Question

我正在尝试编写一个SQL查询来执行以下操作：

给出下表：

    <settings xmlns="http://maven.apache.org/SETTINGS/1.0.0"
          xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
          xsi:schemaLocation="http://maven.apache.org/SETTINGS/1.0.0 http://maven.apache.org/xsd/settings-1.0.0.xsd">

  <localRepository>Z:/maven/repository</localRepository>

  <pluginGroups>
  </pluginGroups>
  <proxies>
  </proxies>

  <servers>
    <server>
      <id>nexus</id>
      <username>user</username>
      <password>password</password>
    </server>
  </servers>

  <mirrors>
  </mirrors>

  <profiles>
  </profiles>

</settings>

选择* 价值互不相同按group_id分组

对于此示例，输出应为：

+----+----------+-----------+
| id | group_id | value     |
+----+----------+-----------+
| 1  | 1        | 0         |
+----+----------+-----------+
| 2  | 1        | 0         |
+----+----------+-----------+
| 3  | 2        | null      |
+----+----------+-----------+
| 4  | 3        | -1        |
+----+----------+-----------+
| 5  | 3        | 1         |
+----+----------+-----------+
| 6  | 4        | something |
+----+----------+-----------+
| 7  | 5        | something |
+----+----------+-----------+

有人知道这是否可能吗？

Answer 1

如果您只想查找具有不同.detach()的{{1}}值，则可以使用以下查询：

require_grad=True

输出您的样本数据：

group_id

如果要获取与该value关联的行，请使用上述查询作为SELECT group_id FROM data GROUP BY group_id HAVING MIN(value) != MAX(value)表达式的子查询：

group_id
3

输出

group_id

Demo on dbfiddle

Answer 2

您可以使用exists来获取不具有相同值的group_id。这是demo。

select
    distinct group_id
from data d1
where exists
(
    select
        group_id
    from data d2
    where d1.group_id = d2.group_id
    and d1.value <> d2.value
)

输出

*--------*
|group_id|
*--------*
|     3  |
*--------*

如果要同时使用group_id和值，请尝试以下操作

select
    group_id,
    value
from data d1
where exists
(
    select
        group_id
    from data d2
    where d1.group_id = d2.group_id
    and d1.value <> d2.value
)

输出：

*------------------*
|group_id |  value |
*------------------*
|     3  |    -1   |
|     3  |     1   |
*------------------*

Answer 3

此查询：

select group_id
from tablename
group by group_id
having count(distinct value) > 1 and count(distinct value) = count(value)

返回所需的group_id，因此将其与运算符IN一起使用：

select * from tablename
where group_id in (
  select group_id
  from tablename
  group by group_id
  having count(distinct value) > 1 and count(distinct value) = count(value)
)

请参见demo。
结果：

| id  | group_id | value |
| --- | -------- | ----- |
| 4   | 3        | -1    |
| 5   | 3        | 1     |

比较具有相同组ID的行中的值？

3 个答案: