Java Regex程序不适用于所有输入

Question

这是我目前的计划。它适用于给定的字符串INPUT。 但不适合其他投入。我希望它能为其他人工作，如果我＆lt; 30然后它也应该从中检测文字，关键字和变量。

请指导我错误并帮我纠正。 提前谢谢。

这是完整的课程。

import java.util.*;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

import javax.swing.JLabel;

public class Test {

    static String[] keyword = {"abstract", "assert", "boolean", "break", "byte", "case", "catch", "char", "class", "continue",
        "default", "do", "double", "else", "enum", "extends", "final", "finally", "float", "for", "if",
        "implements", "import", "instanceof", "int", "interface", "long", "native", "new", "package", "private",
        "protected", "public", "return", "short", "static", "string", "strictfp", "super", "switch", "this",
        "throw", "throws", "transient", "try", "void", "volatile", "while"};

    static boolean labelsrun = false;
    private static String REGEX = "((?=[a-z_0-9]|[a-z])[a-z_0-9]+(?=\\s*=))";
    private static String INPUT = "public static void main(String[]args)\n" + "{\n" + "    String a=\"Malik\";\n"
            + "    int i=1;\n" + "    double j=2;\n" + "}";

    static JLabel identifiers = new JLabel(""), literals = new JLabel(""), keywords = new JLabel("");

    public static void check_Identifiers(String a[]) {
        int count = 0;
        for (int i = 2; i < a.length; i++) {
            Pattern p = Pattern.compile(REGEX, Pattern.CASE_INSENSITIVE);
            Matcher m = p.matcher(a[i]);

            if (m.find()) {
                if (count == 0) {
                    identifiers.setText(identifiers.getText() + " " + m.group());
                    count++;
                } else if (count > 0) {
                    identifiers.setText(identifiers.getText() + ", " + m.group());
                }

            }
        }
        count = 0;
    }

    public static void check_Literals(String a[]) {
        int count = 0;
        String regex = "=.*";
        String b = "";

        for (int i = 2; i < a.length; i++) {
            Pattern p = Pattern.compile(regex);
            Matcher m = p.matcher(a[i]);

            if (m.find()) {
                if (m.group().contains("=")) {
                    b = m.group();
                    b = b.replaceAll("=", "");
                    b = b.replace(";", "");
                }
                if (count == 0) {
                    literals.setText(literals.getText() + " " + b);
                    count++;
                } else if (count > 0) {
                    literals.setText(literals.getText() + ", " + b);
                }
            }
        }
        count = 0;
    }

    public static void check_Keywords(String a[]) {
        int count = 0;
        for (int i = 0; i < a.length; i++) {
            for (int j = 0; j < keyword.length; j++) {
                String regex = ".*\\b" + keyword[j] + "\\b.*";
                if (a[i].matches(regex)) {
                    if (count == 0) {
                        keywords.setText(keywords.getText() + " " + keyword[j]);
                        count++;
                    } else if (count > 0) {
                        keywords.setText(keywords.getText() + ", " + keyword[j]);
                    }

                }
            }
        }
        count = 0;
    }

    public static void main(String[] args) {
        String a[] = INPUT.toLowerCase().split("\n");
        check_Identifiers(a);
        check_Literals(a);
        check_Keywords(a);
    }
}

这是有问题的方法。目前，仅当输入具有=运算符时才有效。但是，它如何使它适用于所有运营商？或者至少> ＆LT;与=

一起

public void check_Literals(String a[]) {
    int count = 0;
    String regex = "=.*";
    String b = "";

    for (int i = 2; i < a.length; i++) {
        Pattern p = Pattern.compile(regex);
        Matcher m = p.matcher(a[i]);

        if (m.find()) {
            if (m.group().contains("=")) {
                b = m.group();
                b = b.replaceAll("=", "");
                b = b.replace(";", "");
            }
            if (count == 0) {

                literals.setText(literals.getText() + " " + b);
                count++;
            } else if (count > 0) {
                literals.setText(literals.getText() + ", " + b);
            }
        }
    }
    count = 0;
}