我尝试从以下查询中加入另一个表:
SELECT * FROM users u,
(SELECT submissions_comments.parent_id from submissions_comments where id = 96) x
WHERE u.id = x.parent_id\G
*************************** 1. row ***************************
id: 17
email: bobcobb@gmail.com
hash: $2a$10$m9egfe.NtKC4QbR869el9.t.w0E7jN2Omk6AXQCbF/ERyfseJVx6S
salt: $2a$10$m9egfe.NtKC4QbR869el9.
username: test
name:
role: normal
about:
created: 2014-01-17 00:53:52
last_login: 2014-01-17 08:53:52
created_ip: 1270
last_login_ip: 1270
remember_me: 1
photo:
confirmed: 0
confirm_code: 814eaadfb411fb1a982c4b4b79b91cc37860c672
public_profile: 1
parent_id: 17
1 row in set (0.00 sec)
基本上我想加入上面的user.id
(在这种情况下是17)到另一个表。下面的查询为我提供了我想要添加到上述查询的内容:
SELECT email_reply_on_comment
FROM email_settings
WHERE id = 17;
+------------------------+
| email_reply_on_comment |
+------------------------+
| 0 |
+------------------------+
我正在尝试各种变体:
SELECT * FROM users u,
(SELECT submissions_comments.parent_id from submissions_comments where id = 96) x,
email_settings.reply_on_comment
WHERE u.id = x.parent_id
LEFT JOIN email_settings
ON email_settings.user_id = x.parent_id;
但没有到达任何地方。
答案 0 :(得分:0)
尝试以下加入:
SELECT u.*, es.email_reply_on_comment
FROM users u
INNER JOIN
(SELECT submissions_comments.parent_id from submissions_comments where id = 96) x
ON u.id = x.parent_id
LEFT JOIN email_settings es
ON es.id = u.id;
<强>参考强>: