我试图使用一个选择语句从某个MySQL表中获得除一个以外的所有列。有什么简单的方法吗?

编辑:在这个表格中有53列(不是我的设计)


当前回答

(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

其他回答

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

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

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'

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。

我同意@Mahomedalid的回答,但我不想做一些准备好的语句,我不想输入所有的字段,所以我有一个愚蠢的解决方案。

去phpmyadmin->sql->select表,它转储查询:复制,替换和完成!:)