在 mysql 上加入多个复杂的表格

Question

我正在尝试查找对问题 "platform" (id 1) 回答 'facebook' 的所有客户

每个客户都有调查问卷，我正在尝试寻找回答 "facebook" 问题的客户 "platform" 问题的 id 是固定的，例如 "platform" 问题的 id table 是 1 ， "age" 是 2 等等。每个问卷都有不同的问题集，例如问卷 id 1 有问题 "platform" 但答案将通过选择框而不是文本框输入（因此具有需要 answer_option table) 和问卷 ID 2 可能没有问题平台等。来自搜索屏幕的参数是问卷 ID 和回答问题的文本 "platform".

目前这就是我目前所拥有的..

SELECT customers.* FROM customers
INNER JOIN questionnaire_answers
ON questionnaire_answers.answerable_id = customers.id
AND questionnaire_answers.answerable_type = 'Customer'
INNER JOIN questionnaires
ON questionnaires.id = questionnaire_answers.questionnaire_id
INNER JOIN questions
ON questions.questionnaire_id = questionnaires.id
INNER JOIN answers
ON answers.question_id = questions.id
INNER JOIN answer_options
ON answer_options.id = answers.answer_option_id
WHERE (questions.id = 1 and answers.answer LIKE '%facebook%');

如果您注意到 question.id 将始终为 1，因为我只是在平台中寻找答案，这是我的屏幕的一个特殊功能。我用 1 - 20 行数据尝试了它，它可以工作，但如果数据变得像 20,000 行一样大，它永远不会 returns 我已经用真实数据尝试过的记录查询永远需要取消它。由于问题 id 已经是常量并且问卷 id 来自搜索参数，我想也许我不需要加入不必要的 table 并且我认为放置索引不会有帮助，因为我有一个通配符参数？（不确定），但我不知道如何。

示例数据和预期结果

Customer:
 id:1,
 family_name: "smith",
 first_name: "smith",
..
Customer:
 id:2,
 family_name: "david",
 first_name: "david",
..

QuestionnaireAnswer:
  id: 20199,
  questionnaire_id: 4,
  answerable_type: "Customer",
  answerable_id: "1",
QuestionnaireAnswer:
  id: 20200,
  questionnaire_id: 5,
  answerable_type: "Customer",
  answerable_id: "2",

Questionnaire:
  id: 4,
  name: "Survey for Psychology Students",
  group_code: "10",
Questionnaire:
  id: 5,
  name: "Survey for Students",
  group_code: "10",

Question:
  id: 1,
  name: "platform",
  answer_type: "08", //selectbox
  default_text: nil,
  sequence: 1,
  question_valid: true,
  disabled: false,
  questionnaire_id: 5,
  ...
Question:
  id: 2,
  name: "school",
  answer_type: "01", //textbox
  default_text: nil,
  sequence: 2,
  question_valid: true,
  disabled: false,
  questionnaire_id: 5,
  ...

Question:
  id: 3, // I said the the id for platform is fixed but this time this question is textbox inputted so it is different
  name: "platform",
  answer_type: "01", //text
  default_text: nil,
  sequence: 1,
  question_valid: true,
  disabled: false,
  questionnaire_id: 5,
  ...
Question:
  id: 4,
  name: "town",
  answer_type: "01", //textbox
  default_text: nil,
  sequence: 2,
  question_valid: true,
  disabled: false,
  questionnaire_id: 4,
  ...

Answer:
  id: 1,
  question_id: 3,
  answer_option_id: nil,
  answer: "facebook",
  questionnaire_answer_id: 19585,
  created_at: Wed, 13 Feb 2019 21:43:15 JST +09:00,
  updated_at: Wed, 13 Feb 2019 21:43:15 JST +09:00,
  member_id: nil>,

我们将不胜感激。

Answer 1

您正在加入几个不必要的表。您没有使用 answer_options 中的任何内容，因此您不需要它。 answers 有 questionnaire_answers 和 questions 的外键，所以你不需要 questionnaires.

SELECT customers.* 
FROM customers AS c
INNER JOIN questionnaire_answers AS qa
ON qa.answerable_id = c.id
INNER JOIN answers AS a
ON a.questionnaire_answers_id = qa.id
INNER JOIN questions AS q
ON q.id = a.question_id
WHERE q.id = 1
    AND a.answer LIKE '%facebook%'
    AND qa.answerable_type = 'Customer'

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