bash 当'set -e'处于活动状态时，Bash获取命令的退出状态？
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Bash get exit status of command when 'set -e' is active?
提问by David Wolever
I generally have -eset in my Bash scripts, but occasionally I would like to run a command and get the return value.
我通常-e在我的 Bash 脚本中设置，但偶尔我想运行一个命令并获取返回值。
Without doing the set +e; some-command; res=$?; set -edance, how can I do that?
不做set +e; some-command; res=$?; set -e舞蹈，我怎么能做到呢？
回答by Adrian Frühwirth
From the bashmanual:
从bash手册：
The shell does not exit if the command that fails is [...] part of any command executed in a && or || list [...].
如果失败的命令是在 && 或 || 中执行的任何命令的 [...] 一部分，则 shell 不会退出 列表 [...]。
So, just do:
所以，只要这样做：
#!/bin/bash
set -eu
foo() {
# exit code will be 0, 1, or 2
return $(( RANDOM % 3 ))
}
ret=0
foo || ret=$?
echo "foo() exited with: $ret"
Example runs:
示例运行：
$ ./foo.sh
foo() exited with: 1
$ ./foo.sh
foo() exited with: 0
$ ./foo.sh
foo() exited with: 2
This is the canonical way of doing it.
这是规范的做法。
回答by KitsuneYMG
as an alternative
作为备选
ans=0
some-command || ans=$?
回答by Phil Miller
Maybe try running the commands in question in a subshell, like this?
也许尝试在子shell中运行有问题的命令，像这样？
res=$(some-command > /dev/null; echo $?)
回答by konsolebox
Use a wrapper function to execute your commands:
使用包装函数来执行您的命令：
function __e {
set +e
"$@"
__r=$?
set -e
}
__e yourcommand arg1 arg2
And use $__rinstead of $?:
并使用$__r代替$?：
if [[ __r -eq 0 ]]; then
echo "success"
else
echo "failed"
fi
Another method to call commands in a pipe, only that you have to quote the pipe. This does a safe eval.
在管道中调用命令的另一种方法，只是您必须引用管道。这是一个安全的评估。
function __p {
set +e
local __A=() __I
for (( __I = 1; __I <= $#; ++__I )); do
if [[ "${!__I}" == '|' ]]; then
__A+=('|')
else
__A+=("\"$$__I\"")
fi
done
eval "${__A[@]}"
__r=$?
set -e
}
Example:
例子：
__p echo abc '|' grep abc
And I actually prefer this syntax:
我实际上更喜欢这种语法：
__p echo abc :: grep abc
Which I could do with
我可以用
...
if [[ ${!__I} == '::' ]]; then
...
