Question

我想转置一行表格。

例如，如果我有一个“人物”表，其中“ID”，“名字”，“姓氏”，“年龄”作为属性，我希望在“人物”表中转换一行以下两个列：

Column_name，Column_value

我可以使用以下方式获取表的列名：

SELECT *
FROM `INFORMATION_SCHEMA`.`COLUMNS` 
WHERE `TABLE_SCHEMA`='databasename'
AND `TABLE_NAME`='tablename';

我尝试使用以下方法获取列名的值：

select attributes.`COLUMN_NAME`, person.attributes.`COLUMN_NAME` as `Column_Value`
from (select * from Persons where ID=1) as person,
(SELECT * FROM `INFORMATION_SCHEMA`.`COLUMNS` 
WHERE `TABLE_SCHEMA`='databasename' and `TABLE_NAME`='tablename');

但第二个参数也是给出了列名而不是值。

如何解决此问题。

Answer 1

您可以使用带有预准备语句的存储过程

DROP PROCEDURE IF EXISTS proc;
DELIMITER $$
CREATE PROCEDURE proc()
BEGIN
  DECLARE query longtext;
  SELECT
    GROUP_CONCAT( CONCAT("(select '",column_name,"' as     col_name,",concat("person.",column_name)," as col_val
    FROM (select * from test.game_action where id=1) as person,(SELECT * FROM `INFORMATION_SCHEMA`.`COLUMNS` 
    WHERE `TABLE_SCHEMA`='test' and `TABLE_NAME`='game_action')as t)")  SEPARATOR ' union ') into query
  FROM (select * from test.game_action where id=1) as person,(SELECT * FROM `INFORMATION_SCHEMA`.`COLUMNS` 
  WHERE `TABLE_SCHEMA`='test' and `TABLE_NAME`='game_action')as t;

 SET @s = query;
 PREPARE stmt FROM @s;
 EXECUTE stmt;
 DEALLOCATE PREPARE stmt;
END

如果需要，可以将id作为参数传递

获取列mysql的值，列名为变量

1 个答案: