判断IP是否合法
2015-07-22 10:52
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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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