如何通过管道输入 Bash while 循环并在循环结束后保留变量
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How to pipe input to a Bash while loop and preserve variables after loop ends
提问by Wakan Tanka
Bash allows to use: cat <(echo "$FILECONTENT")
Bash 允许使用： cat <(echo "$FILECONTENT")
Bash also allow to use: while read i; do echo $i; done </etc/passwd
Bash 还允许使用： while read i; do echo $i; done </etc/passwd
to combine previous two this can be used: echo $FILECONTENT | while read i; do echo $i; done
结合前两个可以使用： echo $FILECONTENT | while read i; do echo $i; done
The problem with last one is that it creates sub-shell and after the while loop ends variable icannot be accessed any more.
最后一个的问题是它创建了子外壳，并且在 while 循环结束后，i无法再访问变量。
My question is:
我的问题是：
How to achieve something like this: while read i; do echo $i; done <(echo "$FILECONTENT")or in other words: How can I be sure that isurvives while loop?
如何实现这样的事情：while read i; do echo $i; done <(echo "$FILECONTENT")或者换句话说：我怎么能确定在iwhile循环中幸存下来？
Please note that I am aware of enclosing while statement into {}but this does not solves the problem (imagine that you want use the while loop in function and return ivariable)
请注意，我知道将 while 语句括起来，{}但这并不能解决问题（想象一下，您想在函数中使用 while 循环并返回i变量）
回答by Jonathan Leffler
The correct notation for Process Substitutionis:
进程替换的正确符号是：
while read i; do echo $i; done < <(echo "$FILECONTENT")
The last value of iassigned in the loop is then available when the loop terminates.
An alternative is:
i当循环终止时，循环中分配的最后一个值可用。另一种选择是：
echo $FILECONTENT |
{
while read i; do echo $i; done
...do other things using $i here...
}
The braces are an I/O grouping operation and do not themselves create a subshell. In this context, they are part of a pipeline and are therefore run as a subshell, but it is because of the |, not the { ... }. You mention this in the question. AFAIK, you can do a return from within these inside a function.
大括号是 I/O 分组操作，本身不会创建子 shell。在这种情况下，他们是一个管道的一部分，因此运行在一个子shell，但它是因为|，不是{ ... }。你在问题中提到了这一点。AFAIK，您可以在函数内部从这些内部返回。
Bash also provides the shoptbuiltin and one of its many options is:
Bash 还提供了shopt内置函数，其众多选项之一是：
lastpipeIf set, and job control is not active, the shell runs the last command of a pipeline not executed in the background in the current shell environment.
lastpipe如果设置，并且作业控制未处于活动状态，shell 将运行当前 shell 环境中未在后台执行的管道的最后一个命令。
Thus, using something like this in a scriptmakes the modfied sumavailable after the loop:
因此，在脚本中使用类似的东西sum可以在循环后使用modfied ：
FILECONTENT="12 Name
13 Number
14 Information"
shopt -s lastpipe # Comment this out to see the alternative behaviour
sum=0
echo "$FILECONTENT" |
while read number name; do ((sum+=$number)); done
echo $sum
Doing this at the command line usually runs foul of 'job control is not active' (that is, at the command line, job control is active). Testing this without using a script failed.
在命令行执行此操作通常会违反“作业控制未激活”（即，在命令行中，作业控制处于活动状态）。在不使用脚本的情况下测试失败。
Also, as noted by Gareth Reesin his answer, you can sometimes use a here string:
此外，正如Gareth Rees在他的回答中所指出的，您有时可以使用here 字符串：
while read i; do echo $i; done <<< "$FILECONTENT"
This doesn't require shopt; you may be able to save a process using it.
这不需要shopt; 您可以使用它来保存进程。
回答by Gareth Rees
Jonathan Leffler explainshow to do what you want using process substitution, but another possibility is to use a here string:
Jonathan Leffler 解释了如何使用流程替换来做您想做的事情，但另一种可能性是使用here 字符串：
while read i; do echo "$i"; done <<<"$FILECONTENT"
This saves a process.
这节省了一个过程。
