php 如何在 MySQL 中使用外键进行查询？
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How can I query using a foreign key in MySQL?
提问by Josh Mountain
Right now I have a small database with two tables that look something like this:
现在我有一个带有两个表的小型数据库，看起来像这样：
users table
====================
id name status_id
1 Bobby 3
2 James 2
and
和
statuses table
=============
id value
1 Waiting
2 Approved
3 Other
status_id is setup as a foreign key constraint to id from the statuses table. My query looks something like this:
status_id 被设置为来自 statuses 表的 id 的外键约束。我的查询看起来像这样：
SELECT *
FROM `users`
WHERE `status_id` = 2";
When I display $row['status_id']it outputs 2but I would like it to display as Approvedinstead, what is the best way to accomplish this?
当我显示$row['status_id']它输出2但我希望它显示Approved为时，实现此目的的最佳方法是什么？
回答by Clodoaldo Neto
SELECT u.*, s.*
FROM users u
inner join statuses s on u.status_id = s.id
WHERE u.status_id = 2
回答by Iberê
What you need is this
你需要的是这个
SELECT *
FROM `users`
JOIN statuses ON statuses.id = users.status_id
WHERE `status_id` = 2";
and then you can refer to
然后你可以参考
$row['value'];
回答by Tobb
The easiest way would be through joins:
最简单的方法是通过联接：
select *
from User u join Status s on u.status_id = s.id;
(if you dont want the status-id at all, you can specify the columns that you do want in the select-clause.)
（如果您根本不需要 status-id，您可以在 select-clause 中指定您需要的列。）
回答by thatidiotguy
Your users table does not have the value of approved in it. It is in your statuses table. When you request status_id you are going to get that value back from that query. You have to do a JOIN ON status_idto make this work out I think. Or do a second query.
您的用户表中没有批准的值。它在您的状态表中。当您请求 status_id 时，您将从该查询中获取该值。JOIN ON status_id我认为你必须做一个才能完成这项工作。或者进行第二次查询。
