0

当我在 phpmyadmin 中运行此查询时:

SELECT * FROM events e LEFT JOIN venues v ON e.vid = v.vid

什么都没发生。没有错误或任何东西,同样的屏幕只是返回给我。

但是当我运行这个时:

SELECT * FROM events LEFT JOIN venues ON events.vid = venues.vid

它工作得很好。我错过了什么吗?

表结构:https ://dl.dropbox.com/u/28104350/help.png

4

2 回答 2

0

试试这个:放"AS"

SELECT * FROM events AS e LEFT JOIN venues AS v ON e.vid = v.vid
于 2012-10-11T09:23:03.310 回答
-1

你应该使用下面的别名

SELECT * FROM events as  e LEFT JOIN venues as  v ON e.vid = v.vid
于 2012-10-13T08:12:04.930 回答