#1064-创建视图时，SQL语法有错误 [英] #1064 - You have an error in your SQL syntax when creating a View

问题描述

有人可以告诉我这段代码出了什么问题吗?

Could someone please tell me what's wrong with this code that's making it spit back an error?

我的代码如下:

CREATE OR REPLACE VIEW vw_training AS 
SELECT training.train_attended, clients.client_firstname, clients.client_lastname, clients.client_swn, clients.client_id, locations.loc_id, locations.loc_title, locationsp.loc_id, locationsp.loc_title, 
FROM training 
JOIN clients ON clients.client_id = training.train_clientid
JOIN locations AS locationsp ON locations.loc_id = training.train_pickup
LEFT JOIN locations ON locations.loc_id = clients.client_winz

这是我回来的错误:

#1064-您的SQL语法有错误；检查手册 对应于您的MySQL服务器版本以使用正确的语法 在'FROM培训附近，在客户端上加入客户端.client_id= training.train_clientid加入第3行

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FROM training JOIN clients ON clients.client_id = training.train_clientid JOIN' at line 3

我正在运行phpmyadmin版本信息:3.5.2.2

I'm running phpmyadmin Version information: 3.5.2.2

我之前将此脚本使用了不同的值

I've used this script with different values before with no issues

推荐答案

在FROM子句之前您还有一个逗号结尾

You have an extra trailing comma before the FROM clause

SELECT ....,
       locationsp.loc_id, 
       locationsp.loc_title, -- <<== remove this trailing comma
FROM   training ...

和另一个将引发此消息的错误:Unknown column 'locations.loc_id' in 'on clause'是使用tablename而不是提供的别名.应该是

and another error that will raise this message: Unknown column 'locations.loc_id' in 'on clause' is the use of tablename and not the alias supplied. it should be,

JOIN locations AS locationsp ON locationsp.loc_id = training.train_pickup
                                    ^^ should use alias here

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