MySql存储过程... group_concat中的错误

Question

我有以下查询在phpMyAdmin中工作

SELECT CONCAT("'",GROUP_CONCAT( fence_id SEPARATOR "','" ),"'") AS fence_ids 
FROM asset_fence af 
INNER JOIN assets a ON a.vehicle_id = af.vehicle_id 
WHERE a.client_id=1

但是同样的查询在存储过程中给出了错误

错误是：

您的SQL语法有错误;检查与MariaDB服务器版本对应的手册，以便在'@fence_ids = SELECT CONCAT（“'”，GROUP_CONCAT（fence_id SEPARATOR“'，'”），“'”附近使用正确的语法）第8行的AS'

存储过程是

DELIMITER $$

CREATE PROCEDURE `prcDeleteClient`(IN `f_client_id` INT, OUT AROWS INT)
BEGIN


    START TRANSACTION;


        @fence_ids = SELECT CONCAT("'",GROUP_CONCAT( fence_id SEPARATOR "','" ),"'") AS fence_ids FROM asset_fence af INNER JOIN assets a ON a.vehicle_id = af.vehicle_id WHERE a.client_id=f_client_id


        DELETE
            asset_fence,
            geo_fence

        FROM 
            geo_fence gf INNER JOIN asset_fence af ON gf.fence_id = af.fence_id

        WHERE
            af.fence_id IN (@fence_ids)



        DELETE
            client,
            assets,
            asset_movement

        FROM 
            asset_movement am INNER JOIN assets a ON am.vehicle_id = a.vehicle_id
            assets a INNER JOIN client c ON a.client_id = c.client_id

        WHERE
            c.client_id=f_client_id


        SET @AROWS = ROW_COUNT();
        SELECT @AROWS as AROWS;

    COMMIT;
END $$

DELIMITER ;

更新程序是

DELIMITER $$

CREATE PROCEDURE `prcDeleteClient`(IN `f_client_id` INT, OUT AROWS INT)
BEGIN


    START TRANSACTION;


        DELETE
            af,
            gf
        FROM 
            geo_fence gf

            INNER JOIN asset_fence af ON gf.fence_id = af.fence_id
            INNER JOIN assets a ON a.vehicle_id = af.vehicle_id

        WHERE
            a.client_id=f_client_id;



        DELETE
            c,
            a,
            am

        FROM 
            asset_movement am INNER JOIN assets a ON am.vehicle_id = a.vehicle_id
            assets a INNER JOIN clients c ON a.client_id = c.client_id

        WHERE
            c.client_id=f_client_id;


        SELECT ROW_COUNT() AS AROWS;

    COMMIT;
END $$

DELIMITER ;

错误是：MULTI DELETE中的未知表'c'

什么是错误，我该如何纠正？

最好的祝福

Answer 1

首先，分配变量的语法是错误的，它需要在变量名之前使用SET命令。

其次，如果要将SELECT查询的结果用作值，则必须在其周围加上括号：

SET @fence_ids = (SELECT ...);

三，使用时：

WHERE af.fence_id IN (@fence_ids)

它会将@fence_ids视为单个ID，而不是ID列表。所以这相当于：

WHERE af.fence_id = @fence_ids

如果要在逗号分隔列表中搜索某些内容，则需要使用FIND_IN_SET：

WHERE FIND_IN_SET(af.fence_id, @fence_ids)

您也不应该在GROUP_CONCAT()中的值周围添加引号。

但是你首先不应该使用GROUP_CONCAT，你应该加入返回所需ID的查询。

   DELETE
        af,
        gf
    FROM 
        geo_fence gf 
    INNER JOIN asset_fence af ON gf.fence_id = af.fence_id
    INNER JOIN assets a ON a.vehicle_id = af.vehicle_id 
    WHERE a.client_id=f_client_id;

您不需要在两个语句中执行此操作：

SET @AROWS = ROW_COUNT();
SELECT @AROWS as AROWS;

你可以这样做：

SELECT ROW_COUNT() AS AROWS;

Answer 2

您正在设置一个值，因此您不需要列别名...删除theAS fence_ids

    SELECT CONCAT("'", GROUP_CONCAT( af.fence_id SEPARATOR "','" ),"'") 
    FROM asset_fence af 
    INNER JOIN assets a ON a.vehicle_id = af.vehicle_id
    WHERE a.client_id=1

你错过了每次删除的命令终结符

   DELETE
        asset_fence,
        geo_fence

    FROM 
        geo_fence gf INNER JOIN asset_fence af ON gf.fence_id = af.fence_id

    WHERE
        af.fence_id IN (@fence_ids);



    DELETE
        client,
        assets,
        asset_movement

    FROM 
        asset_movement am INNER JOIN assets a ON am.vehicle_id = a.vehicle_id
        assets a INNER JOIN client c ON a.client_id = c.client_id

    WHERE
        c.client_id=f_client_id;