我想使用 Java 验证 IPv4 地址。它应该使用点十进制表示法编写,因此它应该有 3 个点 (" .
"),没有字符,点之间有数字,并且数字应该在有效范围内。应该怎么做?
我想使用 Java 验证 IPv4 地址。它应该使用点十进制表示法编写,因此它应该有 3 个点 (" .
"),没有字符,点之间有数字,并且数字应该在有效范围内。应该怎么做?
试试 InetAddressValidator 实用程序类。
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在这里下载:
使用番石榴的
InetAddresses.isInetAddress(ipStr)
请参阅使用 apache 公共库的https://stackoverflow.com/a/5668971/1586965InetAddressValidator
或者你可以使用这个功能 -
public static boolean validate(final String ip) {
String PATTERN = "^((0|1\\d?\\d?|2[0-4]?\\d?|25[0-5]?|[3-9]\\d?)\\.){3}(0|1\\d?\\d?|2[0-4]?\\d?|25[0-5]?|[3-9]\\d?)$";
return ip.matches(PATTERN);
}
您可以使用正则表达式,如下所示:
(([0-1]?[0-9]{1,2}\.)|(2[0-4][0-9]\.)|(25[0-5]\.)){3}(([0-1]?[0-9]{1,2})|(2[0-4][0-9])|(25[0-5]))
这个验证值在范围内。
Android 支持正则表达式。请参阅java.util.regex.Pattern。
class ValidateIPV4
{
static private final String IPV4_REGEX = "(([0-1]?[0-9]{1,2}\\.)|(2[0-4][0-9]\\.)|(25[0-5]\\.)){3}(([0-1]?[0-9]{1,2})|(2[0-4][0-9])|(25[0-5]))";
static private Pattern IPV4_PATTERN = Pattern.compile(IPV4_REGEX);
public static boolean isValidIPV4(final String s)
{
return IPV4_PATTERN.matcher(s).matches();
}
}
为了避免一遍又一遍地重新编译模式,最好进行Pattern.compile()
调用,使其只执行一次。
还有一个未记录的实用程序类sun.net.util.IPAddressUtil
,您实际上不应该使用它,尽管它可能在快速一次性使用的实用程序中很有用:
boolean isIP = IPAddressUtil.isIPv4LiteralAddress(ipAddressString);
在内部,这是InetAddress
用于解析 IP 地址的实用程序类。
请注意,对于像“123”这样的字符串,这将返回 true,这些字符串在技术上是有效的 IPv4 地址,只是不是点十进制表示法。
IPAddress Java 库会做到这一点。链接中提供了 javadoc。免责声明:我是项目经理。
该库透明地支持 IPv4 和 IPv6,因此验证两者的工作方式如下相同,并且它还支持 CIDR 子网。
验证地址是否有效
String str = "1.2.3.4";
IPAddressString addrString = new IPAddressString(str);
try {
IPAddress addr = addrString.toAddress();
...
} catch(AddressStringException e) {
//e.getMessage provides validation issue
}
这适用于 Android,测试IPv4
和IPv6
注意:常用的InetAddressUtils
已弃用。使用新InetAddress
类
public static Boolean isIPv4Address(String address) {
if (address.isEmpty()) {
return false;
}
try {
Object res = InetAddress.getByName(address);
return res instanceof Inet4Address || res instanceof Inet6Address;
} catch (final UnknownHostException exception) {
return false;
}
}
编写一个合适的正则表达式并根据它进行验证。JVM 完全支持正则表达式。
如果是IP4,可以使用正则表达式如下:
^(2[0-5][0-5])|(1\\d\\d)|([1-9]?\\d)\\.){3}(2[0-5][0-5])|(1\\d\\d)|([1-9]?\\d)$
.
