如何分割用逗号没有空格名单 [英] How to split a list by comma not space

问题描述

我要拆与逗号文本，不是空间 在中富列表。假设我有一个CSV文件 CSV_File 里面坐了以下文字：

I want to split a text with comma , not space in for foo in list. Suppose I have a CSV file CSV_File with following text inside it:

Hello,World,Questions,Answers,bash shell,script
...

我用下面的code将其分割成几个单词：

I used following code to split it into several words:

for word in $(cat CSV_File | sed -n 1'p' | tr ',' '\n')
do echo $word
done

它打印：

Hello
World
Questions
Answers
bash
shell
script

但我想它分裂用逗号文本没有空格：

But I want it to split the text by commas not spaces:

Hello
World
Questions
Answers
bash shell
script

我怎么能在bash做到这一点？

How can I achieve this in bash?

推荐答案

使用子shell替换来分析的话撤销所有你正在做把空间一起工作。

Using a subshell substitution to parse the words undoes all the work you are doing to put spaces together.

而不是尝试：

cat CSV_file | sed -n 1'p' | tr ',' '\n' | while read word; do
    echo $word
done

这也增加了并行性。使用子shell作为你的问题迫使整个子shell进程完成之前，你可以重头再来的答案迭代。管道连接到子shell（在我的答案），让他们可以并行工作。这一点很重要，如果只有你当然有文件中的许多行。

That also increases parallelism. Using a subshell as in your question forces the entire subshell process to finish before you can start iterating over the answers. Piping to a subshell (as in my answer) lets them work in parallel. This matters only if you have many lines in the file, of course.

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