将标准输出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，它调用几个长时间运行的进程.出于处理原因，我想将这些调用的输出捕获到变量中.但是，因为这些是长时间运行的进程，我希望 rsync 调用的输出能够实时显示在控制台中，而不是事后显示.

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，我找到了一种方法，但它依赖于输出文本到/dev/stderr.我觉得输出到/dev/stderr 不是一个好的做事方式.

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，我调用了几次 rsync，我想在处理文件时查看文件名，但最后我仍然希望输出在一个变量中，因为我稍后会解析它.

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.

有没有一种更干净"的方法来实现这一点?

Is there a 'cleaner' way of accomplishing this?

如果有区别，我使用的是 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)并在子 shell 中使用 &5(以便您将写入 stdout(&5)1) 父壳):

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

将打印 aaa 两次，一次是因为子shell中的回声，第二次打印变量的值.

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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