简体   繁体   English

如何在 MySQL 查询中有条件地连接表

[英]How to conditionally Join tables in MySQL Query

Hie,嗨,

I have 3 tables namely Accounts , Companies and Freelancers .我有 3 张表,即AccountsCompaniesFreelancers The accounts table holds the email , password and account_type fields where if account_type values is an ENUM('companies','freelancers') . accounts 表包含emailpasswordaccount_type字段,如果account_type值为ENUM('companies','freelancers') I want to know how I can write a query that Joins with Companies Table if account_type fields is equal to companies else should Join with Freelancers我想知道如果account_type字段等于companies ,我如何编写一个加入Companies表的查询,否则应该加入Freelancers

If you want to get all the data to one table,如果你想把所有数据放到一张表中,

  1. you can simply use a union to bind 2 results sets that you get from joining 2 tables.您可以简单地使用union来绑定从连接 2 个表中获得的 2 个结果集。

     SELECT * FROM users Accounts a LEFT JOIN Companies c ON c.account_id = a.id WHERE a.account_type = 'companies' UNION SELECT * FROM users Accounts a LEFT JOIN Freelancers f ON f.account_id = a.id WHERE a.account_type = 'freelancers'
  2. You can join the tables according to the ids by getting the ids to one table.您可以通过将 id 获取到一张表来根据 id 加入表。

     SELECT Accounts.id, Accounts.type, Companies.account_id AS id2, Freelancers.account_id AS id3 FROM Accounts LEFT JOIN Companies ON Accounts.id = Companies.account_id AND Accounts.type = 'companies' LEFT JOIN Freelancers ON Freelancers.account_id = Accounts.id AND Accounts.type = 'freelancers'

I hope this will help you to resolve your issue.我希望这将帮助您解决您的问题。

For example例如

SELECT Accounts.column1, 
       COALESCE(Companies.column2, Freelancers.column3), -- common column
       Companies.column4,                                -- specific column
       Freelancers.column5                               -- specific column
FROM Accounts
LEFT JOIN Companies ON Companies.account_id = Accounts.id 
                   AND Accounts.account_type = 'companies'
LEFT JOIN Freelancers ON Freelancers.account_id = Accounts.id 
                   AND Accounts.account_type = 'freelancers'

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM