Reporting services SSRS-矩阵中的重复值

我是这方面的初学者，请原谅我的无知。我已经搜索了几个小时，但还没有经验来更好地了解到底要搜索什么

我在SSRS中有如下数据集：

    |---------------------|------------------|
    |      Subject        |     Student      |
    |---------------------|------------------|
    |      Biology        |    Student A     |
    |---------------------|------------------|
    |      Chemistry      |    Student A     |
    |---------------------|------------------|
    |      Chemistry      |    Student B     |
    |---------------------|------------------|
    |      Physics        |    Student B     |
    |---------------------|------------------|

我想写一个矩阵，统计所有同时学习化学和物理等学科的学生。目前，我的矩阵只计算单个学科的学生人数，结果如下：

|---------------------|------------------|------------------|------------------|
|                     |      Biology     |    Chemistry     |     Physics      |
|---------------------|------------------|------------------|------------------|
|    Biology          |        20        |        0         |        0         |
|---------------------|------------------|------------------|------------------|
|    Chemistry        |        0         |        36        |        0         |
|---------------------|------------------|------------------|------------------|
|    Physics          |        0         |        0         |        16        |
|---------------------|------------------|------------------|------------------|

---

我遗漏了什么？

首先，您的数据集需要将主题合并为Subject1和Subject2，例如，通过在Student字段中将表连接到自身

---

然后，在矩阵中，使用按Subject1分组的行组和按Subject2分组的列组。

首先，您的数据集需要将主题合并为Subject1和Subject2，例如，通过在Student字段中将表连接到自身

然后，在矩阵中，使用按主题1分组的行组和按主题2分组的列组。

这很简单： 您可以将学生和主题之间的地图与其自身连接起来，但前提是学生相同，但主题不同

然后在报告中对studentId进行计数，不管它是来自map1还是map2，因为它总是相同的

;WITH subjects AS 
(
              SELECT 1 AS subject_id, 'Biology' AS subject_name 
    UNION ALL SELECT 2 AS subject_id, 'Chemistry' AS subject_name 
    UNION ALL SELECT 3 AS subject_id, 'Physics' AS subject_name 
)
,students AS 
(
              SELECT 1 AS student_id, 'Student A' AS student_name 
    UNION ALL SELECT 2 AS student_id, 'Student B' AS student_name 
    UNION ALL SELECT 3 AS student_id, 'Student C' AS student_name 
    UNION ALL SELECT 4 AS student_id, 'Student D' AS student_name 
)
,map_student_subject AS 
(
              SELECT 1 AS student_id, 1 AS subject_id 
    UNION ALL SELECT 1 AS student_id, 2 AS subject_id 
    UNION ALL SELECT 2 AS student_id, 2 AS subject_id 
    UNION ALL SELECT 2 AS student_id, 3 AS subject_id 
    UNION ALL SELECT 3 AS student_id, 3 AS subject_id 
    UNION ALL SELECT 4 AS student_id, 3 AS subject_id 
)
SELECT 
     map1.student_id AS map1_student_id 
    ,map1_student.student_name AS map1_student_name 
    ,map1.subject_id AS map1_subject_id
    ,map1_subject.subject_name AS map1_subject_name 

    ,map2.student_id AS map2_student_id 
    ,map2_student.student_name AS map2_student_name 
    ,map2.subject_id AS map2_subject_id
    ,map2_subject.subject_name AS map2_subject_name 
FROM map_student_subject AS map1 

INNER JOIN map_student_subject AS map2 
    ON map2.student_id = map1.student_id 
    AND map2.subject_id <> map1.subject_id 

LEFT JOIN students AS map1_student 
    ON map1_student.student_id = map1.student_id 

LEFT JOIN students AS map2_student 
    ON map2_student.student_id = map2.student_id 

LEFT JOIN subjects AS map1_subject 
    ON map1_subject.subject_id = map1.subject_id 

LEFT JOIN subjects AS map2_subject 
    ON map2_subject.subject_id = map2.subject_id 

结果:

如果你想列出那些只有一门学科的学生，你需要建立一个联盟，因为只要去掉这个联盟，就不可能计算出一门学科的实际学生人数：

;WITH subjects AS 
(
              SELECT 1 AS subject_id, 'Biology' AS subject_name 
    UNION ALL SELECT 2 AS subject_id, 'Chemistry' AS subject_name 
    UNION ALL SELECT 3 AS subject_id, 'Physics' AS subject_name 
)
,students AS 
(
              SELECT 1 AS student_id, 'Student A' AS student_name 
    UNION ALL SELECT 2 AS student_id, 'Student B' AS student_name 
    UNION ALL SELECT 3 AS student_id, 'Student C' AS student_name 
    UNION ALL SELECT 4 AS student_id, 'Student D' AS student_name 
)
,map_student_subject AS 
(
              SELECT 1 AS student_id, 1 AS subject_id 
    UNION ALL SELECT 1 AS student_id, 2 AS subject_id 
    UNION ALL SELECT 2 AS student_id, 2 AS subject_id 
    UNION ALL SELECT 2 AS student_id, 3 AS subject_id 
    UNION ALL SELECT 3 AS student_id, 3 AS subject_id 
    UNION ALL SELECT 4 AS student_id, 3 AS subject_id 
)
SELECT 
     map1.student_id AS map1_student_id 
    ,map1_student.student_name AS map1_student_name 
    ,map1.subject_id AS map1_subject_id
    ,map1_subject.subject_name AS map1_subject_name 

    ,map2.student_id AS map2_student_id 
    ,map2_student.student_name AS map2_student_name 
    ,map2.subject_id AS map2_subject_id
    ,map2_subject.subject_name AS map2_subject_name 
FROM map_student_subject AS map1 

INNER JOIN map_student_subject AS map2 
    ON map2.student_id = map1.student_id 
    AND map2.subject_id <> map1.subject_id 

LEFT JOIN students AS map1_student 
    ON map1_student.student_id = map1.student_id 

LEFT JOIN students AS map2_student 
    ON map2_student.student_id = map2.student_id 

LEFT JOIN subjects AS map1_subject 
    ON map1_subject.subject_id = map1.subject_id 

LEFT JOIN subjects AS map2_subject 
    ON map2_subject.subject_id = map2.subject_id 


UNION ALL 


SELECT 
     students.student_id  AS map1_student_id 
    ,students.student_name  AS map1_student_name 
    ,MAX(subjects.subject_id) AS map1_subject_id
    ,MAX(subjects.subject_name) AS map1_subject_name 

    ,students.student_id  AS map2_student_id 
    ,students.student_name  AS map2_student_name 
    ,MAX(subjects.subject_id) AS map2_subject_id
    ,MAX(subjects.subject_name) AS map2_subject_name 
FROM map_student_subject 
LEFT JOIN students ON students.student_id = map_student_subject.student_id 
LEFT JOIN subjects ON subjects.subject_id = map_student_subject.subject_id 


GROUP BY 
     students.student_id 
    ,students.student_name 

HAVING COUNT(*) = 1

这很简单： 您可以将学生和主题之间的地图与其自身连接起来，但前提是学生相同，但主题不同

然后在报告中对studentId进行计数，不管它是来自map1还是map2，因为它总是相同的

;WITH subjects AS 
(
              SELECT 1 AS subject_id, 'Biology' AS subject_name 
    UNION ALL SELECT 2 AS subject_id, 'Chemistry' AS subject_name 
    UNION ALL SELECT 3 AS subject_id, 'Physics' AS subject_name 
)
,students AS 
(
              SELECT 1 AS student_id, 'Student A' AS student_name 
    UNION ALL SELECT 2 AS student_id, 'Student B' AS student_name 
    UNION ALL SELECT 3 AS student_id, 'Student C' AS student_name 
    UNION ALL SELECT 4 AS student_id, 'Student D' AS student_name 
)
,map_student_subject AS 
(
              SELECT 1 AS student_id, 1 AS subject_id 
    UNION ALL SELECT 1 AS student_id, 2 AS subject_id 
    UNION ALL SELECT 2 AS student_id, 2 AS subject_id 
    UNION ALL SELECT 2 AS student_id, 3 AS subject_id 
    UNION ALL SELECT 3 AS student_id, 3 AS subject_id 
    UNION ALL SELECT 4 AS student_id, 3 AS subject_id 
)
SELECT 
     map1.student_id AS map1_student_id 
    ,map1_student.student_name AS map1_student_name 
    ,map1.subject_id AS map1_subject_id
    ,map1_subject.subject_name AS map1_subject_name 

    ,map2.student_id AS map2_student_id 
    ,map2_student.student_name AS map2_student_name 
    ,map2.subject_id AS map2_subject_id
    ,map2_subject.subject_name AS map2_subject_name 
FROM map_student_subject AS map1 

INNER JOIN map_student_subject AS map2 
    ON map2.student_id = map1.student_id 
    AND map2.subject_id <> map1.subject_id 

LEFT JOIN students AS map1_student 
    ON map1_student.student_id = map1.student_id 

LEFT JOIN students AS map2_student 
    ON map2_student.student_id = map2.student_id 

LEFT JOIN subjects AS map1_subject 
    ON map1_subject.subject_id = map1.subject_id 

LEFT JOIN subjects AS map2_subject 
    ON map2_subject.subject_id = map2.subject_id 

结果:

如果你想列出那些只有一门学科的学生，你需要建立一个联盟，因为只要去掉这个联盟，就不可能计算出一门学科的实际学生人数：

;WITH subjects AS 
(
              SELECT 1 AS subject_id, 'Biology' AS subject_name 
    UNION ALL SELECT 2 AS subject_id, 'Chemistry' AS subject_name 
    UNION ALL SELECT 3 AS subject_id, 'Physics' AS subject_name 
)
,students AS 
(
              SELECT 1 AS student_id, 'Student A' AS student_name 
    UNION ALL SELECT 2 AS student_id, 'Student B' AS student_name 
    UNION ALL SELECT 3 AS student_id, 'Student C' AS student_name 
    UNION ALL SELECT 4 AS student_id, 'Student D' AS student_name 
)
,map_student_subject AS 
(
              SELECT 1 AS student_id, 1 AS subject_id 
    UNION ALL SELECT 1 AS student_id, 2 AS subject_id 
    UNION ALL SELECT 2 AS student_id, 2 AS subject_id 
    UNION ALL SELECT 2 AS student_id, 3 AS subject_id 
    UNION ALL SELECT 3 AS student_id, 3 AS subject_id 
    UNION ALL SELECT 4 AS student_id, 3 AS subject_id 
)
SELECT 
     map1.student_id AS map1_student_id 
    ,map1_student.student_name AS map1_student_name 
    ,map1.subject_id AS map1_subject_id
    ,map1_subject.subject_name AS map1_subject_name 

    ,map2.student_id AS map2_student_id 
    ,map2_student.student_name AS map2_student_name 
    ,map2.subject_id AS map2_subject_id
    ,map2_subject.subject_name AS map2_subject_name 
FROM map_student_subject AS map1 

INNER JOIN map_student_subject AS map2 
    ON map2.student_id = map1.student_id 
    AND map2.subject_id <> map1.subject_id 

LEFT JOIN students AS map1_student 
    ON map1_student.student_id = map1.student_id 

LEFT JOIN students AS map2_student 
    ON map2_student.student_id = map2.student_id 

LEFT JOIN subjects AS map1_subject 
    ON map1_subject.subject_id = map1.subject_id 

LEFT JOIN subjects AS map2_subject 
    ON map2_subject.subject_id = map2.subject_id 


UNION ALL 


SELECT 
     students.student_id  AS map1_student_id 
    ,students.student_name  AS map1_student_name 
    ,MAX(subjects.subject_id) AS map1_subject_id
    ,MAX(subjects.subject_name) AS map1_subject_name 

    ,students.student_id  AS map2_student_id 
    ,students.student_name  AS map2_student_name 
    ,MAX(subjects.subject_id) AS map2_subject_id
    ,MAX(subjects.subject_name) AS map2_subject_name 
FROM map_student_subject 
LEFT JOIN students ON students.student_id = map_student_subject.student_id 
LEFT JOIN subjects ON subjects.subject_id = map_student_subject.subject_id 


GROUP BY 
     students.student_id 
    ,students.student_name 

HAVING COUNT(*) = 1

---

您能否分享一些关于当前设计和分组的更多信息，以获得该结果？设计视图的一些屏幕截图/图像可能非常有用。您能否分享一些关于当前设计的更多信息，以及用于获得该结果的分组？设计视图的一些截图/图像可能非常有用。