有很多方法可以实现,但正则表达式更有效。
看看下面的代码:
public static void main(String[] args) {
String ipStr1 = "255.245.188.123"; // valid IP address
String ipStr2 = "255.245.188.273"; // invalid IP address - 273 is greater than 255
validateIP(ipStr1);
validateIP(ipStr2);
}
public static void validateIP(String ipStr) {
String regex = "\\b((25[0–5]|2[0–4]\\d|[01]?\\d\\d?)(\\.)){3}(25[0–5]|2[0–4]\\d|[01]?\\d\\d?)\\b";
System.out.println(ipStr + " is valid? " + Pattern.matches(regex, ipStr));
}
正则表达式是解决这个问题最有效的方法。看看下面的代码。除了有效性,它还检查它所属的IP地址类别以及它是否是保留的IP地址
Pattern ipPattern;
int[] arr=new int[4];
int i=0;
//Method to check validity
private String validateIpAddress(String ipAddress) {
Matcher ipMatcher=ipPattern.matcher(ipAddress);
//Condition to check input IP format
if(ipMatcher.matches()) {
//Split input IP Address on basis of .
String[] octate=ipAddress.split("[.]");
for(String x:octate) {
//Convert String number into integer
arr[i]=Integer.parseInt(x);
i++;
}
//Check whether input is Class A IP Address or not
if(arr[0]<=127) {
if(arr[0]==0||arr[0]==127)
return(" is Reserved IP Address of Class A");
else if(arr[1]==0&&arr[2]==0&&arr[3]==0)
return(" is Class A Network address");
else if(arr[1]==255&&arr[2]==255&&arr[3]==255)
return( " is Class A Broadcast address");
else
return(" is valid IP Address of Class A");
}
//Check whether input is Class B IP Address or not
else if(arr[0]>=128&&arr[0]<=191) {
if(arr[2]==0&&arr[3]==0)
return(" is Class B Network address");
else if(arr[2]==255&&arr[3]==255)
return(" is Class B Broadcast address");
else
return(" is valid IP Address of Class B");
}
//Check whether input is Class C IP Address or not
else if(arr[0]>=192&&arr[0]<=223) {
if(arr[3]==0)
return(" is Class C Network address");
else if(arr[3]==255)
return(" is Class C Broadcast address");
else
return( " is valid IP Address of Class C");
}
//Check whether input is Class D IP Address or not
else if(arr[0]>=224&&arr[0]<=239) {
return(" is Class D IP Address Reserved for multicasting");
}
//Execute if input is Class E IP Address
else {
return(" is Class E IP Address Reserved for Research and Development by DOD");
}
}
//Input not matched with IP Address pattern
else
return(" is Invalid IP Address");
}
public static void main(String[] args) {
Scanner scan= new Scanner(System.in);
System.out.println("Enter IP Address: ");
//Input IP Address from user
String ipAddress=scan.nextLine();
scan.close();
IPAddress obj=new IPAddress();
//Regex for IP Address
obj.ipPattern=Pattern.compile("((([0-1]?\\d\\d?|2[0-4]\\d|25[0-5])\\.){3}([0-1]?\\d\\d?|2[0-4]\\d|25[0-5]))");
//Display output
System.out.println(ipAddress+ obj.validateIpAddress(ipAddress));
}
使用正则表达式在两行中获取有效的 ip 地址请检查代码的注释会话正则表达式如何工作以获取数字范围。
public class regexTest {
public static void main(String[] args) {
String IP = "255.001.001.255";
System.out.println(IP.matches(new MyRegex().pattern));
}
}
/*
* /d - stands for any number between 0 to 9
* /d{1,2} - preceding number that 0 to 9 here , can be of 1 digit to 2 digit . so minimum 0 and maximum 99
* | this stands for or operator
*
* () this is applied on a group to get the single value of outcome
* (0|1)\d{2} = first digit is either 0 or 1 and other two digits can be any number between ( 0 to 9)
* 2[0-4]\d - first digit is 2 , second digit can be between 0 to 4 and last digit can be 0 to 9
* 25[0-5] - first two digit will be 25 and last digit will be between 0 to 5
*
* */
class MyRegex {
String zeroTo255 = "(\\d{1,2}|(0|1)\\d{2}|2[0-4]\\d|25[0-5])";
public String pattern = zeroTo255 + "\\." + zeroTo255 + "\\." + zeroTo255 + "\\." + zeroTo255;;
}
请查看sun.net.util 中存在的IPAddressUtil OOTB 类,这应该对您有所帮助。
如果您不关心范围,以下表达式将有助于验证从 1.1.1.1 到 999.999.999.999
"[1-9]{1,3}\\.[1-9]{1,3}\\.[1-9]{1,3}\\.[1-9]{1,3}"
public static boolean isIpv4(String ipAddress) {
if (ipAddress == null) {
return false;
}
String ip = "^(1\\d{2}|2[0-4]\\d|25[0-5]|[1-9]\\d|[1-9])\\."
