Bash - 逐行打印句子

Question

我还有另一个bash问题。如果我有像......这样的文件。

Today is sunny. I like the sun. It is awesome!

我希望打印出每个字符，直到它到达?，.或!。这也可能是一个特殊的班轮吗？我希望打印输出看起来像。

Today is sunny.
I like the sun.
It is awesome!

Answer 1

echo Today is sunny. I like the sun. It is awesome! | sed 's/[.!?] */&\n/g'

Answer 2

您只需指定记录分隔符：

awk 'BEGIN {RS="[.!?] *"} {print}'

Answer 3

awk中关闭的东西（不完全是需要但很有趣）：

echo 'Today is sunny. I like the sun. It is awesome!'|awk '1' RS='[.,?]'

规范的awk方式：

echo 'Today is sunny. I like the sun. It is awesome!'|gawk 'a=gensub(/(\.|!|\?) */, "\\1\n", "g"){print a}'

Answer 4

使用sed的解决方案：

sed 's/\([\?\!\.]\)\s*/\1\n/g'

Answer 5

echo Today is sunny. I like the sun. It is awesome! | awk '{gsub(/([?,.!])/,"&\n");print}'