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判断IP是否合法

2015-07-22 10:52 931 查看
现在不需要自己写正则表达式去匹配IP是否合法,只需要调用一个简单的函数就可以解决问题。

String ip = "xxx.xxx.xxx.xxx";
boolean isIpString = InetAddress.isNumeric(ip);
返回true,表示合法;返回false,表示不是合法的IP。

/**
     * Returns true if the string is a valid numeric IPv4 or IPv6 address (such as "192.168.0.1").
     * This copes with all forms of address that Java supports, detailed in the {@link InetAddress}
     * class documentation.
     *
     * @hide used by frameworks/base to ensure that a getAllByName won't cause a DNS lookup.
     */
    public static boolean isNumeric(String address) {
        InetAddress inetAddress = parseNumericAddressNoThrow(address);
        return inetAddress != null && disallowDeprecatedFormats(address, inetAddress) != null;
    }

 private static InetAddress parseNumericAddressNoThrow(String address) {
        // Accept IPv6 addresses (only) in square brackets for compatibility.
        if (address.startsWith("[") && address.endsWith("]") && address.indexOf(':') != -1) {
            address = address.substring(1, address.length() - 1);
        }
        StructAddrinfo hints = new StructAddrinfo();
        hints.ai_flags = AI_NUMERICHOST;
        InetAddress[] addresses = null;
        try {
            addresses = Libcore.os.getaddrinfo(address, hints);
        } catch (GaiException ignored) {
        }
        return (addresses != null) ? addresses[0] : null;
    }

 private static InetAddress disallowDeprecatedFormats(String address, InetAddress inetAddress) {
        // Only IPv4 addresses are problematic.
        if (!(inetAddress instanceof Inet4Address) || address.indexOf(':') != -1) {
            return inetAddress;
        }
        // If inet_pton(3) can't parse it, it must have been a deprecated format.
        // We need to return inet_pton(3)'s result to ensure that numbers assumed to be octal
        // by getaddrinfo(3) are reinterpreted by inet_pton(3) as decimal.
        return Libcore.os.inet_pton(AF_INET, address);
    }
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