我已阅读弹性beanstalk定期任务手册here, 并在cron.yaml文件中查看此代码之后:
def dir_files_map(start_dir):
import os
dd = {}
# create dictionary where the key is folder root path
# and the values are the files in that folder
# filter files based on endswiths(string) clause
for root, subfolders, filenames in os.walk(start_dir):
for f in filenames:
if f.endswith('.b'):
dd.setdefault(root,[]).append(f)
return dd
我不确定如何执行以下命令:
version: 1
cron:
- name: "backup-job" # required - unique across all entries in this file
url: "/backup" # required - does not need to be unique
schedule: "0 */12 * * *" # required - does not need to be unique
网址应该是" /var/www/html/myfolder/task.php"或" php /var/www/html/myfolder/task.php"?
答案 0 :(得分:0)
url
值是执行任务时工作者实例上的SQS守护程序将调用的值。在您的情况下,守护程序将连接到localhost
并发出以下请求:
POST /backup HTTP/1.1
Host: localhost
User-Agent: aws-sqsd/1.1
Content-Type: application/json
...
如here所述。
换句话说,您的脚本应该在执行时运行:
curl -X POST http://localhost/backup
来自工人实例的。