Question

我可能会混淆其他语言的SQL语法，这就是为什么它不起作用，我的子查询肯定是不正确的。

有以下2个表：

TEST_RESULTS
Student_ID Test_ID Test_Result
    A1       234       90  
    B2       234       80  
    C3       345       85
    D4       234       95
    A1       345       95
    C3       456       95


TEST_DESCRIPTION 
Test_ID  Test_Description Passing_Score
  234         Test A            85
  345         Test B            90
  456         Test C            95

我想计算每次测试的传递率并按它排序。

我正在寻找的输出：

Test_ID   Test_Description  students_taking students_passing   rate
456             Test C         1                1               1
234             Test A         3                2               0.666666667
345             Test B         2                1               0.5

这是我的查询

SELECT td.Test_ID, td.Test_Description, COUNT(tr.Student_ID) as 
students_taking, students_passing, students_passing/students_taking as rate
FROM 
(SELECT td.Test_ID, td.Test_Description, COUNT(tr.Student_ID) as 
students_passing
FROM TEST_RESULTS tr
JOIN TEST_DESCRIPTION td
on tr.Test_ID = td.Test_ID
WHERE tr.Test_Result > td.)
GROUP BY td.Test_ID, td.Test_Description
ORDER BY rate DESC, td.Test_ID, td.Test_Description

我从select中选择是错误的，因为我的查询没有结果。

Answer 1

我正在使用CTE，LEFT JOIN来获得所需的结果。试试这个查询 -

;WITH CTE
AS
(
    SELECT
        TD.TEST_ID,
        TEST_DESCRIPTION,
        COUNT(TR.STUDENT_ID) AS STUDENTS_TAKING,
        COUNT(CASE WHEN TR.TEST_RESULT >= TD.PASSING_SCORES THEN 
            TR.STUDENT_ID END) AS STUDENTS_PASSING
    FROM TEST_DESCRIPTION TD
    LEFT JOIN TEST_RESULTS TR
    ON TD.TEST_ID = TR.TEST_ID
    GROUP BY TD.TEST_ID,
        TEST_DESCRIPTION
)
SELECT 
    TEST_ID,
    TEST_DESCRIPTION,
    STUDENTS_TAKING,
    STUDENTS_PASSING,
    STUDENTS_PASSING / CONVERT (DECIMAL(4,2),STUDENTS_TAKING) AS RATE
FROM CTE
ORDER BY TEST_DESCRIPTION

Answer 2

SELECT td.Test_ID, td.Test_Description,
    students_taking  = counts.students_taking,
    students_passing = counts.students_passing,
    rate             = counts.rate
FROM TEST_DESCRIPTION td 
OUTER APPLY (
    SELECT 
        students_taking    = COUNT(1),
        students_passing   = COUNT(CASE WHEN tr.Test_Result > td.Passing_score THEN 1 ELSE NULL END),
        rate               = IIF(COUNT(1) <> 0, COUNT(CASE WHEN tr.Test_Result > td.Passing_score THEN 1 ELSE NULL END) / CAST(COUNT(1) AS FLOAT), 0)
    FROM TEST_RESULTS tr
    WHERE tr.Test_ID = td.Test_ID
) counts
ORDER BY counts.rate DESC, td.Test_ID

Answer 3

请查看以下查询 -

SELECT TD.TEST_ID,
TD.TEST_DESCRIPTION,
STUDENT_TAKING,
STUDENT_PASSING,
RATE 
FROM TEST_DESCRIPTION TD,
  (SELECT TR.TEST_ID,COUNT(TR.STUDENT_ID) "STUDENT_TAKING",
  COUNT(CASE WHEN TEST_RESULT>=PASSING_SCORE THEN STUDENT_ID END) STUDENT_PASSING,
  TO_NUMBER(TO_CHAR(COUNT(CASE WHEN TEST_RESULT>=PASSING_SCORE THEN STUDENT_ID END)/COUNT(TR.STUDENT_ID),'9999.99')) RATE
  FROM TEST_RESULTS TR,TEST_DESCRIPTION TD 
  WHERE TR.TEST_ID=TD.TEST_ID
  GROUP  BY TR.TEST_ID)SUB
  WHERE SUB.TEST_ID=TD.TEST_ID ORDER BY RATE DESC;

SQL：按两个表之间的计算速率排序

3 个答案: