我有两张表如下：
表1：common_member
+-----------+-----------+
| uid | username |
+-----------+-----------+
| 1 | haha |
| 2 | walao |
| 3 | alamak |
| 4 | hero |
| 5 | theone |
| 6 | nobody |
+-----------+-----------+
表2：
labour_slog+--------------+-------------+--------------+-------------+
| uid | slaveid | masterid | bytime |
+--------------+-------------+--------------+-------------+
| 1 | 2 | 3 | 123456 |
| 4 | 5 | 6 | 456789 |
+--------------+-------------+--------------+-------------+
我将数据作为脚本获取，如下所示：
$queryLabourSlog = DB::query("SELECT * FROM ".DB::table('labour_slog')." ORDER BY id desc");
while($rowLabourSlog = DB::fetch($queryLabourSlog)) {
$user_list_slog[] = $rowLabourSlog;
};
array_multisort($idss, SORT_DESC, $user_list_slog);
在我的html中，我使用
<!--{loop $user_list_slog $value}-->{$value[uid]} on {$value[bytime]} forced hire {$value[masterid]}'s employee {$value[slaveid]}.<!--{/loop}-->
html将显示：
1 on 123456 forced hire 3's employee 2.
4 on 456789 forced hire 6's employee 5.
如何加入
Table 1的用户名数据以获得如下的循环显示？haha on 123456 forced hire alamak's employee walao.
hero on 456789 forced hire nobody's employee theone.
谢谢。
最佳答案
SELECT table_slog.*, u1.`name`, u2.`name` FROM `table_slog` LEFT JOIN `common_member` AS u1 ON `table_slog`.`uid`=u1.`uid` LEFT JOIN `common_member` AS u2 ON `table_slog`.`slaveid`=u2.`uid` LEFT JOIN `common_member` AS u3 ON `table_slog`.`masterid`=u3.uid ORDER BY `id` DESC
关于php - MySQL表2中的数据，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/33610222/