Bash - 逐行打印句子

Question

我還有另一個bash問題。如果我有一個文件一樣......

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

我想打印出每個字符，直到遇到一個?，.，或!。這也可以是一個awk單線程？我想讓打印輸出看起來像。

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

Answer 1

你只需要指定記錄分隔符：

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

Answer 2

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

Answer 3

一些接近在AWK（不完全想需要的，但有趣的）：

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

的規範awk的方式：

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

Answer 4

的解決方案使用sed：

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

Answer 5

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