我有第一张表已注册，此表包含：
uid (int)
email (varchar)
telephone (varchar)
看起来像这样
++++++++++++++++++++++++++++++++++++++++++++++++
+ uid + email + telephone +
++++++++++++++++++++++++++++++++++++++++++++++++
+ 1 + david@gmail.com + 333-111-5555 +
++++++++++++++++++++++++++++++++++++++++++++++++
+ 2 + erick@gmail.com + 333-222-5555 +
++++++++++++++++++++++++++++++++++++++++++++++++
+ 3 + james@gmail.com + 333-333-5555 +
++++++++++++++++++++++++++++++++++++++++++++++++
+ 4 + svene@gmail.com + 333-444-5555 +
++++++++++++++++++++++++++++++++++++++++++++++++
+ 5 + simon@gmail.com + 333-555-5555 +
++++++++++++++++++++++++++++++++++++++++++++++++
+ 6 + peter@gmail.com + 333-666-5555 +
++++++++++++++++++++++++++++++++++++++++++++++++
+ 7 + mikka@gmail.com + 333-777-5555 +
++++++++++++++++++++++++++++++++++++++++++++++++
我有第二个表引荐，此表包含以下列：
id (int)
uid0 (int)
uid1 (int)
第二个表保存id为autoincrement，然后uid0为父用户id，uid1为referer的用户id
它可以包含这些值，例如
+++++++++++++++++++++++
+ id + uid0 + uid1 +
+++++++++++++++++++++++
+ 1 + 2 + 3 +
+++++++++++++++++++++++
+ 2 + 2 + 5 +
+++++++++++++++++++++++
我知道如何在foreach循环中回显来自已注册表的所有已注册的电子邮件和电话。
但是，我需要加入并回复所有与特定用户（uid0）相关的推荐人（uid1）电子邮件。
+++++++++++++++++++++++
+ id + uid0 + uid1 +
+++++++++++++++++++++++
+ 1 + 2 + 3 +
+++++++++++++++++++++++
+ 2 + 2 + 5 +
+++++++++++++++++++++++
+ 3 + 4 + 1 +
+++++++++++++++++++++++
+ 4 + 4 + 6 +
+++++++++++++++++++++++
+ 5 + 4 + 7 +
+++++++++++++++++++++++
因此，最终的回声结果将如下所示：
email telephone referers
erick@gmail.com 333-222-5555 james@gmail.com, simon@gmail.com
svene@gmail.com 333-444-5555 david@gmail.com, peter@gmail.com, mikka@gmail.com
我怎样才能做这个“线程化”的foreach循环？
提前致谢。
答案 0 :(得分：2)
您可以尝试使用group concat;
进行子选择类似的东西：
select r.email, r.telephone from registered as r,
(SELECT
GROUP_CONCAT(r.email)
FROM registered
GROUP uid1 where uid0 = r.uid)