为什么不能在存储过程中使用变量作为表名? [英] Why can't I use a variable as the table name in a stored procedure?

查看:361
本文介绍了为什么不能在存储过程中使用变量作为表名?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我试图在存储过程中将变量用作表名,并将其用作字符串文字而不是实际的表名.为什么是这样?除了在PHP中执行此操作外,还有其他方法可以做到这一点吗?

I'm trying to use a variable as the table name within a stored procedure and it's using it as a string literal instead of as the actual table name. Why is this? Is there another way I should be doing this (aside from just doing it in PHP)?

DROP PROCEDURE IF EXISTS settonull;

DELIMITER //

CREATE PROCEDURE settonull()
BEGIN
  DECLARE done INT DEFAULT FALSE;
  DECLARE _tablename VARCHAR(255);
  DECLARE _columnname VARCHAR(255);
  DECLARE cur1 CURSOR FOR SELECT CONCAT(TABLE_SCHEMA, '.', TABLE_NAME) AS table_name, COLUMN_NAME AS column_name FROM information_schema.COLUMNS WHERE IS_NULLABLE = 'YES' AND TABLE_SCHEMA = 'blip_notify' AND table_name = 'notify_queue' LIMIT 1;

  DECLARE CONTINUE HANDLER FOR NOT FOUND SET done = TRUE;

  OPEN cur1;

  read_loop: LOOP
    FETCH cur1 INTO _tablename, _columnname;

    IF done THEN
      LEAVE read_loop;
    END IF;

    UPDATE _tablename SET _columnname = NULL WHERE LENGTH(TRIM(_columnname)) = 0;

  END LOOP;

  CLOSE cur1;
END//

DELIMITER ;

CALL settonull();

输出:

0 row(s) affected, 1 warning(s)

Execution Time : 0 sec
Transfer Time  : 1.094 sec
Total Time     : 1.095 sec

Note Code : 1305
PROCEDURE settonull does not exist
---------------------------------------------------

0 row(s) affected

Execution Time : 0.002 sec
Transfer Time  : 1.011 sec
Total Time     : 1.014 sec
---------------------------------------------------

Query: call settonull()

Error Code: 1146
Table 'blip_notify._tablename' doesn't exist

Execution Time : 0 sec
Transfer Time  : 0 sec
Total Time     : 0.003 sec
---------------------------------------------------

推荐答案

您需要使用动态sql.是的,嗯.

You need to use dynamic sql. ya, ugh.

SET @s = CONCAT('UPDATE ', _tablename, ' SET ', _columnname, ' = NULL WHERE LENGTH(TRIM(', _columnname, ')) = 0' );
PREPARE stmt FROM @s;
EXECUTE stmt;

这篇关于为什么不能在存储过程中使用变量作为表名?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