Java Pattern/ Matcher

问题

This is a sample text: \1f\1e\1d\020028. I cannot modify the input text, I am reading long string of texts from a file.

---

I want to extract the following: \1f, \1e, \1d, \02

For this, I have written the following regular expression pattern: "\\[a-fA-F0-9]"

I am using Pattern and Matcher classes, but my matcher is not able find the pattern using the mentioned regular expression. I have tested this regex with the text on some online regex websites and surprisingly it works there.

Where am I going wrong?

Original code:

public static void main(String[] args) {
    String inputText = "\1f\1e\1d\02002868BF03030000000000000000S023\1f\1e\1d\03\0d";
    inputText        = inputText.replace("\\", "\\\\");

    String regex     = "\\\\[a-fA-F0-9]{2}";

    Pattern p = Pattern.compile(regex);
    Matcher m = p.matcher(inputText);

    while (m.find()) {
        System.out.println(m.group());
    }
}

Output: Nothing is printed

回答1:

_{(answer changed after OP added more details)}

Your string

String inputText = "\1f\1e\1d\02002868BF03030000000000000000S023\1f\1e\1d\03\0d";

Doesn't actually contains any \ literals because according to Java Language Specification in section 3.10.6. Escape Sequences for Character and String Literals \xxx will be interpreted as character indexed in Unicode Table with octal (base/radix 8) value represented by xxx part.

Example \123 = 1*8² + 2*8¹ + 3*8⁰ = 1*64 + 2*8 + 3*1 = 64+16+3 = 83 which represents character S

If string you presented in your question is written exactly the same in your text file then you should write it as

String inputText = "\\1f\\1e\\1d\\02002868BF03030000000000000000S023\\1f\\1e\\1d\\03\\0d";

(with escaped \ which now will represent literal).

---

(older version of my answer)

It is hard to tell what exactly you did wrong without seeing your code. You should be able to find at least \1, \1, \1, \0 since your regex can match one \ and one hexadecimal character placed after it.

Anyway this is how you can find results you mentioned in question:

String text = "\\1f\\1e\\1d\\020028";
Pattern p = Pattern.compile("\\\\[a-fA-F0-9]{2}");
//                                          ^^^--we want to find two hexadecimal 
//                                               characters after \
Matcher m = p.matcher(text);
while (m.find())
    System.out.println(m.group());

Output:

\1f
\1e
\1d
\02

回答2:

You need to read the file properly and replace '\' characters with '\\'. Assume that there is file called test_file in your project with this content:

\1f\1e\1d\02002868BF03030000000000000000S023\1f\1e\1d\03\0d

Here is the code to read the file and extract values:

public static void main(String[] args) throws IOException, URISyntaxException {        
    Test t = new Test();
    t.test();
}

public void test() throws IOException {        
    BufferedReader br =
        new BufferedReader(
            new InputStreamReader(
                getClass().getResourceAsStream("/test_file.txt"), "UTF-8"));
    String inputText;

    while ((inputText = br.readLine()) != null) {
        inputText = inputText.replace("\\", "\\\\");

        Pattern pattern = Pattern.compile("\\\\[a-fA-F0-9]{2}");
        Matcher match = pattern.matcher(inputText);

        while (match.find()) {
            System.out.println(match.group());
        }
    }
}

回答3:

Try adding a . at the end, like:

\\[a-fA-F0-9].

回答4:

If you don't want to modify the input string, you could try something like:

static public void main(String[] argv) {

            String s = "\1f\1e\1d\020028";
            Pattern regex = Pattern.compile("[\\x00-\\x1f][0-9A-Fa-f]");
            Matcher match = regex.matcher(s);
            while (match.find()) {
                    char[] c = match.group().toCharArray();
                    System.out.println(String.format("\\%d%s",c[0]+0, c[1])) ;
            }
    }

Yes, it's not perfect, but you get the idea.

来源：https://stackoverflow.com/questions/26767866/java-pattern-matcher