条款中的未知栏目

Question

我有一个简单的查询:

SELECT u_name AS user_name FROM users WHERE user_name = "john";

我得到Unknown Column 'user_name' in where clause.'user_name'即使之后,我还不能在声明的其他部分提及select 'u_name as user_name'吗？

Answer 1

SQL从右到左向后评估.因此,在select子句之前解析和评估where子句.因此,尚未发生u_name到user_name的别名.

Answer 2

关于什么:

SELECT u_name AS user_name FROM users HAVING user_name = "john";

Answer 3

请参阅以下MySQL手册页:http://dev.mysql.com/doc/refman/5.0/en/select.html

"可以使用AS alias_name为select_expr指定别名.别名用作表达式的列名,可以在GROUP BY,ORDER BY或HAVING子句中使用."

(......)

不允许在WHERE子句中引用列别名,因为在执行WHERE子句时可能尚未确定列值.请参见第B.5.4.4节"列别名的问题".

Answer 4

select u_name as user_name from users where u_name = "john";

可以这样想,你的where子句首先计算,以确定需要返回哪些行(或连接的行).执行where子句后,将为其运行select子句.

为了更好的方式,想象一下:

select distinct(u_name) as user_name from users where u_name = "john";

没有第二个,你不能参考前半部分.总是首先评估的地方,然后是select子句.

Answer 5

如果您尝试执行如下所示的查询(查找具有至少一个附件的所有节点),您使用SELECT语句创建一个实际上不存在于数据库中的新字段,并尝试使用该结果的别名你将遇到同样的问题:

SELECT nodes.*, (SELECT (COUNT(*) FROM attachments 
WHERE attachments.nodeid = nodes.id) AS attachmentcount 
FROM nodes
WHERE attachmentcount > 0;

您将在WHERE子句中收到错误"未知列'attachmentcount'".

解决方案实际上非常简单 - 只需用产生别名的语句替换别名,例如:

SELECT nodes.*, (SELECT (COUNT(*) FROM attachments 
WHERE attachments.nodeid = nodes.id) AS attachmentcount 
FROM nodes 
WHERE (SELECT (COUNT(*) FROM attachments WHERE attachments.nodeid = nodes.id) > 0;

您仍然会返回别名,但现在SQL不应该使用未知别名.

Answer 6

您必须使用该条款alias的WHERE条款不欢迎您定义的内容HAVING

SELECT u_name AS user_name FROM users HAVING user_name = "john";

或者您可以直接使用原始列名称 WHERE

SELECT u_name AS user_name FROM users WHERE u_name = "john";

与由于子查询或任何计算而导致用户定义别名的结果相同,它将由HAVING子句访问,而不是由子句访问WHERE

SELECT u_name AS user_name ,
(SELECT last_name FROM users2 WHERE id=users.id) as user_last_name
FROM users  WHERE u_name = "john" HAVING user_last_name ='smith'

Answer 7

更正:

SELECT u_name AS user_name FROM users WHERE u_name = 'john';

Answer 8

或者:

SELECT u_name AS user_name
FROM   users
WHERE  u_name = "john";

要么:

SELECT user_name
from
(
SELECT u_name AS user_name
FROM   users
)
WHERE  u_name = "john";

如果RDBMS支持谓词推入内联视图,后者应该与前者相同.

Answer 9

不需要选择正确的名称.如果你给了表,你可以从别名中选择,但是你可以使用它.

Answer 10

你不能。user_name 在返回之前不存在。