当我在phpmyadmin中运行此查询时:
SELECT * FROM events e LEFT JOIN venues v ON e.vid = v.vid
没有任何反应。没有任何错误或任何内容,只需将相同的屏幕返回给我。
但是当我运行时:
SELECT * FROM events LEFT JOIN venues ON events.vid = venues.vid
它运作得很好。我错过了什么吗?
答案 0 :(得分:0)
试试这个:
放"AS"
SELECT * FROM events AS e LEFT JOIN venues AS v ON e.vid = v.vid
答案 1 :(得分:-1)
您应该使用 别名 ,如下所示
SELECT * FROM events as e LEFT JOIN venues as v ON e.vid = v.vid