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これが私のデータベース構造の写真です：

観測記録-データベース構造

ここのユーザーの助けを借りて、私は以下を使用して非常に複雑なSQLステートメントをまとめることができましたGROUP_CONCAT。

SELECT
     t1.Name AS Teacher_Name,
     t2.Name AS Observer_Name,
     o.Datetime AS Datetime,
     o.Type AS Type,
     o.Year_Group AS Year_Group,
     o.Class_Name AS Class_Name,
     c.Title AS Course_Name,
     GROUP_CONCAT(l.Title) AS Focus,
     o.Achievement_Grade AS Achievement_Grade,
     o.Behaviour_Grade AS Behaviour_Grade,
     o.Teaching_Grade AS Teaching_Grade,
     GROUP_CONCAT(cl1.Title) AS Positive,
     GROUP_CONCAT(cl2.title) AS Development,
     o.Notes AS Notes
FROM observations o
LEFT JOIN teachers t1 
    ON o.Teacher_ID = t1.Teacher_ID
LEFT JOIN teachers t2 
    ON o.Observer_ID = t2.Teacher_ID
LEFT JOIN courses c 
    ON o.Course_ID = c.Course_ID
LEFT JOIN  foci f
    ON o.ID = f.Observation_ID
LEFT JOIN focus_labels l
    on f.focus_id = l.id
LEFT JOIN  criteria c1
    ON o.ID = c1.Observation_ID
LEFT JOIN criteria_labels cl1
    on c1.Criteria_ID = cl1.ID AND c1.Type = 'P'
LEFT JOIN  criteria c2
    ON o.ID = c2.Observation_ID AND c2.Type = 'D'
LEFT JOIN criteria_labels cl2
    on c2.Criteria_ID = cl2.ID
GROUP BY o.id
ORDER BY `Datetime` DESC";

の連結数が最も多いフィールドに応じて、、、およびがそれぞれ繰り返されるFocusというPositive事実を除けば、これは問題なく機能しているように見えます。Development

たとえば、PositivehasPace,Progress,Attainmentが、FocusのみAppraisalの場合、3回繰り返されます（Appraisal,Appraisal,Appraisal）。

観察記録-複雑な結合構造

GROUP私はこれを調べました、そしてそれは私がこれらのそれぞれをする必要があるからかもしれないと思いますGROUP_CONCAT JOINs。しかし、どうすればいいのかわかりません。

誰か助けてもらえますか？前もって感謝します、

4

1 に答える 1

1

GROUP_CONCATには、重複を削除するために適用できるDISTINCT属性があります。

SELECT
     t1.Name AS Teacher_Name,
     t2.Name AS Observer_Name,
     o.Datetime AS Datetime,
     o.Type AS Type,
     o.Year_Group AS Year_Group,
     o.Class_Name AS Class_Name,
     c.Title AS Course_Name,
     GROUP_CONCAT(DISTINCT l.Title) AS Focus,
     o.Achievement_Grade AS Achievement_Grade,
     o.Behaviour_Grade AS Behaviour_Grade,
     o.Teaching_Grade AS Teaching_Grade,
     GROUP_CONCAT(cl1.Title) AS Positive,
     GROUP_CONCAT(cl2.title) AS Development,
     o.Notes AS Notes
FROM observations o
LEFT JOIN teachers t1 
    ON o.Teacher_ID = t1.Teacher_ID
LEFT JOIN teachers t2 
    ON o.Observer_ID = t2.Teacher_ID
LEFT JOIN courses c 
    ON o.Course_ID = c.Course_ID
LEFT JOIN  foci f
    ON o.ID = f.Observation_ID
LEFT JOIN focus_labels l
    on f.focus_id = l.id
LEFT JOIN  criteria c1
    ON o.ID = c1.Observation_ID
LEFT JOIN criteria_labels cl1
    on c1.Criteria_ID = cl1.ID AND c1.Type = 'P'
LEFT JOIN  criteria c2
    ON o.ID = c2.Observation_ID AND c2.Type = 'D'
LEFT JOIN criteria_labels cl2
    on c2.Criteria_ID = cl2.ID
GROUP BY o.id
ORDER BY `Datetime` DESC";
于 2012-09-18T19:52:51.433 に答える