是否可以从 bash 脚本中设置超时？
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Is it possible to set time out from bash script?
提问by maihabunash
Sometimes my bash scripts are hanging and hold without clear reason
有时我的 bash 脚本没有明确的原因就挂了
So they actually can hang for ever ( script process will run until I kill it )
所以他们实际上可以永远挂起（脚本进程会一直运行直到我杀死它）
Is it possible to combine in the bash script time out mechanism in order to exit from the program after for example ? hour?
是否可以结合 bash 脚本超时机制，以便在例如之后退出程序？小时？
回答by rici
If you have Gnu coreutils, you can use the timeoutcommand:
如果您有 Gnu coreutils，则可以使用以下timeout命令：
timeout 1800s ./myscript
To check if the timeout occurred check the status code:
要检查是否发生超时，请检查状态代码：
timeout 1800s ./myscript
if (($? == 124)); then
echo "./myscript timed out after 30 minutes" >>/path/to/logfile
exit 124
fi
回答by Michael Jaros
This Bash-only approach encapsulates all the timeout code inside your script by running a function as a background job to enforce the timeout:
这种 Bash-only 方法通过运行一个函数作为后台作业来强制超时，将所有超时代码封装在脚本中：
#!/bin/bash
Timeout=1800 # 30 minutes
function timeout_monitor() {
sleep "$Timeout"
kill ""
}
# start the timeout monitor in
# background and pass the PID:
timeout_monitor "$$" &
Timeout_monitor_pid=$!
# <your script here>
# kill timeout monitor when terminating:
kill "$Timeout_monitor_pid"
Note that the function will be executed in a separate process. Therefore the PID of the monitored process ($$) must be passed. I left out the usual parameter checking for the sake of brevity.
请注意，该函数将在单独的进程中执行。因此$$必须传递被监视进程的PID ( )。为简洁起见，我省略了通常的参数检查。
