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。

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。

我有一个建议，但不是解决办法。 如果您的一些列有较大的数据集，那么您应该尝试使用以下方法

SELECT *, LEFT(col1, 0) AS col1, LEFT(col2, 0) as col2 FROM table

我想添加另一个观点来解决这个问题，特别是如果你有少量的列要删除。

您可以使用像MySQL Workbench这样的DB工具来为您生成选择语句，因此您只需手动删除生成语句的那些列，并将其复制到SQL脚本中。

在MySQL Workbench中，生成它的方法是:

右键单击表->发送到Sql编辑器->选择所有语句。

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'

只做

SELECT * FROM table WHERE whatever

然后用您最喜欢的编程语言php删除该列

while (($data = mysql_fetch_array($result, MYSQL_ASSOC)) !== FALSE) {
   unset($data["id"]);
   foreach ($data as $k => $v) { 
      echo"$v,";
   }      
}