+ "(1\\d{2}|2[0-4]\\d|25[0-5]|[1-9]\\d|\\d)\\."
+ "(1\\d{2}|2[0-4]\\d|25[0-5]|[1-9]\\d|\\d)\\."
+ "(1\\d{2}|2[0-4]\\d|25[0-5]|[1-9]\\d|\\d)$";
Pattern pattern = Pattern.compile(ip);
Matcher matcher = pattern.matcher(ipAddress);
return matcher.matches();
}
/**
* Check if ip is valid
*
* @param ip to be checked
* @return <tt>true</tt> if <tt>ip</tt> is valid, otherwise <tt>false</tt>
*/
private static boolean isValid(String ip) {
String[] bits = ip.split("\\.");
if (bits.length != 4) {
return false;
}
for (String bit : bits) {
try {
if (Integer.valueOf(bit) < 0 || Integer.valueOf(bit) > 255) {
return false;
}
} catch (NumberFormatException e) {
return false; /* contains other other character */
}
}
return true;
}
apache-httpcomponents 的库
// ipv4 is true
assertTrue(InetAddressUtils.isIPv4Address("127.0.0.1"));
// not detect the ipv6
assertFalse(InetAddressUtils.isIPv4Address("2001:0db8:85a3:0000:0000:8a2e:0370:7334"));
maven lib(2019 年 9 月更新)
<!-- https://mvnrepository.com/artifact/org.apache.httpcomponents/httpclient -->
<dependency>
<groupId>org.apache.httpcomponents</groupId>
<artifactId>httpclient</artifactId>
<version>4.5.10</version>
</dependency>
我的解决方案(支持前导 0):
String pattern="^[0-9](\\d{1,2}|1?[0-9][0-9]|2?[0-4][0-9]|25?[0-5])?\\.(\\d{1,2}|1?[0-9][0-9]|2?[0-4][0-9]|25[0-5])?\\.(\\d{1,2}|1?[0-9][0-9]|2?[0-4][0-9]|25[0-5])?\\.(\\d{1,2}|1?[0-9][0-9]|2?[0-4][0-9]|25[0-5])?$";
private static final String IPV4_PATTERN_ALLOW_LEADING_ZERO =
"^([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." +
"([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." +
"([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." +
"([01]?\\d\\d?|2[0-4]\\d|25[0-5])$";
private static boolean findValidIP(String ipAddress) {
String[] ipAddressSplited = ipAddress.split("\\.");
int correctFragments = 0;
for (String ctr : ipAddressSplited) {
int value = -10;
try {
value = Integer.parseInt(ctr);
} catch (NumberFormatException exception) {
value = -1;
}
if (value >= 0 && value <= 255 && ipAddressSplited.length == 4) correctFragments++;
}
System.out.println(correctFragments == 4 ? "Valid IP Address - " + ipAddress : "Invalid IP Address - " + ipAddress);
return correctFragments == 4;
}
public void setIpAddress(String ipAddress) { if(ipAddress.matches("^(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9] [0-9]?)\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]? )$")) //ipv4的正则表达式 this.ipAddress = ipAddress; else System.out.println("IP 地址不正确"); }
使用正则表达式非常简单(但请注意 ,与使用 Apache Commons Utility 的worpet 的答案相比,这效率低得多且难以阅读)
private static final Pattern PATTERN = Pattern.compile( "^(([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\.){3}([01]?\\d\\d?|2[0-4]\\d|25[0-5])$"); public static boolean validate(final String ip) { return PATTERN.matcher(ip).matches(); }
基于后Mkyong