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

如果你使用MySQL工作台,你可以右键单击你的表,点击发送到sql编辑器,然后选择所有语句,这将创建一个所有字段都被列出的语句,如下所示:

SELECT `purchase_history`.`id`,
    `purchase_history`.`user_id`,
    `purchase_history`.`deleted_at`
FROM `fs_normal_run_2`.`purchase_history`;
SELECT * FROM fs_normal_run_2.purchase_history;

现在你可以删除那些你不想要的。

其他回答

即使要查询所有列,也最好指定要查询的列。

因此,我建议您在语句中写下每一列的名称(不包括您不想要的列)。

SELECT
    col1
    , col2
    , col3
    , col..
    , col53

FROM table

据我所知,没有。你可以这样做:

SELECT col1, col2, col3, col4 FROM tbl

并手动选择所需的列。然而,如果你想要很多列,那么你可能只需要做一个:

SELECT * FROM tbl 

忽略你不想要的。

针对你的特殊情况,我建议:

SELECT * FROM tbl

除非你只想要几列。如果你只想要四列,那么:

SELECT col3, col6, col45, col 52 FROM tbl

这很好,但如果您想要50个列,那么任何使查询变得(太?)难以阅读的代码。

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'

是的,尽管根据表的不同,I/O可能会很高,但我找到了一个解决方案。

SELECT *
INTO #temp
FROM table

ALTER TABLE #temp DROP COlUMN column_name

SELECT *
FROM #temp

你可以:

SELECT column1, column2, column4 FROM table WHERE whatever

没有得到列3,尽管您可能在寻找一个更一般的解?