如何查找并打印 bash 中的所有 AWK 匹配项？

Question

我在变量中存储了很多文本。

text="This is sentence! this is not sentence! This is sentence. this is not sencence."

我正在通过此命令查找句子：

echo $text | awk 'match([=12=],/([A-Z])([^!?.]*)([!?.])/) { print substr([=12=],RSTART,RLENGTH) }'

我的输出是：

This is sentence!

预期输出：

This is sentence!
This is sentence.

更多样本： 课文中有语法正确和错误的句子。正确的句子由开头和结尾字符 (.?!) 的大写字母标识。我只想打印正确的句子。

text="incorrect sentence! this is not sentence! This is sentence. this is not sencence. This is correct sentence."

预期输出：

This is sentence.
This is correct sentence.

我能找到第一个匹配项，但不是所有匹配项。感谢您的帮助:)

Answer 1

您需要 while() 和 match():

$ echo $text | awk '
{
    while(match([=10=],/([A-Z])([^!?.]*)([!?.])/)) {   # while there are matches
        print substr([=10=],RSTART,RLENGTH)            # output them
        [=10=]=substr([=10=],RSTART+RLENGTH)               # and move forward
    }
}'

输出：

This is sentence!
This is sentence.

Answer 2

您可以将 GNU awk 用于 multi-char RS:

$ echo "$text" | awk -v RS='[A-Z][^!?.]*[!?.]' 'RT{print RT}'
This is sentence!
This is sentence.

或 FPAT 的 GNU awk:

$ echo "$text" | awk -v FPAT='[A-Z][^!?.]*[!?.]' '{for (i=1; i<=NF; i++) print $i}'
This is sentence!
This is sentence.

或 -o 的 GNU grep:

$ echo "$text" | grep -o '[A-Z][^!?.]*[!?.]'
This is sentence!
This is sentence.

如果一个句子可以包含换行符，则只有上述第一个有效。

如何查找并打印 bash 中的所有 AWK 匹配项？

How to find and print all AWK matches in bash?

awk

echo