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我生成了一个查询，列出了我的源数据中的一些记录（在底部）

我的表格设置由 3 个表格组成，看起来像

tu_unit_tbl

| unit_id | unit_name | unit_subject |

sp_student_tbl

| st_id |  st_fname  | st_lname |

sp_test_tbl

| ts_id |  ts_st_id  |  ts_unit_id | ts_session |

到目前为止，我的查询运行良好，但我需要在其上添加一个计算字段：

SELECT  U.unit_subject,
        U.unit_name,
        S.st_fname + ' ' + S.st_lname as sname,
        T.ts_st_id,
        T.ts_session
FROM sp_test_tbl T
JOIN sp_student_tbl S on T.ts_st_id = st_id
JOIN tu_unit_tbl U on unit_id = ts_unit_id

WHERE ts_st_id = 184318
GROUP BY unit_name, U.unit_subject, ts_session,  st_lname, st_fname, ts_st_id

它产生这个：

如果你注意到，有 3 行。前两个具有相同的“单位名称”。我正在寻找添加一个字段来计算这些行的出现次数

unit_subject | unit_name                            | sname    | ts_st_id | ts_session
--------------------------------------
Mathematics  | Algebra, Patterns, and Relationships | Frog Man | 184318   | 2012-07-31 15:22:42.000
Mathematics  | Algebra, Patterns, and Relationships | Frog Man | 184318   | 2012-07-31 15:25:47.000
Mathematics  | Data, Statistics, and Probability    | Frog Man | 184318   | 2012-07-31 15:25:59.000

我想得到什么

unit_subject | unit_name                            | sname    | **COUNT OF UNIT_NAME**
---------------------------------------------------------------------------------------
Mathematics  | Algebra, Patterns, and Relationships | Frog Man | **2**
Mathematics  | Data, Statistics, and Probability    | Frog Man | **1**

请注意，现在只有 2 条记录，因为前 2 条已合并计算。

当然，这是一个汇总结果，我想在一个新字段中计算这些汇总项（称为单元名称计数）

我尝试放入一个 Count(*) 字段，甚至JOIN是尝试对这些字段进行计数的子查询并没有给我一个聚合计数，而是计算我聚合的每个项目！

如何计算结果中的重复单位名称？

这是原始数据：（重复数据表示学生的测试会话。每条记录都是会话的一部分）

select ts_st_id,
       ts_session
from sp_test_tbl
where ts_st_id = 184318

ts_st_id | ts_session

184318 2012-07-31 15:22:42.000

184318 2012-07-31 15:22:42.000

184318 2012-07-31 15:22:42.000

184318 2012-07-31 15:22:42.000

184318 2012-07-31 15:22:42.000

184318 2012-07-31 15:22:42.000

184318 2012-07-31 15:22:42.000

184318 2012-07-31 15:22:42.000

184318 2012-07-31 15:22:42.000

184318 2012-07-31 15:22:42.000

184318 2012-07-31 15:22:42.000

184318 2012-07-31 15:22:42.000

184318 2012-07-31 15:22:42.000

184318 2012-07-31 15:25:47.000

184318 2012-07-31 15:25:47.000

184318 2012-07-31 15:25:47.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

184318 2012-07-31 15:25:59.000

我最近的尝试（由 praveen 推荐）似乎可行，但会出错。

WITH CTE (unit_subject,
        unitName,
        sname,
        ts_st_id,
        ts_session)
AS
(
    SELECT  U.unit_subject,
        U.unit_name as unitName,
        S.st_fname + ' ' + S.st_lname as sname,
        T.ts_st_id,
        T.ts_session
    FROM sp_test_tbl T
    JOIN sp_student_tbl S on T.ts_st_id = st_id
    JOIN tu_unit_tbl U on unit_id = ts_unit_id

    WHERE ts_st_id = 1234
    GROUP BY unit_name, U.unit_subject, ts_session,  st_lname, st_fname, ts_st_id
)

select count(unitName) over (partition by unit_name)

Go

生成以下错误消息：

Msg 207, Level 16, State 1, Line 22
Invalid column name 'unit_name'.
Msg 207, Level 16, State 1, Line 22
Invalid column name 'unitName'.
4

1 回答 1

0

使用公用表表达式做到了。谢谢普雷文！

 WITH CTE 
    AS
    (
        SELECT 
            U.unit_subject,
            U.unit_name,
            S.st_fname + ' ' + S.st_lname as sname,
            T.ts_st_id,
            st_grade,
            T.ts_session
        FROM sp_test_tbl T
        JOIN sp_student_tbl S on T.ts_st_id = st_id
        JOIN tu_unit_tbl U on unit_id = ts_unit_id

        WHERE ts_st_id = 1234
        GROUP BY unit_name, U.unit_subject, ts_session,  st_lname, st_fname, ts_st_id, st_grade
    )

    select distinct unit_subject, unit_name, sname, ts_st_id, st_grade, count(unit_name) over (partition by unit_name) as TestSessions
    from CTE
于 2012-08-15T23:25:33.940 回答