unix - 监控无法启动进程

Question

我有一个简单的脚本，试图让 Monit 进行监控。后 some digging around我发现了这个小金块:start program = "su - myuser -c '/home/user/myscript.rb start' "我认为这应该可行，但查看日志文件它说:

[PDT Oct 30 02:47:17] info     : 'simple_script' start: su
[PDT Oct 30 02:47:17] error    : Error: Could not execute su

同样，较早的尝试似乎只读取空格前的部分，因此:

start program = "/home/user/.rvm/rubies/ruby-1.9.2-p290/bin/ruby /home/user/simple_script_daemon.rb stop"

结果是…

[PDT Oct 30 03:09:49] info     : 'simple_script' start: /home/user/.rvm/rubies/ruby-1.9.2-p290/bin/ruby

这不会像第一个例子那样失败，但似乎仍然只执行空格之前的部分。

这是我的完整声明:

check process simple_script
    with pidfile /home/user/simple_script.rb.pid
    start program = "su - user -c '/home/user/simple_script_daemon.rb start' "
    stop program = "su - user -c '/home/user/simple_script_daemon.rb stop' "
    group simple_script

如果您知道可能会发生什么，我很乐意听取您的意见!

Answer 1

我想你可以尝试这样的事情:

check process simple_script
    with pidfile /home/user/simple_script.rb.pid
    start program = "/home/user/simple_script_daemon.rb start" as uid user and gid user
    stop program = "/home/user/simple_script_daemon.rb stop" as uid user and gid user
    group simple_script

如 monit doc 中所述.