将stdout捕获到变量，但仍将其显示在控制台中 [英] Capture stdout to a variable but still display it in the console

问题描述

我有一个bash脚本，它调用了几个长时间运行的进程.由于处理原因，我想将这些调用的输出捕获到变量中.但是，由于这些进程是长期运行的进程，因此我希望rsync调用的输出在实时中而不是在事后显示在控制台中.

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，我找到了 的一种实现方法，但它依赖于输出文本到/dev/stderr.我觉得输出到/dev/stderr并不是一个好方法.

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)

在上面的示例中，我多次调用rsync，并希望在处理文件时看到它们的名称，但是最后我仍然希望将输出结果包含在变量中，因为稍后将对其进行解析.

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.

是否有一种更清洁"的方法来实现这一目标?

Is there a 'cleaner' way of accomplishing this?

如果有所作为，我正在使用Ubuntu 12.04，bash 4.2.24.

If it makes a difference, I am using Ubuntu 12.04, bash 4.2.24.

推荐答案

在您的shell中复制& 1(在我的示例中为5)，在子shell中使用& 5(以便您将写入stdout(& 1)的父外壳):

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

将两次打印aaa，一次是由于子外壳中的回显，而第二次将打印变量的值.

Will print aaa two times, ones because of the echo in the subshell, and second time print the value of the variable.

在您的代码中:

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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