如何在shell脚本中陷阱1个信号？

Question

我想尝试诱捕信号1但失败

#!/bin/bash
# capture an interrupt # 0
trap 'echo "Exit 0 signal detected..."' 0
trap 'echo "Exit 1 signal detected..."' SIGHUP

# display something
echo "This is a checkpoint 1"
exit 1

echo "This is checkpoint 2"
# exit shell script with 0 signal
exit 0

Output--
kithokit@15:02:55 trunk (master) $ ./test.sh 
This is a checkpoint 1
Exit 0 signal detected...
kithokit@15:03:44 trunk (master) $ 

.

即使是退出1，它始终陷入陷阱0，任何一个人都知道如何解决这个问题？

感谢

Answer 1

exit 1不发送稍微升级。它退出返回代码（AKA退出状态）1.要发送SIGHUP使用kill：

#!/bin/bash
# capture an interrupt # 0
trap 'echo "Signal 0 detected..."' 0
trap 'echo "SIGHUP detected..."' SIGHUP

# display something
echo "This is a checkpoint 1"
kill -1 $$

echo "This is checkpoint 2"
# exit shell script with 0 signal
exit 0

.

$$是当前过程的ID。因此，kill -1 $$将信号1（sighup）发送到当前过程。上述脚本的输出为：

This is a checkpoint 1
SIGHUP detected...
This is checkpoint 2
Signal 0 detected...

.

如何检查退出返回码

如果目标是检查返回码（也称为退出状态）而不是捕获特殊信号，那么我们需要做的就是检查退出时的状态变量生成$?：

#!/bin/bash
# capture an interrupt # 0
trap 'echo "EXIT detected with exit status $?"' EXIT

echo "This is checkpoint 1"
# exit shell script with 0 signal
exit "$1"
echo "This is checkpoint 2"

.

在命令行运行时，这会产生：

$ status_catcher 5
This is checkpoint 1
EXIT detected with exit status 5
$ status_catcher 208
This is checkpoint 1
EXIT detected with exit status 208

.

请注意，trap语句可以调用bash函数，该函数可以包括不同复杂的语句来处理不同的返回代码。