庆典：捕获标准输出到一个变量，但仍然在控制台中显示它 [英] bash: Capture stdout to a variable but still display it in the console
问题描述
我有一个bash脚本调用几个长期运行的流程。我想这些调用的输出捕捉到处理的原因变量。但是，因为这些都是长时间运行的进程，我想rsync的输出要求要显示在控制台的实时的，而不是事后。
I have a bash script which calls several long-running processes. I want to capture the output of those calls into variables for processing reasons. However, because these are long running processes, I would like the output of the rsync calls to be displayed in the console in real-time and not after the fact.
为此，我有发现做的一种方式，但它依靠输出文本到/ dev /标准错误。我觉得输出到/ dev /标准错误不是一个做事的好方法。
To this end, I have found a way of doing it but it relies on outputting the text to /dev/stderr. I feel that outputting to /dev/stderr is not a good way of doing things.
VAR1=$(for i in {1..5}; do sleep 1; echo $i; done | tee /dev/stderr)
VAR2=$(rsync -r -t --out-format='%n%L' --delete -s /path/source1/ /path/target1 | tee /dev/stderr)
VAR3=$(rsync -r -t --out-format='%n%L' --delete -s /path/source2/ /path/target2 | tee /dev/stderr)
在上面的例子中，我打电话rsync的几次，我希望看到他们正在处理的文件名，但最后我还是想在一个变量的输出，因为我将在后面分析它。
In the example above, I am calling rsync a few times and I want to see the file names as they are processed, but in the end I still want the output in a variable because I will be parsing it later.
有没有实现这一点的'清洁'的方式？
Is there a 'cleaner' way of accomplishing this?
如果它的确与众不同，我使用Ubuntu 12.04，bash的4.2.24。
If it makes a difference, I am using Ubuntu 12.04, bash 4.2.24.
推荐答案
复制和放大器;在shell 1（在我〔实施例5），并用和5中的子shell（这样你会写入stdout（安培; 1）母公司壳）：
Duplicate &1 in your shell (in my examle to 5) and use &5 in the subshell (so that you will write to stdout (&1) of the parent shell):
exec 5>&1
FF=$(echo aaa|tee >(cat - >&5))
echo $FF
可否打印AAA两次，是因为在子外壳的回波，及第二时间的那些打印变量的值
Will print aaa two times, ones because of the echo in the subshell, and second time print the value of the variable.
在您的code：
exec 5>&1
VAR1=$(for i in {1..5}; do sleep 1; echo $i; done | tee >(cat - >&5))
# use the value of VAR1
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