bash 在另一个脚本中获取最后一个 shell 命令的退出代码
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Getting exit code of last shell command in another script
提问by sebastiangeiger
I am trying to beef up my notify script. The way the script works is that I put it behind a long running shell command and then all sorts of notifications get invoked after the long running script finished.
我正在尝试加强我的通知脚本。脚本的工作方式是我将它放在一个长时间运行的 shell 命令后面，然后在长时间运行的脚本完成后调用各种通知。
For example:
例如：
sleep 100; my_notify
It would be nice to get the exit code of the long running script, the problem is that calling my_notifycreates a new process that does not have access to the $?variable.
获得长时间运行脚本的退出代码会很好，问题是调用my_notify会创建一个无法访问该$?变量的新进程。
Compare:
相比：
~ $: ls nonexisting_file; echo "exit code: $?"; echo "PPID: $PPID"
ls: nonexisting_file: No such file or directory
exit code: 1
PPID: 6203
vs.
对比
~ $: ls nonexisting_file; my_notify
ls: nonexisting_file: No such file or directory
exit code: 0
PPID: 6205
The my_notifyscript has the following in it:
该my_notify脚本包含以下内容：
#!/bin/sh
echo "exit code: $?"
echo "PPID: $PPID"
I am looking for a way to get the exit code of the previous command without changing the structure of the command too much. I am aware of the fact that if I change it to work more like time, e.g. my_notify longrunning_command...my problem would be solved, but I actually like that I can tack it at the end of a command and I fear complications of this second solution.
我正在寻找一种方法来获取上一个命令的退出代码，而不会过多地改变命令的结构。我知道如果我将其更改为更像time，例如，my_notify longrunning_command...我的问题将得到解决，但我实际上喜欢这样，我可以在命令的末尾添加它，并且我担心第二个解决方案会出现并发症。
Can this be done or is it fundamentally incompatible with the way that shells work?
这可以完成还是从根本上与 shell 的工作方式不兼容？
My shell is zshbut I would like it to work with bashas well.
我的外壳是，zsh但我希望它也能使用bash。
回答by qqx
You'd really need to use a shell function in order to accomplish that. For a simple script like that it should be pretty easy to have it working in both zsh and bash. Just place the following in a file:
您确实需要使用 shell 函数来实现这一点。对于像这样的简单脚本，让它在 zsh 和 bash 中工作应该很容易。只需将以下内容放入文件中：
my_notify() {
echo "exit code: $?"
echo "PPID: $PPID"
}
Then source that file from your shell startup files. Although since that would be run from within your interactive shell, you may want to use $$ rather than $PPID.
然后从您的 shell 启动文件中获取该文件。虽然因为它会在你的交互式 shell 中运行，你可能想要使用 $$ 而不是 $PPID。
回答by Ignacio Vazquez-Abrams
It is incompatible. $?onlyexists within the current shell; if you want it available in subprocesses then you must copy it to an environment variable.
这是不兼容的。$?只存在于当前shell中；如果您希望它在子进程中可用，则必须将其复制到环境变量中。
The alternative is to write a shell function that uses it in some way instead.
另一种方法是编写一个 shell 函数，以某种方式使用它。
回答by BIBS
One method to implement this could be to use EOF tag and a master script which will create your my_notify script.
实现这一点的一种方法可能是使用 EOF 标记和一个主脚本，它将创建您的 my_notify 脚本。
#!/bin/bash
if [ -f my_notify ] ; then
rm -rf my_notify
fi
if [ -f my_temp ] ; then
rm -rf my_temp
fi
retval=`ls non_existent_file &> /dev/null ; echo $?`
ppid=$PPID
echo "retval=$retval"
echo "ppid=$ppid"
cat >> my_notify << 'EOF'
#!/bin/bash
echo "exit code: $retval"
echo " PPID =$ppid"
EOF
sh my_notify
You can refine this script for your purpose.
您可以根据自己的目的优化此脚本。
