创建/使用带有laravel迁移的存储过程

e5nqia27  于 2021-06-18  发布在  Mysql
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这个有点松鼠!
我构建了一个工作正常的存储过程。但是,当我将其放入laravel(雄辩的)迁移时,迁移会生成存储过程,但当调用它时,会给出错误1292(截断的不正确的双精度值)。在mysql workbench中,我右键单击迁移生成的存储过程,复制create语句,删除存储过程,然后从它自己的create语句中重建它。。。。还有。。。。。一切正常!
我对任何猜测都持开放态度。很明显,我希望这个存储过程是迁移的一部分,而不是需要手动构建。提前感谢:

DELIMITER $$

CREATE PROCEDURE sp_cancelTasksOnApplicationCancel(IN loanAppId INT(20))

BEGIN

            SELECT loanAppId;

            #SET @taskids := 0;

            update tasks set task_status = 26, task_status_note = 'Application Cancelled or Denied' 

where loan_app_id = loanAppId AND task_status IN (22, 23, 24) AND ( SELECT @taskids := CONCAT_WS(',', id, @taskids) );        

# SELECT @taskids;

SET @cleanids = SUBSTRING(@taskids, 1, CHAR_LENGTH(@taskids) -1);

# SELECT @cleanids;

            SET @pos = 0;

            SET @len = 0;

            WHILE LOCATE(',', @cleanids, @pos+1) > 0 DO

                            SET @len = LOCATE(',', @cleanids,@pos+1) - @pos;

                            SET @id = SUBSTRING(@cleanids, @pos, @len);

                                            #select (@id);

                                            IF (@id <> 0) THEN

                                            insert into status_historys 

                                                            (loan_app_id, `type`, type_recid, type_keyfield,type_status_field, type_status_date_field, type_status_note_field, type_status_userid_field, `status`, status_date, status_note, status_userid, created_at, updated_at) 

            values(loanAppId, 'tasks', @id, 'id', 'task status', 'tasks status datetime', 'tasks status note', 'tasks status userid', 26, now(), 'Application Cancelled or Denied', 1, now(), now()); 

                                            END IF;

                            SET @pos = LOCATE(',', @taskids, @pos+@len) + 1;

            END WHILE;

END $$

DELIMITER ;

############## 更新

问题在于以下两行。即使在命令行中,它也会对不正确的双精度值发出警告。

SET @taskids := '0';

update tasks set task_status = 22, task_status_note = 'Just Testing' 
where loan_app_id = loanAppId AND task_status IN (22, 23, 24) AND ( SELECT 
@taskids := CONCAT_WS(',', cast(id as char), @taskids) );
yftpprvb

yftpprvb1#

确定-主要问题是updatewhere子句中的select关键字。我相信它一直在重新声明@taskids用户变量,如果存储过程是从工作台生成的,mysql会发出警告,但是如果存储过程是通过迁移生成的,mysql就会失败。
所以正确的代码是


# snip#

# Note string '0' rather than 0

SET @taskids := '0';

# note casting of id and removing the select keyword before the first @taskids

update tasks set task_status = 22, task_status_note = 'Just Testing' 
where loan_app_id = loanAppId AND task_status IN (22, 23, 24) AND (  
@taskids := CONCAT_WS(',', cast(id as char), @taskids) );

# snip#

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