如何从准备好的语句中获取标量结果? [英] How to get scalar result from prepared statement?
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问题描述
是否可以将准备好的语句的结果设置为变量?我正在尝试创建以下存储过程,但是失败了:
Is it possible to set the result from a prepared statement into a variable? I am trying to create the following stored procedure but it is failing:
第31行的错误1064(42000):您的SQL语法有错误;查看与您的MySQL服务器版本相对应的手册以获取正确的语法,以在'stmt USING @ m,@ c,@ a;
ERROR 1064 (42000) at line 31: 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 'stmt USING @m, @c, @a;
DROP PROCEDURE IF EXISTS deleteAction;
DELIMITER $$
CREATE PROCEDURE deleteAction(
IN modul CHAR(64),
IN controller CHAR(64),
IN actn CHAR(64))
MODIFIES SQL DATA
BEGIN
PREPARE stmt FROM 'SELECT id
FROM actions
WHERE `module` = ?
AND `controller` = ?
AND `action` = ?';
SET @m = modul;
SET @c = controller;
SET @a = actn;
SET @i = EXECUTE stmt USING @m, @c, @a;
DEALLOCATE PREPARE stmt;
DELETE FROM acl WHERE action_id = @i;
DELETE FROM actions WHERE id = @i;
END
$$
DELIMITER ;
推荐答案
这似乎很奇怪,但是您可以在准备好的语句字符串中直接分配变量:
It may seem strange, but you can assign the variable directly in the prepared statement string:
PREPARE stmt FROM 'SELECT @i := id FROM ...';
-- ...
EXECUTE stmt USING @m, @c, @a;
-- @i will hold the id returned from your query.
测试用例:
CREATE TABLE actions (id int, a int);
INSERT INTO actions VALUES (1, 100);
INSERT INTO actions VALUES (2, 200);
INSERT INTO actions VALUES (3, 300);
INSERT INTO actions VALUES (4, 400);
INSERT INTO actions VALUES (5, 500);
DELIMITER $$
CREATE PROCEDURE myProc(
IN p int
)
MODIFIES SQL DATA
BEGIN
PREPARE stmt FROM 'SELECT @i := id FROM actions WHERE `a` = ?';
SET @a = p;
EXECUTE stmt USING @a;
SELECT @i AS result;
DEALLOCATE PREPARE stmt;
END
$$
DELIMITER ;
结果:
CALL myProc(400);
+---------+
| result |
+---------+
| 4 |
+---------+
1 row in set (0.00 sec)
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