在MySQL迁移脚本中,我试图删除表的所有外键,而不知道约束本身的名称。我需要这样做,因为我只能知道特定数据库安装中的约束名称,但是脚本也必须在其他在当前脚本编码时名称未知的安装中工作。
这是我对存储过程的第一个猜测,但它不起作用。尤其是抱怨“?”在
DELIMITER $$
DROP PROCEDURE IF EXISTS drop_foreign_key_documents $$
CREATE PROCEDURE drop_foreign_key_documents ( )
BEGIN
WHILE (SELECT COUNT(*) AS index_exists FROM `information_schema`.`TABLE_CONSTRAINTS` c WHERE `c`.`CONSTRAINT_TYPE` LIKE '%FOREIGN%' AND `c`.`TABLE_NAME`='documents' and `c`.`TABLE_SCHEMA`='mydb') > 0 DO
SET @ctype = '%FOREIGN%';
SET @tname = 'documents';
SET @dbname = 'mydb';
SET @n = 'select `CONSTRAINT_NAME` INTO @cname FROM `information_schema`.`TABLE_CONSTRAINTS` c WHERE `c`.`CONSTRAINT_TYPE` LIKE ? AND `c`.`TABLE_NAME`=? and `c`.`TABLE_SCHEMA`=? LIMIT 0,1';
PREPARE stmt FROM @n;
EXECUTE stmt USING @ctype,@tname,@dbname;
SELECT @cname;
SET @s = 'ALTER TABLE `documents` DROP FOREIGN KEY ?';
PREPARE stmtd FROM @s;
EXECUTE stmtd USING @cname;
END WHILE;
END $$
DELIMITER ;
CALL drop_foreign_key_documents;MySQL的输出是:
@cname
documents_ibfk_13
ERROR 1064 (42000) at line 23: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '?' at line 1现在我可以想象它为什么会抱怨: ALTER语句不能使用'?‘因为约束名称不是WHERE子句的一部分,所以不能用位置参数替换它。我不知道的是如何构建一个参数化的ALTER语句。
发布于 2013-08-14 09:10:36
显而易见的答案:
SET @s = CONCAT('ALTER TABLE `documents` DROP FOREIGN KEY ', @cname);
PREPARE stmtd FROM @s;
EXECUTE stmtd;https://stackoverflow.com/questions/18227275
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