Question

我有三张牌桌：do script，osascript -e 'tell app "Terminal" to do script "/usr/local/bin/python3.6 /path/to/testscript.py"'和answers我已加入他们以获得我想要的结果

questions

现在我还有一个名为users的表。该表基本上存储upvote，downvote用于用户的答案。

如果用户提出了答案，即列SELECT answers.answer_content as answer_content, answers.id as answer_id, answers.created_at as created_at, questions.id as question_id, questions.question_title as question_title, questions.question_slug as question_slug, users.id as user_id, users.name as user_name, users.user_slug as user_slug, FROM answers JOIN questions on questions.id = answers.question_id JOIN users on users.id = answers.user_id WHERE questions.question_active = 1 and answers.answer_active = 1，upvote_answers，user_id（1或0 - 分别为upvote或downvote）

所以我的问题是：我想再获得一个布尔结果的列，该列应该显示upvoted或downvoted，或者对于当前登录用户的每个答案没有任何内容

我尝试像这样编写查询

answer_id

当用户登录时，他应该能够看到他是否已经投票或低估了答案。

Answer 1

考虑到给定用户只能对任何答案投票一次，这应该可以解决问题：

SELECT answers.answer_content as answer_content,
       answers.id as answer_id,
       answers.created_at as created_at,
       questions.id as question_id, 
       questions.question_title as question_title, 
       questions.question_slug as question_slug, 
       users.id as user_id, 
       users.name as user_name, 
       users.user_slug as user_slug,
       upvote_answers.upvote as upvote
FROM 
    answers
    JOIN questions on questions.id = answers.question_id
    JOIN users on users.id = answers.user_id
    LEFT JOIN upvote_answers ON
        upvote_answers.answer_id = answers.id AND
        upvote_answers.user_id = answers.user_id
WHERE 
    questions.question_active = 1 and 
    answers.answer_active = 1

如果LEFT JOIN子句匹配，则会生成结果，但如果不匹配，则会带来NULL个值（对于其表）。

Sql join查询并再选择一列

1 个答案: