如何使用嵌套子查询SQL获取嵌套关系实体？

Question

我有这 4 个表格：记分卡、部分、主题、答案。

每个记分卡可以有多个部分。每个部分可以有很多主题，每个主题可以有很多答案。

我希望以下列形式检索数据。获取记分卡 id 等于某个 id 的所有部分，我通过左连接来做到这一点，到目前为止它一直有效。然后对于每个部分，我也想获得它的所有主题和答案。

=> 预期输出

[{section1, 
topics:[{topicName, answers:[{answerName}]}]}, 
{section2, 
topics:[{topicName, answers:[{answerName}]}]}]

我调整了我的查询并正确地连接了表格，但输出数据没有像我原来那样正确组织。

查询：

    SELECT sec."id",
       sec."name" AS "sectionName",
       to_json(topics) as "topics"
       
FROM "scorecard" sc 
LEFT JOIN "section" sec ON sc."id" = sec."scorecardId"
LEFT JOIN  (SELECT tp.*, to_json(answers.*) as "answers"
            FROM "topic" tp
            LEFT JOIN (SELECT ans."name" as "answerName", ans."topicId"
            FROM "answer" ans) answers ON answers."topicId" = tp."id"
            ) topics
            ON topics."sectionId" = sec."id"
WHERE sc."id" =

当前输出：

    [  id: 85,
      sectionName: 'Consultation',
      topics: {
        id: 109,
        name: 'Was the information correct?',
        answers: [{ answerName: 'Yes', topicId: 109 },        answers: { answerName: 'NO', topicId: 109 }
]
      },
      id: 85,
      sectionName: 'Consultation',
      topics: {
        id: 109,
        name: 'Was the information correct?',
        answers: { answerName: 'NO', topicId: 109 }
      }]

我的预期输出：

[  {id: 85,
  sectionName: 'Consultation',
  topics: [{
    id: 109,
    name: 'Was the information correct?',
    answers: [{ answerName: 'Yes', topicId: 109 },
             { answerName: 'NO', topicId: 109 }]
  }]
} ]

我也尝试加入所有实体：

SELECT sec."id",
       sec."name" AS "sectionName",
       tp."name" AS topics, ans."name" AS answers
FROM "scorecard" sc 
LEFT JOIN "section" sec ON sc."id" = sec."scorecardId"
LEFT JOIN "topic" tp ON tp."sectionId" = sec."id"
LEFT JOIN "answer" ans ON ans."topicId"= tp."id"
WHERE sc."id" =

输出没有按应有的方式组织。

[
 {
    id: 91,
    sectionName: 'Politeness',
    topics: 'Was the agent polite with the client?',
    answers: 'No'
  },
  {
    id: 91,
    sectionName: 'Politeness',
    topics: 'Was the agent polite with the client?',
    answers: 'Not everytime'
  },
  {
    id: 91,
    sectionName: 'Politeness',
    topics: 'Was the agent polite with the client?',
    answers: 'Yes'
  }
]

Answer 1

您缺少 table topics

的 JOIN 条件

SELECT sec."id",
       sec."name" AS "sectionName",
       topics
FROM "scorecard" sc 
LEFT JOIN "section" sec ON sc."id" = sec."scorecardId"
LEFT JOIN (SELECT tp.* 
      FROM "topic" tp 
      WHERE tp."sectionId" = sec."id"
           (SELECT tp.* 
            FROM "topic" tp 
            WHERE tp."sectionId" = sec."id"
           )
      ) as topics  ON topics.??? = ???  -- Here you should enter the relation
                                        --  from topics to your section.
WHERE sc."id" = 30

编辑：

根据你的（当前）输入，我创建了这个 DBFIDDLE

输出：

[
    {
        "id": 85,
        "sectionName": "Consulation",
        "topics": {
            "id": 109,
            "name": "Was the information correct?",
            "sectionId": 85,
            "answers": {
                "answerName": "Yes",
                "topicId": 109
            }
        }
    },
    {
        "id": 85,
        "sectionName": "Consulation",
        "topics": {
            "id": 109,
            "name": "Was the information correct?",
            "sectionId": 85,
            "answers": {
                "answerName": "NO",
                "topicId": 109
            }
        }
    }
]

Answer 2

您可以连接所有四个表以获得结果。

SELECT sec."id",
       sec."name" AS "sectionName",
       tp."name" AS topics, ans."name" AS answers
FROM "scorecard" sc 
LEFT JOIN "section" sec ON sc."id" = sec."scorecardId"
LEFT JOIN "topic" tp on tp."sectionId" = sec."id"
LEFT JOIN "answer" ans on ans."topicId"=tp."id"
WHERE sc."id" = 30

感谢@Luuk 创建数据库fiddle link.

  select json_agg(x) AS result
  from 
  (
    SELECT sec."id",
        sec."name" AS "sectionName",
        json_agg(topics) as "topics"
        
   FROM "scorecard" sc 
   LEFT JOIN "section" sec ON sc."id" = sec."scorecardId"
   LEFT JOIN  (SELECT tp.id,tp.name,tp."sectionId", json_agg(answers.*) as "answers"
             FROM "topic" tp
             LEFT JOIN (SELECT ans."name" as "answerName", ans."topicId"
             FROM "answer" ans) answers ON answers."topicId" = tp."id"
             group by tp.id,tp.name,tp."sectionId"
             ) topics
             ON topics."sectionId" = sec."id"
   WHERE sc."id" = 85
   group by sec."id", sec."name"
 )x

输出：

result
[{"id":85,"sectionName":"Consulation","topics":[{"id":109,"name":"Was the information correct?","sectionId":85,"answers":[{"answerName":"Yes","topicId":109}, <br> {"answerName":"NO","topicId":109}]}]}]

*db<>fiddle here149)

如何使用嵌套子查询SQL获取嵌套关系实体？

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