xargs:用管道替换命令$(...)不起作用 [英] xargs: command substitution $(...) with pipe doesn't work

问题描述

我正在尝试编写简短的脚本以及以下命令:

I'm trying to write short script, and the following command:

echo "aaa111 bbb111" | xargs -I {} echo {} | sed 's/111/222/g'

返回aaa222 bbb222，这就是我所期望的.

returns aaa222 bbb222, which is what I expect.

我期望下一个命令:

echo "aaa111 bbb111" | xargs -I {} echo $(echo {} | sed 's/111/222/g')

返回相同，但返回aaa111 bbb111！为什么会这样?

to return the same, but it returns aaa111 bbb111! Why is that?

UPD:我要实现的目标:

我有很多文件，例如pic30-coff-gcc，pic30-coff-ag等，并且我需要为每个文件建立符号链接，例如pic30-gcc-> pic30-coff-gcc等.

I have many files like pic30-coff-gcc, pic30-coff-ag, etc, and I need to make a symlink for each file, like pic30-gcc -> pic30-coff-gcc, etc.

所以我写了这个:

ls|grep 'coff-'|xargs -I {} ln -s {} $(echo {} | sed 's/coff-//g')

它不起作用:对于每个文件，它报告该文件存在.我检查了这样的命令:

It doesn't work: for each file, it reports that file exists. I checked the command like this:

ls|grep 'coff-'|xargs -I {} echo "ln -s {} $(echo {} | sed 's/coff-//g')"

是的，sed部分不起作用:

ln -s pic30-coff-gcc pic30-coff-gcc
ln -s pic30-coff-gcc-4.0.3 pic30-coff-gcc-4.0.3
...

但是，如果我只输入

echo "ln -s pic30-coff-gcc $(echo pic30-coff-gcc | sed 's/coff-//g')"

有效:

ln -s pic30-coff-gcc pic30-gcc

然后我已经用aaa111编写了测试命令，但它也不起作用.还是不明白，为什么.

Then I've written test command with aaa111, and it doesn't work too. Still can't understand, why.

推荐答案

似乎您所需要的只是一个简单的循环:

It seems that all you need is a simple loop:

for file in *coff*; do
  ln -s "${file}" "${file/coff-/}"
done

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