Question

我在某些邮件作业上运行php artisan queue:work --tries=3，但在日志文件中却不断收到此错误：

[2018-11-02 03:22:02] local.ERROR: SQLSTATE[22003]: Numeric value out of range: 1264 Out of range value for column 'attempts' at row 1 (SQL: update职位set保留{at = 1541128922,尝试= 256 where id = 767) {"exception":"[object] (Illuminate\\Database\\QueryException(code: 22003): SQLSTATE[22003]: Numeric value out of range: 1264 Out of range value for column 'attempts' at row 1 (SQL: update职位set保留职位= 1541128922,尝试= 256 where id = 767) at /var/www/html/project/vendor/laravel/framework/src/Illuminate/Database/Connection.php:664, Doctrine\\DBAL\\Driver\\PDOException(code: 22003): SQLSTATE[22003]: Numeric value out of range: 1264 Out of range value for column 'attempts' at row 1 at /var/www/html/project/vendor/doctrine/dbal/lib/Doctrine/DBAL/Driver/PDOStatement.php:144, PDOException(code: 22003): SQLSTATE[22003]: Numeric value out of range: 1264 Out of range value for column 'attempts' at row 1 at /var/www/html/project/vendor/doctrine/dbal/lib/Doctrine/DBAL/Driver/PDOStatement.php:142) [stacktrace]

该作业实际上是在作业表中创建的，但尚未处理。我该如何解决这个问题？

Answer 1

请增加attempts字段数据值

Answer 2

SQLSTATE [22003]数字值超出范围：1264行1的列“尝试”超出范围值

IT看来您的attemets字段设置为TINYINT的最大值255，因为这一点我可以知道。

 SQL: updatejobssetreserved_at= 1541128922,attempts= 256 whereid= 767

并且因为我知道TinyInt可以容纳255（因为我只是用Google搜索它了..哈哈）。它实际上可能定义为INT(1) UNSIGEND

所以您想将其更改为> TINYINT就像一个NOTSOTINYINT .... :)我的意思是SMALLINT

我被骗了

https://dev.mysql.com/doc/refman/5.5/en/integer-types.html

队列未处理

2 个答案: