bash 从子shell返回值并输出到局部变量
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return value from subshell and output to local variables
提问by Mephi_stofel
I've found the strange behaviour for me, which I can't explain. The following code is work OK:
我发现了我无法解释的奇怪行为。以下代码工作正常：
function prepare-archive {
blah-blah-blah...
_SPEC_FILE=$(check-spec-file "$_GIT_DIR/packaging/")
exit $?
blah-blah-blah...
}
means I get value which I expect:
意味着我得到了我期望的价值：
bash -x ./this-script.sh:
++ exit 1
+ _SPEC_FILE='/home/likern/Print/Oleg/print-service/packaging/print-service.spec
/home/likern/Print/Oleg/print-service/packaging/print-service2.spec'
+ exit 1
As soon as I add localdefinition to variable:
一旦我local向变量添加定义：
local _SPEC_FILE=$(check-spec-file "$_GIT_DIR/packaging/")
I get following:
我得到以下信息：
bash -x ./this-script.sh:
++ exit 1
+ local '_SPEC_FILE=/home/likern/Print/Oleg/print-service/packaging/print-service.spec
/home/likern/Print/Oleg/print-service/packaging/print-service2.spec'
+ exit 0
$:~/MyScripts$ echo $?
0
Question:Why? What has happened? Can I catch output from subshell to localvariable and check subshell's return value reliably?
问题：为什么？发生了什么事？我可以从 subshell 捕获输出到local变量并可靠地检查 subshell 的返回值吗？
P.S.: prepare-archiveis called in the main shell script. The first exitis the exitfrom check-spec-filefunction, the second from prepare-archivefunction - this function itself is executed from main shell script. I return value from check-spec-fileby exit 1, then pass this value to exit $?. Thus I expect they should be the same.
PS：prepare-archive在主shell脚本中调用。第一个exit是exitfromcheck-spec-file函数，第二个是 fromprepare-archive函数——这个函数本身是从主 shell 脚本执行的。我从check-spec-fileby返回值exit 1，然后将此值传递给exit $?. 因此，我希望它们应该是相同的。
采纳答案by Janito Vaqueiro Ferreira Filho
From the bash manual, Shell Builtin Commandssection:
从 bash 手册，Shell Builtin Commands部分：
local:
[...]The return status is zero unless local is used outside a function, an invalid name is supplied, or name is a readonly variable.
Hope this helps =)
希望这有帮助 =)
回答by vilpan
To capture subshell's exit status, declare the variable as local before the assignment, for example, the following script
要捕获子shell的退出状态，在赋值前将变量声明为local，例如下面的脚本
#!/bin/sh
local_test()
{
local local_var
local_var=$(echo "hello from subshell"; exit 1)
echo "subshell exited with $?"
echo "local_var=$local_var"
}
echo "before invocation local_var=$local_var in global scope"
local_test
echo "after invocation local_var=$local_var in global scope"
produces the following output
产生以下输出
before invocation local_var= in global scope
subshell exited with 1
local_var=hello from subshell
after invocation local_var= in global scope
回答by paul.da.programmer
As I use bash subshell parenthesis to group many echo commands I hit this strange problem. In my case all I needed was to pass one value back to the calling shell so I just used the exit command
当我使用 bash subshell 括号对许多 echo 命令进行分组时，我遇到了这个奇怪的问题。就我而言，我所需要的只是将一个值传递回调用 shell，所以我只使用了 exit 命令
RET=0
echo RET: $RET
(echo hello
echo there
RET=123
echo RET: $RET
exit $RET)
RET=$?
echo RET: $RET
gives the following output
给出以下输出
RET: 0
hello
there
RET: 123
RET: 123
without the exit command you will get this which is confusing:
如果没有退出命令，你会得到这个令人困惑的：
RET: 0
hello
there
RET: 123
RET: 0
