使用通配符验证IP地址

Question

String ip = "1.1.&.&";
String WILDCARD_CHARACTER = "&";
String REGEX_IP_ADDRESS_STRING = "(?:(?:"
        + WILDCARD_CHARACTER
        + "|25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\.){3}(?:"
        + WILDCARD_CHARACTER + "|25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)";
Pattern p = Pattern.compile(REGEX_IP_ADDRESS_STRING1);
Matcher m = p.matcher(ip);
System.out.println("Does it match? " + m.matches());

使用上面编码的IP验证工作完美.但我想对通配符进行一些修改,这会导致问题.

目前的情况:

• 192.1.&.&------>真的
• 192.1.0.1 ------>真的
• &------>错误
• 192.1.&------>错误

预期:

• 192.1.&.&------>错误
• 192.1.0.1 ------>真的
• &------>真的
• 192.1.&------>真的

即我想在通配符之后通配所有输入.

正则表达式的哪些修改可以帮助我实现这一目标？任何人都可以帮忙吗？

Answer 1

我建议如下(我&在这个正则表达式中使用了一个字面值;当然你可以将它改为你的+ WILDCARD_CHARACTER构造):

Pattern regex = Pattern.compile(
    "^       # Anchor the match at the start of the string\n" +
    "(?:     # Match either...\n" +
    " &      # the wildcard character\n" +
    " |      # or a number between 0 and 255\n" +
    " (?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\n" +
    " \\.    # followed by a dot, followed by...\n" +
    " (?:    # ...either...\n" +
    "  &     # the wildcard character\n" +
    "  |     # or a number etc. etc.\n" +
    "  (?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\n" +
    "  \\.\n" +
    "  (?:\n" +
    "   &\n" +
    "   |\n" +
    "   (?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\n" +
    "   \\.\n" +
    "   (?:\n" +
    "    &\n" +
    "    |\n" +
    "    (?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\n" +
    "   )\n" +
    "  )\n" +
    " )\n" +
    ")\n" +
    "$        # Anchor the match at the end of the string", 
    Pattern.COMMENTS);