Question

我正试图找到一种从电子书样本中查找语法条款的方法。这是输入的样子：

This is a test my friend, this is just a test; I'm going to do some shopping:`what do you need?`
Nothing, he said.

期望的输出：

This is a test my friend
this is just a test
I'm going to do shopping
what do you need
Nothing
he said

关于如何实现这一目标的任何想法？

非常感谢！

Answer 1

您可以像这样使用gnu-awk：

awk -v RS='[\n.,;:`?]+' -v ORS='\n' '{$1=$1} 1' file
This is a test my friend
this is just a test
I'm going to do some shopping
what do you need
Nothing
he said

Answer 2

这很接近：

grep -o '[[:alpha:][:space:]]\+' file

但是它翻译了＃34;我＆＃39; m＆＃34;到换行符。鉴于您的示例标点符号，这有效：

grep -o '[^,;:`?.]\+' file

这会在标点字符后保留空格。要删除它，请将输出传递给

| sed 's/^ //'

Answer 3

将它传递给tr。

cat input | tr ',' '\n'

Grep语法条款

3 个答案: