Question

在使用以下SQL代码时，我似乎找不到合适的方法来正确连接正在创建的2个CTE，以获取正确的（或至少是我期望的）记录数量。

TEST 1＆TEST 2仅返回每个CTE独立产生的记录； n 和 m 。

我希望可以使用TEST 3（一个左联接）或TEST 4（一个内联接）来检索 n 记录，但是我使用哪种类型的联接（并且我尝试了所有等联接）；我不断收到 m 条记录。


WITH
CTE_ALL_INSPECTIONS AS
(
    SELECT
            InspectorId,
            InspectionetailId
            ...
        FROM ...
        WHERE ...
),

CTE_ALL_INSPECTORS AS
(
    SELECT DISTINCT
            InspectorId,
            COUNT(InspectionDetailId) Inspections
        FROM CTE_ALL_INSPECTIONS
        GROUP BY InspectorId
)

/*
TEST 1:
*/
/*
    SELECT * FROM CTE_ALL_INSPECTORS
*/

/*
TEST 2:
*/
/*
    SELECT * FROM CTE_ALL_INSPECTIONS
*/

/*
TEST 3:
*/
/*
    SELECT
            CTE_ALL_INSPECTIONS.*,
            CTE_ALL_INSPECTORS.Inspections
        FROM
            CTE_ALL_INSPECTORS
            LEFT JOIN CTE_ALL_INSPECTIONS
                ON CTE_ALL_INSPECTIONS.InspectorId = CTE_ALL_INSPECTORS.InspectorId
*/

/*
TEST 4:
*/
/*
    SELECT
            CTE_ALL_INSPECTIONS.*,
            CTE_ALL_INSPECTORS.Inspections
        FROM
            CTE_ALL_INSPECTORS
            INNER JOIN CTE_ALL_INSPECTIONS
                ON CTE_ALL_INSPECTIONS.InspectorId = CTE_ALL_INSPECTORS.InspectorId
*/

如何获取所需的记录？或我还能如何加入这些表格？

Answer 1

使用您的TEST 3，但将... Inspections作为最左边的查询：

SELECT CTE_ALL_INSPECTIONS.*
    , CTE_ALL_INSPECTORS.Inspections 
FROM CTE_ALL_INSPECTIONS
    LEFT JOIN CTE_ALL_INSPECTORS
        ON CTE_ALL_INSPECTIONS.InspectorId = CTE_ALL_INSPECTORS.InspectorId

这是说：“将所有检查以及所有来自检查员的计数都给我。”

加入这些CTE的方式是什么？

1 个答案: