Question

说我有这两个bash脚本：

的/ tmp /捕集器：

#!/bin/bash
trap 'echo trapper: ignoring USR1' USR1
"$(dirname $0)"/usr1er & p=$!
sleep 1
echo trapper: now killing usr1er
kill $p
echo trapper: sleeping
sleep 1
echo trapper: reached end of trapper

的/ tmp / usr1er：

#!/bin/bash
trap 'echo "usr1er: EXIT received, sending USR1"; kill -USR1 0' EXIT
while sleep 1;do echo usr1er: sleeping;done

trapper应该捕获USR1而忽略它。它启动usr1er，用USR1信号杀死其进程组。现在，如果我从交互式shell开始将trapper作为脚本自己启动，它会杀死usr1er并正常退出：

$ /tmp/trapper; echo done
trapper: now killing usr1er
trapper: sleeping
usr1er: EXIT received, sending USR1
/tmp/trapper: line 9: 16596 Terminated              "$(dirname $0)"/usr1er
trapper: ignoring USR1
trapper: reached end of trapper
done

如果我尝试$(/tmp/trapper)，它会退出整个shell。同样，如果我创建一个调用/tmp/trapper的单独脚本，例如/tmp/outer：

#!/bin/bash
"$(dirname $0)"/trapper
echo outer: reached end of outer

它会在没有打印“到达外部的末端”的情况下被杀死：

$ /tmp/outer
trapper: now killing usr1er
trapper: sleeping
usr1er: EXIT received, sending USR1
User defined signal 1
/tmp/trapper: line 9: 23544 Terminated              "$(dirname $0)"/usr1er
User defined signal 1
trapper: ignoring USR1
trapper: reached end of trapper

为什么？

Answer 1

似乎$() 不使用单独的进程组/ PGID（显然使C-c工作）启动进程。

此外，任何非交互式shell也不会为其子项启动单独的PGID（除非您使用set -m打开作业控制）：

$ bash -c '/tmp/trapper;echo done'
trapper: now killing usr1er
trapper: sleeping
usr1er: EXIT received, sending USR1
User defined signal 1
$ /tmp/trapper: line 9: 17522 Terminated              "$(dirname $0)"/usr1er
trapper: ignoring USR1
trapper: reached end of trapper

请注意＆＃34;完成＆＃34;如果没有打印，那么没有捕获USR1的外部bash脚本会被捕获，而捕获者会一直生存到最后。

您可以将ps -o %p%r%c -p$$放入，检查每个流程的PGID 脚本：

$ /tmp/outer
  PID  PGID COMMAND
27630 27630 outer
  PID  PGID COMMAND
27633 27630 trapper
  PID  PGID COMMAND
27635 27630 usr1er
trapper: now killing usr1er
trapper: sleeping
usr1er: EXIT received, sending USR1
User defined signal 1
$ /tmp/trapper: line 9: 27635 Terminated              "$(dirname $0)"/usr1er
trapper: ignoring USR1
trapper: reached end of trapper

Answer 2

尝试此更改

的/ tmp / usr1er：

#!/bin/bash
trap 'echo "usr1er: EXIT received, sending USR1"; kill -USR1 $PPID' TERM
while sleep 1;do echo usr1er: sleeping;done

处理TERM信号而不是EXIT＆amp;已将USR1发送至$PPID而非0

我在调用的脚本中有“陷阱”echo忽略'USR1'，为什么调用脚本会被杀死？

2 个答案: