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

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;

其他回答

我同意列出所有列的“简单”解决方案,但这可能会造成负担,而且打字错误可能会浪费大量时间。我使用“getTableColumns”函数检索适合粘贴到查询中的列的名称。然后我要做的就是删除我不想要的。

CREATE FUNCTION `getTableColumns`(tablename varchar(100)) 
          RETURNS varchar(5000) CHARSET latin1
BEGIN
  DECLARE done INT DEFAULT 0;
  DECLARE res  VARCHAR(5000) DEFAULT "";

  DECLARE col  VARCHAR(200);
  DECLARE cur1 CURSOR FOR 
    select COLUMN_NAME from information_schema.columns 
    where TABLE_NAME=@table AND TABLE_SCHEMA="yourdatabase" ORDER BY ORDINAL_POSITION;
  DECLARE CONTINUE HANDLER FOR NOT FOUND SET done = 1;
  OPEN cur1;
  REPEAT
       FETCH cur1 INTO col;
       IF NOT done THEN 
          set res = CONCAT(res,IF(LENGTH(res)>0,",",""),col);
       END IF;
    UNTIL done END REPEAT;
  CLOSE cur1;
  RETURN res;

您的结果返回一个以逗号分隔的字符串,例如…

col1, col2 col3 col4, ... col53

如果它总是相同的一列,那么你可以创建一个不包含它的视图。

否则,不,我不这么认为。

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。

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