MySQL如何在存储过程中的PREPARE和EXECUTE之后获得结果？

Question

我目前的代码是：

DELIMITER \\

CREATE PROCEDURE sample (IN _car VARCHAR(15))
    BEGIN
        DECLARE _a INTEGER;
        SET @s = CONCAT('SELECT COUNT(*) FROM train WHERE ', _car, '<=0;');
        PREPARE stmt1 FROM @s;
        EXECUTE stmt1;
        DEALLOCATE PREPARE stmt1;

END\\

但我想把SELECT语句的答案捕获到我的_a变量。

我尝试将代码更改为

SET @s = CONCAT('SELECT COUNT(*) INTO', _a,' FROM train WHERE ', _car, '<=0;');

但那没用。

请帮帮忙？

解决！

DELIMITER \\

CREATE PROCEDURE sample (IN _car VARCHAR(15))
    BEGIN
        DECLARE _a INTEGER;
        SET @var = NULL;
        SET @s = CONCAT('SELECT COUNT(*) INTO @var FROM train WHERE ', _car, '<=0;');
        PREPARE stmt1 FROM @s;
        EXECUTE stmt1;
        SELECT @var;
        DEALLOCATE PREPARE stmt1;
END\\

：d

Answer 1

如上所述here，您需要在原始语句声明中包含变量赋值。所以你的陈述是这样的：

SELECT COUNT(*) FROM train WHERE ?<=0 INTO _a

然后你将用：

执行它

EXECUTE stmt1 using _car;

获得结果：

select _a;

让我知道它是否有效。