Question

我有一个简单的bash脚本，我已经编写它来简化我正在做的一些工作。它需要做的只是启动一个进程process_1作为后台进程然后启动另一个进程process_2。完成process_2后，我需要终止process_1。

process_1启动一个程序，除非它收到kill信号，否则实际上不会停止，或者当我自己运行它时，CTRL + C.该程序通过{program} {args} > output_file

输出到一个文件中

process_2可能会花费任意时间，具体取决于给出的参数。

代码：

#!/bin/bash

#Call this on exit to kill all background processes
function killJobs () {
    #Check process is still running before killing
    if kill -0 "$PID"; then
            kill $PID
    fi
}

...检查给定的参数是否有效......

#Start process_1
eval "./process_1 ${Arg1} ${Arg2} ${Arg3}" &
PID=$!

#Lay a trap to catch any exits from script
trap killJobs TERM INT

#Start process_2 - sleep for 5 seconds before and after
#Need space between process_1 and process_2 starting and stopping
sleep 5
eval "./process_2 ${Arg1} ${Arg2} ${Arg3} ${Arg4} 2> ${output_file}"
sleep 5

#Make sure background job is killed on exit
killJobs

我的脚本结束后检查其输出文件是否仍在更新，检查process_1已被终止。

如果我运行脚本然后按CTRL + C，脚本将终止，process_1也将被终止，输出文件将不再更新。

如果我在没有干预process_2的情况下让脚本运行完成，并且脚本都终止，但是当我检查process_1的输出时，它仍在更新。

为了检查这个问题，我在process_1启动后放置一个echo语句，在killJobs的if语句中放置另一个，所以只有在调用kill $PID时才会回显。

执行此操作我可以看到两种退出方式都启动process_1，然后输入if语句来终止它。然而，杀戮实际上并没有在正常退出的情况下杀死进程。也不会产生错误消息。

Answer 1

您正在设置eval而不是process_1，这会设置$！到脚本本身的PID，而不是process_1。改为：

#!/bin/bash

#Call this on exit to kill all background processes
function killJobs () {
    #Check process is still running before killing
    if kill -0 "$PID"; then
            kill $PID
    fi
}
...Check given arguments are valid...

#Start process_1
./process_1 ${Arg1} ${Arg2} ${Arg3} &
PID=$!

#Lay a trap to catch any exits from script
trap killJobs TERM INT

#Start process_2 - sleep for 5 seconds before and after
#Need space between process_1 and process_2 starting and stopping
sleep 5
./process_2 ${Arg1} ${Arg2} ${Arg3} ${Arg4} 2> ${output_file}
sleep 5

#Make sure background job is killed on exit
killJobs

如果正确退出，请不要杀死进程

1 个答案: