bash 将命令输出保存在变量上并检查退出状态
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/36921658/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Save command output on variable and check exit status
提问by Nasr
The following command will resolve google ip
以下命令将解析google ip
> ip=`dig +short google.com`
> echo $ip
> 216.58.210.238
Sometimes (especially when internet connection is lost) this command fail with this error
有时（尤其是当互联网连接丢失时）此命令会因此错误而失败
> ;; connection timed out; no servers could be reached
When the command fail and i use $#the output is 0 for the assigment
当命令失败并且我使用$#输出为 0 进行分配时
> ip=`dig +short google.com`
> echo $#
> 0
> echo $ip # Command failed
> ;; connection timed out; no servers could be reached
How could i save the output of command in variable, also check if the command succeeded
我如何将命令的输出保存在变量中，还要检查命令是否成功
采纳答案by sjsam
You should use $?instead of $#.
您应该使用 $?而不是$#.
$? - contains the return value from the last script.
$# - contains the total number of arguments passed to a script or function
Do something like below :
执行以下操作：
if $?
then
echo "Success" # Do something here
else
echo "Fail" # Fallback mode
fi
Edit :In reply to thiscomment answering the question in title :
编辑：回复此评论，回答标题中的问题：
Save command output on variable and check exit status
将命令输出保存在变量上并检查退出状态
As @chepnerpointed out in thiscomment, an assignment doesn't meddle with the exit status of command. So :
正如@chepner在此评论中指出的那样，赋值不会干扰命令的退出状态。所以 ：
ip=$(dig +short google.com) # Just changed the legacy backticks to $()
[ $? -eq 0 ] && echo "dig succeeded"
回答by Rany Albeg Wein
You can avoid accessing $?, and simply:
您可以避免访问$?, 并且只需：
if ip=$(dig +short google.com); then
# Success.
else
# Failure.
fi
Example:
例子：
The following function will print "fail" and return 1.
以下函数将打印“失败”并返回 1。
print_and_fail() { printf '%s' fail; return 1; }
Thus, if we do the following:
因此，如果我们执行以下操作：
if foo=$(print_and_fail); then printf '%s\n' "$foo";fi
We'll get no output, yet store print_and_failoutput to $foo- in this case, "fail".
我们不会得到任何输出，但将print_and_fail输出存储到$foo- 在这种情况下，“失败”。
But, take a look at the following function, which will print "success" and return 0.
但是，看看下面的函数，它将打印“成功”并返回 0。
print_and_succeed() { printf '%s' success; return 0; }
Let's see what happens now:
让我们看看现在会发生什么：
$ if foo=$(print_and_succeed); then printf '%s\n' "$foo";fi
$ success
回答by David C. Rankin
You can check the return or use command-substitutionand check the resulting variable. e.g.
您可以检查返回或使用命令替换并检查结果变量。例如
$ ip=$(dig +short google.com)
$ [ -n "$ip" ] && echo "all good, ip = $ip"
(you can do the reverse check for failure with -z
（您可以对失败进行反向检查 -z
回答by Swiss
I recommend against the use of $?as much as possible, as it is fragile and easy to overlook when refactoring. If someone adds a line in between the command and $?, the script will break in non-obvious ways.
我建议$?尽量不要使用，因为它很脆弱，在重构时很容易被忽视。如果有人在命令和 之间添加一行$?，脚本将以不明显的方式中断。
If you want to both assign the output to a variable and check the exit code at the same time, you can just do both in your conditional block:
如果您想同时将输出分配给变量并检查退出代码，您可以在条件块中同时执行：
if foo="$(echo "foo"; true)"; then
echo "$foo"
fi
echo "$foo"
回答by John Mark Mitchell
Since you are using Bash, you can use something like the following script which can capture stdout, stderr and the return code https://gist.github.com/jmmitchell/c4369acb8e9ea1f984541f8819c4c87b
由于您使用的是 Bash，您可以使用类似于以下脚本的内容，该脚本可以捕获 stdout、stderr 和返回代码https://gist.github.com/jmmitchell/c4369acb8e9ea1f984541f8819c4c87b
For easy reference I have copied the script here:
为了方便参考，我在这里复制了脚本：
# #!/bin/bash
#
# based on posts from stackexchange:
# http://stackoverflow.com/a/26827443/171475
# http://stackoverflow.com/a/18086548/171475
# http://stackoverflow.com/a/28796214/171475
function example_function {
printf 'example output to stdout %d\n' {1..10}
echo >&2 'example output to stderr'
return 42
}
##############################
### using the dot operator ###
if [ "${BASH_VERSINFO}" -lt 4 ]; then
printf '%s\n' "The source version of this script requires Bash v4 or higher."
else
# stdout & stderr only
source <({ cmd_err=$({ mapfile -t cmd_out < <(example_function); } 2>&1; declare -p cmd_out >&2); declare -p cmd_err; } 2>&1)
printf "\n%s\n" "SOURCE VERSION : STDOUT & STDERR ONLY"
printf "%s\n" "${cmd_out[@]}"
printf "%s\n" "${cmd_err}"
unset cmd_out
unset cmd_err
# stdout & stderr only as well as return code:
source <({ cmd_err=$({ mapfile -t cmd_out< <( \
example_function \
; cmd_rtn=$?; declare -p cmd_rtn >&3); } 3>&2 2>&1; declare -p cmd_out >&2); declare -p cmd_err; } 2>&1)
printf "\n%s\n" "SOURCE VERSION : STDOUT, STDERR & RETURN CODE"
printf '%s\n' "${cmd_out[@]}"
# alternative version
# declare -p cmd_out
printf '%s\n' "${cmd_err}"
printf '%s\n' "${cmd_rtn}"
unset cmd_out
unset cmd_err
unset cmd_rtn
fi
##############################
######### using exec #########
# stdout & stderr only
eval "$({ cmd_err=$({ cmd_out=$( \
example_function \
); } 2>&1; declare -p cmd_out >&2); declare -p cmd_err; } 2>&1)"
printf "\n%s\n" "EVAL VERSION : STDOUT & STDERR ONLY"
printf '%s\n' "${cmd_out}"
printf '%s\n' "${cmd_err}"
printf '%s\n' "${cmd_rtn}"
unset cmd_out
unset cmd_err
# stdout & stderr only as well as return code:
eval "$({ cmd_err=$({ cmd_out=$( \
example_function \
); cmd_rtn=$?; } 2>&1; declare -p cmd_out cmd_rtn >&2); declare -p cmd_err; } 2>&1)"
printf "\n%s\n" "EVAL VERSION : STDOUT, STDERR & RETURN CODE"
printf '%s\n' "${cmd_out}"
printf '%s\n' "${cmd_err}"
printf '%s\n' "${cmd_rtn}"
unset cmd_out
unset cmd_err
unset cmd_rtn
