MySQL表2中的数据

Swee Hong 发表于 Dev

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我有2张桌子，如下所示：

表格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      |
+--------------+-------------+--------------+-------------+

我通过以下脚本获取数据：

$queryLabourSlog = DB::query("SELECT * FROM ".DB::table('labour_slog')." ORDER BY id desc");
    while($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的用户名数据以显示如下所示的循环？

haha on 123456 forced hire alamak's employee walao.
hero on 456789 forced hire nobody's employee theone.

谢谢。

斯凡德勒

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

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