[英]How to find and print all AWK matches in bash?
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."
我正在通过这个命令寻找句子:
echo $text | awk 'match($0,/([A-Z])([^!?.]*)([!?.])/) { print substr($0,RSTART,RLENGTH) }'
我的输出是:
This is sentence!
预期输出:
This is sentence!
This is sentence.
更多示例:文本中有语法正确和错误的句子。 正确的句子由开头的大写字母和结尾字符 (.?!) 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.
我能够找到第一个匹配项,但不是全部。 感谢您的帮助 :)
您可以将 GNU awk 用于多字符 RS:
$ echo "$text" | awk -v RS='[A-Z][^!?.]*[!?.]' 'RT{print RT}'
This is sentence!
This is sentence.
或用于 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 GNU grep for -o
:
$ echo "$text" | grep -o '[A-Z][^!?.]*[!?.]'
This is sentence!
This is sentence.
如果句子可以包含换行符,则只有上述第一个才有效。
你需要一个while()
和match()
:
$ echo $text | awk '
{
while(match($0,/([A-Z])([^!?.]*)([!?.])/)) { # while there are matches
print substr($0,RSTART,RLENGTH) # output them
$0=substr($0,RSTART+RLENGTH) # and move forward
}
}'
输出:
This is sentence!
This is sentence.
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