在尝试@Mahomedalid和@Junaid的解决方案时，我发现了一个问题。所以我想分享一下。如果列名中有空格或连字符(如check-in)，则查询将失败。简单的解决方法是在列名周围使用反标记。修改后的查询如下

SET @SQL = CONCAT('SELECT ', (SELECT GROUP_CONCAT(CONCAT("`", COLUMN_NAME, "`")) FROM
INFORMATION_SCHEMA.COLUMNS WHERE TABLE_NAME = 'users' AND COLUMN_NAME NOT IN ('id')), ' FROM users');
PREPARE stmt1 FROM @SQL;
EXECUTE stmt1;

(不要在大桌子上尝试，结果可能是……令人惊讶的!)

临时表

DROP TABLE IF EXISTS temp_tb;
CREATE TEMPORARY TABLE ENGINE=MEMORY temp_tb SELECT * FROM orig_tb;
ALTER TABLE temp_tb DROP col_a, DROP col_f,DROP col_z;    #// MySQL
SELECT * FROM temp_tb;

DROP语法可能因数据库而异

我同意只选择*是不够的，如果你不需要，正如在其他地方提到的，是一个BLOB，你不希望有这个开销。

我会用所需的数据创建一个视图，然后您可以轻松地选择*——如果数据库软件支持它们的话。否则，将大量数据放到另一个表中。

也许我有一个解决Jan Koritak指出的矛盾的方法

SELECT CONCAT('SELECT ',
( SELECT GROUP_CONCAT(t.col)
FROM
(
    SELECT CASE
    WHEN COLUMN_NAME = 'eid' THEN NULL
    ELSE COLUMN_NAME
    END AS col 
    FROM INFORMATION_SCHEMA.COLUMNS 
    WHERE TABLE_NAME = 'employee' AND TABLE_SCHEMA = 'test'
) t
WHERE t.col IS NOT NULL) ,
' FROM employee' );

表:

SELECT table_name,column_name 
FROM INFORMATION_SCHEMA.COLUMNS 
WHERE TABLE_NAME = 'employee' AND TABLE_SCHEMA = 'test'

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

table_name  column_name
employee    eid
employee    name_eid
employee    sal

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

查询结果:

'SELECT name_eid,sal FROM employee'

视图在这种情况下工作得更好吗?

CREATE VIEW vwTable
as  
SELECT  
    col1  
    , col2  
    , col3  
    , col..  
    , col53  
FROM table

我的主要问题是在连接表时获得了许多列。虽然这不是您问题的答案(如何从一个表中选择除某些列之外的所有列)，但我认为值得一提的是，您可以指定表。从特定表中获取所有列，而不是仅指定。

下面是一个很有用的例子:

select users.*, phone.meta_value as phone, zipcode.meta_value as zipcode

from users

left join user_meta as phone
on ( (users.user_id = phone.user_id) AND (phone.meta_key = 'phone') )

left join user_meta as zipcode
on ( (users.user_id = zipcode.user_id) AND (zipcode.meta_key = 'zipcode') )

结果是用户表中的所有列，以及从元表中连接的两个附加列。