SQL 查詢 - 匹配至少 5 個值中的 4 個值的條件

Question

我是SQL新手，我需要幫助編寫查詢（附照片圖）

概括：

IG School 系統，要求學生在特定的學習時段中選擇 5 個學習科目，那么查詢結果應該是：

1) 已經有所選科目的可用組；

或者

2) 至少符合 4 個所選科目的組，並且還返回沒有匹配的科目。

表格：

Sessions -----> Like Semesters
Groups   -----> Like classes that has schedule

我已經寫了一個查詢，但對於只有 4 個匹配主題的可用組沒有條件，只返回匹配所有 5 個主題的組（問題），如下所示：

    SELECT Group_Subjects.Group_ID, Groups.Name as Groups_Name,  Subject_Schedule.ID as Subject_Schedule_ID,Subjects.ID Subjects_ID,
    Subjects.Name AS Subjects_Name,Sessions.Name AS Sessions_Name
    FROM   Groups 
    INNER JOIN Group_Subjects ON Groups.ID = Group_Subjects.Group_ID 
    INNER JOIN Subject_Schedule ON Group_Subjects.Subject_Schedule_ID = Subject_Schedule.ID 
    INNER JOIN Subjects ON Subject_Schedule.Subject_ID = Subjects.ID
    INNER JOIN Sessions ON dbo.Groups.Session_ID = dbo.Sessions.ID   
    where Groups.Session_ID=13 and  Subjects.ID in (16,117,125,126,127)
    order by Group_Subjects.Group_ID,Subjects_Name

照片圖

Answer 1

您可以將選定組的結果分組（分組依據）並按主題的出現次數（計數（主題）> = 4）進行過濾，如下所示：

SELECT Group_Subjects.Group_ID, Groups.Name as Groups_Name,  Subject_Schedule.ID as Subject_Schedule_ID,Subjects.ID Subjects_ID,
    Subjects.Name AS Subjects_Name,Sessions.Name AS Sessions_Name
    , **count (Subjects.ID)** 
    FROM   Groups 
    INNER JOIN Group_Subjects ON Groups.ID = Group_Subjects.Group_ID 
    INNER JOIN Subject_Schedule ON Group_Subjects.Subject_Schedule_ID = Subject_Schedule.ID 
    INNER JOIN Subjects ON Subject_Schedule.Subject_ID = Subjects.ID
    INNER JOIN Sessions ON dbo.Groups.Session_ID = dbo.Sessions.ID   
    where  Subjects.ID in (16,117,125,126,127)
    **group by** Group_Subjects.Group_ID, Groups.Name as Groups_Name,  Subject_Schedule.ID as 
    Subject_Schedule_ID,Subjects.ID Subjects_ID,
    Subjects.Name AS Subjects_Name,Sessions.Name AS Sessions_Name
    **having count (Subjects.ID) >= 4** 

Answer 2

如果我理解得很好（盡管我不是 100% 確定），您的要求包括： 1. 選擇與 5 個給定“主題”中的至少 4 個匹配的“組”。 2. 不要包括匹配少於 5 個給定“主題”中的 4 個的“組”。

為此，您可以運行以下查詢：

SELECT 
    gr.name, sub.name
FROM
    groups           gr  JOIN
    group_subject    gs  ON gs.group_id = gr.id JOIN
    subject_schedule ss  ON ss.id = gs.subject_schedule_id JOIN
    subject          sub ON sub.id = ss.subject_id  
WHERE
    EXISTS (
        SELECT 
            g.id, COUNT(*) num_subjects
        FROM
            groups           g  JOIN
            group_subject    gs  ON gs.group_id = g.id JOIN
            subject_schedule ss  ON ss.id = gs.subject_schedule_id JOIN
            subject          sub ON sub.id = ss.subject_id  
        WHERE
            sub.id IN ('A', 'B', 'C', 'D', 'E') AND
            g.id = gr.id
        GROUP BY gr.id
        HAVING num_subjects >=4
    )
ORDER BY 
    gr.name, sub.name;

除此之外， subject_schedule似乎指的是sessions 。 同時， groups也指sessions 。
如果數據不一致，這可能會導致問題。 出於這個原因，我沒有在我的查詢中包含“會話”，因為其他表包含您的用例所需的所有信息。

我已經創建了一個模式的示例模型來測試該查詢（MySQL）：

CREATE TABLE groups (
    id                   VARCHAR(5) PRIMARY KEY,
    name                 TEXT,
    session_id           VARCHAR(7) REFERENCES sessions
);
CREATE TABLE group_subject (
    id                   INTEGER PRIMARY KEY,
    group_id             VARCHAR(5) REFERENCES groups,
    subject_schedule_id  INTEGER REFERENCES subject_schedule,
    description          TEXT
);
CREATE TABLE subject_schedule (
    id                   INTEGER PRIMARY KEY,
    name                 TEXT,
    session_id           VARCHAR(7) REFERENCES sessions,
    subject_id           VARCHAR(1) REFERENCES subject
);
CREATE TABLE subject (
    id                   VARCHAR(1) PRIMARY KEY,
    name                 TEXT
);
CREATE TABLE sessions (
    id                   VARCHAR(7) PRIMARY KEY,
    name                 TEXT
);

希望能幫助到你。