如何为弹性beanstalk工作者定义入站规则,只接受ips成为我的SQS队列

时间:2017-09-18 08:19:29

标签: amazon-web-services amazon-ec2 elastic-beanstalk worker amazon-elastic-beanstalk

在我的工作人员(AWS弹性beanstalk工作者)上,我收到了很多这样的请求:

151.100.143.22 (-) - - [14/Sep/2017:10:11:55 +0000] "HEAD http://34.224.82.145:80/mysql/pMA/ HTTP/1.1" 404 - "-" "Mozilla/5.0 Jorgee"
151.100.143.22 (-) - - [14/Sep/2017:10:11:56 +0000] "HEAD http://34.224.82.145:80/sql/phpmanager/ HTTP/1.1" 404 - "-" "Mozilla/5.0 Jorgee"
151.100.143.22 (-) - - [14/Sep/2017:10:11:56 +0000] "HEAD http://34.224.82.145:80/sql/php-myadmin/ HTTP/1.1" 404 - "-" "Mozilla/5.0 Jorgee"
151.100.143.22 (-) - - [14/Sep/2017:10:11:56 +0000] "HEAD http://34.224.82.145:80/sql/phpmy-admin/ HTTP/1.1" 404 - "-" "Mozilla/5.0 Jorgee"
151.100.143.22 (-) - - [14/Sep/2017:10:11:57 +0000] "HEAD http://34.224.82.145:80/sql/sql/ HTTP/1.1" 404 - "-" "Mozilla/5.0 Jorgee"
151.100.143.22 (-) - - [14/Sep/2017:10:11:57 +0000] "HEAD http://34.224.82.145:80/sql/myadmin/ HTTP/1.1" 404 - "-" "Mozilla/5.0 Jorgee"
151.100.143.22 (-) - - [14/Sep/2017:10:11:57 +0000] "HEAD http://34.224.82.145:80/sql/webadmin/ HTTP/1.1" 404 - "-" "Mozilla/5.0 Jorgee"
151.100.143.22 (-) - - [14/Sep/2017:10:11:58 +0000] "HEAD http://34.224.82.145:80/sql/sqlweb/ HTTP/1.1" 404 - "-" "Mozilla/5.0 Jorgee"
151.100.143.22 (-) - - [14/Sep/2017:10:11:58 +0000] "HEAD http://34.224.82.145:80/sql/websql/ HTTP/1.1" 404 - "-" "Mozilla/5.0 Jorgee"
151.100.143.22 (-) - - [14/Sep/2017:10:11:58 +0000] "HEAD http://34.224.82.145:80/sql/webdb/ HTTP/1.1" 404 - "-" "Mozilla/5.0 Jorgee"
151.100.143.22 (-) - - [14/Sep/2017:10:11:59 +0000] "HEAD http://34.224.82.145:80/sql/sqladmin/ HTTP/1.1" 404 - "-" "Mozilla/5.0 Jorgee"
151.100.143.22 (-) - - [14/Sep/2017:10:11:59 +0000] "HEAD http://34.224.82.145:80/sql/sql-admin/ HTTP/1.1" 404 - "-" "Mozilla/5.0 Jorgee"

我的服务器停止工作(我需要重新启动)所以,我添加了入站规则:HTTP TCP 80 127.0.0.0/8它的工作原理除了现在,我的工作人员没有成功监听/写入数据来到我的默认SQS队列

这是我向你展示的最后一个access_log,cronjob工作但不是主队列(队列/收到):

127.0.0.1 (-) - - [18/Sep/2017:16:31:00 +0000] "POST /workers/cron/products/getProductForId HTTP/1.1" 200 41 "-" "aws-sqsd/2.3"
127.0.0.1 (-) - - [18/Sep/2017:16:31:00 +0000] "POST /workers/cron/999 HTTP/1.1" 200 30 "-" "aws-sqsd/2.3"
127.0.0.1 (-) - - [18/Sep/2017:16:31:27 +0000] "POST /queue/received HTTP/1.1" 500 495 "-" "aws-sqsd/2.3"
127.0.0.1 (-) - - [18/Sep/2017:16:31:59 +0000] "POST /workers/cron/999 HTTP/1.1" 200 30 "-" "aws-sqsd/2.3"
127.0.0.1 (-) - - [18/Sep/2017:16:32:00 +0000] "POST /workers/cron/products/getProductForId HTTP/1.1" 200 41 "-" "aws-sqsd/2.3"

您知道吗,我如何为弹性beanstalk工作者定义入站规则,只接受ips成为我的SQS队列

由于

0 个答案:

没有答案