使用 sed 在两个模式之间打印

Question

所以我在deleting the text between two patterns 和printing the text between two delimiters 上找到了很多东西，但我没有找到任何关于使用 bash 函数在两个模式之间打印文本的内容。

如果我有：

"Alas poor Yorik, I knew him well"

我想打印模式“差”和“好”（不包括）之间的所有内容：

" Yorik, I knew him "

如何使用 sed 或 awk 之类的方法来实现这一点？

Answer 1

dtpwmbp:~ pwadas$ echo "Alas poor Yorik, I knew him well" | sed -e 's/^.*poor //g;s/ well.*$//g'
Yorik, I knew him
dtpwmbp:~ pwadas$ echo "Alas poor Yorik, I knew him well" | awk '{sub(/.*poor /,"");sub(/ well.*/,"");print;}'
Yorik, I knew him

与文件输入一起使用：

dtpwmbp:~ pwadas$ echo "Alas poor Yorik, I knew him well" > infile
dtpwmbp:~ pwadas$ cat infile 
Alas poor Yorik, I knew him well
dtpwmbp:~ pwadas$ cat infile | sed -e 's/^.*poor //g;s/ well.*$//g'
Yorik, I knew him
dtpwmbp:~ pwadas$ sed -e 's/^.*poor //g;s/ well.*$//g' < infile
Yorik, I knew him
dtpwmbp:~ pwadas$ cat infile | awk '{sub(/.*poor /,"");sub(/ well.*/,"");print;}'
Yorik, I knew him
dtpwmbp:~ pwadas$ awk '{sub(/.*poor /,"");sub(/ well.*/,"");print;}' < infile 
Yorik, I knew him