Question

现在，我有这个问题：

SELECT COUNT(*) AS Count, SUM(Ask) AS Ask, SUM(Cost) AS Cost, Provider, Factura FROM store_items 
    WHERE (
      Provider NOT IN(SELECT Provider FROM store_provider_invoices)
      AND Factura NOT IN(SELECT Factura FROM store_provider_invoices)
    ) 
    OR Factura NOT IN(SELECT Factura FROM store_provider_invoices) 
    GROUP BY Provider, Factura

这很好用，并返回以下数组：

Array ( 
    [0] => Array ( 
      [Count] => 1 
      [ID] => 13 
      [Ask] => 20.00 
      [Cost] => 10.00 
      [Provider] => 5 
      [Factura] => 8 
    ) 
    [1] => Array ( 
      [Count] => 1 
      [ID] => 18 
      [Ask] => 125.01 
      [Cost] => 110.01 
      [Provider] => 5 
      [Factura] => 34 
    ) 
    [3] => Array ( 
      [Count] => 3 
      [ID] => 14 
      [Ask] => 210.00 
      [Cost] => 150.00 
      [Provider] => 6 
      [Factura] => 5 
    )
)

我想要做的是返回与store_items表中的查询匹配的所有ID，例如：

Array ( 
    [0] => Array ( 
      [ID] => Array (
        [0] => 101
      )
      [Count] => 1 
      [Ask] => 20.00 
      [Cost] => 10.00 
      [Provider] => 5 
      [Factura] => 8 
    ) 
    [1] => Array ( 
      [ID] => Array (
        [0] => 102
      )
      [Count] => 1 
      [Ask] => 125.01 
      [Cost] => 110.01 
      [Provider] => 5 
      [Factura] => 34 
    ) 
    [3] => Array ( 
      [ID] => Array (
        [0] => 103
        [1] => 104
        [2] => 105
      )
      [Count] => 3 
      [Ask] => 210.00 
      [Cost] => 150.00 
      [Provider] => 6 
      [Factura] => 5 
    )
)

因此，例如，在上面的最后一个数组元素中，不仅返回Count of 3，还返回它计算的每一行的ID。

Answer 1

您无法真正获得嵌套结果，但您可以使用 GROUP_CONCAT(DISTINCT store_items.ID ORDER BY ID) AS siIDs 获取以逗号分隔的id值列表。

您还可以修改分隔符

http://dev.mysql.com/doc/refman/5.7/en/group-by-functions.html#function_group-concat

如何在使用GROUP BY

1 个答案: