linux – 防止bash脚本在处理SIGINT后终止

概述我正在为应用程序编写一个bash包装器.该包装器负责更改用户,运行软件和记录其输出. 我也希望它传播SIGINT信号. 到目前为止,这是我的代码： #!/bin/bashset -e; set -ufunction child_of { ps --ppid $1 -o "pid" --no-headers | head -n1}function handle_int { 我正在为应用程序编写一个bash包装器.该包装器负责更改用户,运行软件和记录其输出.
我也希望它传播SIGINT信号.

到目前为止,这是我的代码：

#!/bin/bashset -e; set -ufunction child_of {    ps --ppID  -o "pID" --no-headers | head -n1}function handle_int {    echo "Received SIGINT"    kill -int $(child_of $SU_PID)}su myuser -p -c "bash /opt/loop.sh 2>&1 | tee -i >(logger -t mytag)" &SU_PID=$!trap "handle_int" SIGINTwait $SU_PIDecho "This is the end."

我的问题是,当我向这个包装器发送一个SIGINT时,handle_int被调用但是脚本结束了,而我希望它继续等待$SU_PID.

有没有办法捕获int信号,做一些事情,然后阻止脚本终止？

解决方法 你有一个问题：在Ctrl-C之后,“这就是结束.”预计但它永远不会到来,因为脚本已经过早退出.原因是在set -e下运行时wait(意外地)返回非零值.

根据“男子打击”：

If bash is waiting for a command to complete and receives a signal for which a trap has been set,the trap
will not be executed until the command completes. When bash is waiting for an asynchronous command via the
wait builtin,the reception of a signal for which a trap has been set will cause the wait builtin to return
immediately with an exit status greater than 128,immediately after which the trap is executed.

您应该将set wait包装在set e中,以便在等待异步命令时处理捕获的信号后程序可以继续运行.

像这样：

# wait function that handles trapped signal on asynchronous commands.function safe_async_wait {  set +e  wait  # returns >128 on asynchronous commands  set -e}#...safe_async_wait $SU_PID

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