1

这是我的查询（我用通用的表名替换了表名）。我正在尝试对两个不同的查询进行联合，以便按日期对它们进行分组，以便日期相似的结果显示为一行。

尝试执行时出现“每个派生表必须有自己的别名”错误。我打错了什么？

我对此进行了研究，但找不到答案。每个选定的字段都有一个别名？还是第一个 SELECT 中的问题？

SELECT SUM(val), id, dat, title FROM (

                      SELECT table1.product_id as id, SUM(table1.qty) as val, DATE_FORMAT(table1.created, '%Y-%m-1') as dat, table2.title as title 
                      FROM table1
                      LEFT JOIN table3 ON table1.event_id = table3.id
                      LEFT JOIN table2 ON table1.product_id = table2.id
                      WHERE table1.user_id = $user_id AND table3.active != 2 AND table3.temp = 0 AND table2.active != 2

                      GROUP BY dat

                      UNION ALL

                      SELECT table4.product_id as id, SUM(table4.qty) as val, DATE_FORMAT(table4.created, '%Y-%m-1') as dat, table2.title as title 
                      FROM table4
                      LEFT JOIN table5 ON table4.festival_id = table5.id
                      LEFT JOIN table2 ON table4.product_id = table2.id
                      WHERE table4.user_id = $user_id AND table5.active != 2 AND table2.active != 2

                      GROUP BY dat

                      )
                      GROUP BY id
                      ORDER BY dat ASC

这是我正在尝试做的事情：

我原来的结果：

Array
(
[0] => stdClass Object
    (
        [id] => 1
        [val] => 1
        [dat] => 2012-05-1
        [title] => Test Product
    )

[1] => stdClass Object
    (
        [id] => 1
        [val] => 8
        [dat] => 2012-06-1
        [title] => Test Product
    )

[2] => stdClass Object
    (
        [id] => 2
        [val] => 4
        [dat] => 2012-06-1
        [title] => Test Product 2
    )

[3] => stdClass Object
    (
        [id] => 3
        [val] => 6
        [dat] => 2012-06-1
        [title] => Test Product 3
    )

[4] => stdClass Object
    (
        [id] => 1
        [val] => 10
        [dat] => 2012-05-1
        [title] => Test Product
    )

[5] => stdClass Object
    (
        [id] => 1
        [val] => 8
        [dat] => 2012-06-1
        [title] => Test Product
    )

[6] => stdClass Object
    (
        [id] => 2
        [val] => 3
        [dat] => 2012-06-1
        [title] => Test Product 2
    )

[7] => stdClass Object
    (
        [id] => 3
        [val] => 3
        [dat] => 2012-06-1
        [title] => Test Product 3
    )

)

因此，如果他们有相似的日期和 ID，我需要这些只是一个结果。像这样：

    Array
(
[0] => stdClass Object
    (
        [id] => 1
        [val] => 11
        [dat] => 2012-05-1
        [title] => Test Product
    )

[1] => stdClass Object
    (
        [id] => 1
        [val] => 8
        [dat] => 2012-06-1
        [title] => Test Product
    )

[2] => stdClass Object
    (
        [id] => 2
        [val] => 7
        [dat] => 2012-06-1
        [title] => Test Product 2
    )

[3] => stdClass Object
    (
        [id] => 3
        [val] => 9
        [dat] => 2012-06-1
        [title] => Test Product 3
    )

)

如果您需要其他任何东西，请告诉我。提前致谢。

4

1 回答 1

1

尝试这个：

SELECT SUM(val), id, dat, title FROM (

                  SELECT table1.product_id as id, SUM(table1.qty) as val, DATE_FORMAT(table1.created, '%Y-%m-1') as dat, table2.title as title 
                  FROM table1
                  LEFT JOIN table3 ON table1.event_id = table3.id
                  LEFT JOIN table2 ON table1.product_id = table2.id
                  WHERE table1.user_id = $user_id AND table3.active != 2 AND table3.temp = 0 AND table2.active != 2

                  GROUP BY dat

                  UNION ALL

                  SELECT table4.product_id as id, SUM(table4.qty) as val, DATE_FORMAT(table4.created, '%Y-%m-1') as dat, table2.title as title 
                  FROM table4
                  LEFT JOIN table5 ON table4.festival_id = table5.id
                  LEFT JOIN table2 ON table4.product_id = table2.id
                  WHERE table4.user_id = $user_id AND table5.active != 2 AND table2.active != 2

                  GROUP BY dat

                  ) AS t
                  GROUP BY id, dat
                  ORDER BY dat ASC

正如错误所暗示的，每个视图/派生表都必须有一个别名..

编辑：这将为您提供具有不同 id/dat 对的记录。看来这就是你所追求的。

于 2012-06-19T15:05:46.527 回答