Java：正则表达式可识别句子中的标点符号并将其删除

Question

I have the following string:

String input = "Remove from em?ty sentence 1? Remove from sentence 2! But not from ip address 190.168.10.110!";

I want to remove punctuation marks at the right places. My output needs to be:

String str = "Remove from em?ty sentence 1 Remove from sentence 2 But not from ip address 190.168.10.110";

I am using the following code:

while (stream.hasNext()) { 
    token = stream.next();
    char[] tokenArray = token.toCharArray();
    token = token.trim();

    if(token.matches(".*?[0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}[\\.\\?!]+")){
        System.out.println("case2");
        stream.previous();
        int len = token.length()-1;
        for(int i = token.length()-1; i>7; i--){
            if(tokenArray[i]=='.'||tokenArray[i]=='?'||tokenArray[i]=='!'){
                --len;
            }
            else
                break;
        }
        stream.set(token.substring(0, len+1));
    }
    else if(token.matches(".*?\\b[a-zA-Z_0-9]+\\b[\\.\\?!]+")){
        System.out.println("case1");
        stream.previous();
        str = token.replaceAll("[\\.\\?!]+", "");
        stream.set(str);

        System.out.println(stream.next());                          
    }
}

'Tokens' are getting sent from the 'input' string. Can you please indicate what i am doing wrong in terms of regex or the logic?

A punctuation is considered one when it ends a sentence, is not present within an ip address, not within words such as !true , emp?ty (leave them alone). Also may be followed by a space or end of string.

Answer 1

You can use this pattern:

\\p{Punct}(?=\\s|$)

and replace it with nothing.

example:

String subject = "Remove from em?ty sentence 1? Remove from sentence 2! But not from ip address 190.168.10.110!";
String regex = "\\p{Punct}(?=\\s|$)";
String result = subject.replaceAll(regex, "");
System.out.println(result);

Answer 2

String input = "Remove from em?ty sentence 1? Remove from sentence 2! But not from ip address 190.168.10.110!";
System.out.println(input.replaceAll("[?!]", ""));

Gave output:

Remove from emty sentence 1 Remove from sentence 2 But not from ip address 190.168.10.110

Answer 3

为什么不使用

string.replaceAll("[?!] ", ""));

Answer 4

I would do it the other way around.

if(token.matches("[\\.\\!\\:\\?\\;] "){
token.replace("");
}

Now, I am assuming that the punctuation marks would have a trailing space. It leaves out only the last punctuation, mark in the sentence, which you can remove separately.

Answer 5

Something like this might work. It rules out everything, then takes what punctuation is
significant to you. [,.!?]

Just replace with $1

    # ([^\pL\pN\s]*[\pL\pN](?:[\pL\pN_-]|\pP(?=[\pL\pN\pP_-]))*)|[,.!?]
    # "([^\\pL\\pN\\s]*[\\pL\\pN](?:[\\pL\\pN_-]|\\pP(?=[\\pL\\pN\\pP_-]))*)|[,.!?]"

    (                              # (1 start)
         [^\pL\pN\s]* [\pL\pN] 
         (?:
              [\pL\pN_-] 
           |  \pP 
              (?= [\pL\pN\pP_-] )
         )*
    )                              # (1 end)
 |  
    [,.!?]