|-------|------|----------|-----|
|     1 | 2018 |      aaa |  90 |
|     1 | 2018 |      bbb | 220 |
|     1 | 2018 |      ccc | 100 |

SQLFiddle

由于您提到要在两个不同的条件下获得两个不同的总数，因此可以将JOIN与一些精心设计的子查询结合使用：

SELECT a.cd_hs, EXTRACT(YEAR FROM a.dates), a.name_cnt, COALESCE(b.total_a, 0) as "Total A", COALESCE(c.total_b, 0) as "Total B"
FROM foo a
LEFT JOIN (
  SELECT b.cd_hs, b.name_cnt, EXTRACT(YEAR FROM b.dates), SUM(value_) as total_a
  FROM foo b
  WHERE b.cd_cnt NOT IN (1, 2)
  GROUP BY b.cd_hs, b.name_cnt, EXTRACT(YEAR from b.dates)
) b ON a.cd_hs = b.cd_hs AND a.name_cnt = b.name_cnt
LEFT JOIN (
  SELECT c.cd_hs, c.name_cnt, EXTRACT(YEAR FROM c.dates), SUM(value_) as total_b
  FROM foo c
  WHERE c.cd_cnt IN (1, 2)
  GROUP BY c.cd_hs, c.name_cnt, EXTRACT(YEAR from c.dates)
) c ON a.cd_hs = c.cd_hs AND a.name_cnt = c.name_cnt

这种特殊的解决方案易于阅读，可以使您获得正确的最终结果，但很可能无法以当前形式进行扩展。

结果：

| cd_hs | date_part | name_cnt | Total A | Total B |
|-------|-----------|----------|---------|---------|
|     1 |      2018 |      aaa |       0 |      90 |
|     1 |      2018 |      bbb |       0 |     220 |
|     1 |      2018 |      ccc |     100 |       0 |
|     1 |      2018 |      aaa |       0 |      90 |
|     1 |      2018 |      bbb |       0 |     220 |
|     1 |      2018 |      ccc |     100 |       0 |

SQLFiddle

用1列查询不同条件

1 个答案: