在Java中验证IPv4地址
我想使用Java验证IPv4地址。应该使用编写，因此应该有3个点（“在Java中验证IPv4地址,java,ip-address,Java,Ip Address,我想使用Java验证IPv4地址。应该使用编写，因此应该有3个点（“”），点之间没有字符、数字，并且数字应该在有效范围内。应该怎么做？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。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])) 这一个验证值是否在范围内 Android支持正则表达式。看 为了
(([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]))
为了避免反复编译模式，最好放置
pattern.compile（）基于post如果它是IP4，则可以使用正则表达式，如下所示：
^（2[0-5][0-5]）|（1\\d\\d）|（[1-9]？\\d）\\）{3}（2[0-5][0-5]）|（1\\d\\d）|（[1-9]？\\d）$请查看sun.net.util中的IPAddressUtil OOTB类，这应该会对您有所帮助 还有一个未记录的实用程序类
sun.net.util.IPAddressUtilboolean isIP = IPAddressUtil.isIPv4LiteralAddress(ipAddressString);
InetAddressInetAddressValidatorpublic static boolean validate(final String ip) {
String PATTERN = "^((0|1\\d?\\d?|2[0-4]?\\d?|25[0-5]?|[3-9]\\d?)\\.){3}(0|1\\d?\\d?|2[0-4]?\\d?|25[0-5]?|[3-9]\\d?)$";
return ip.matches(PATTERN);
}
这是针对Android的，测试的是
IPv4IPv6InetAddressUtilsInetAddresspublic 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;
}
}
如果不关心范围，则以下表达式将有助于从1.1.1.1验证到999.999.999.999
"[1-9]{1,3}\\.[1-9]{1,3}\\.[1-9]{1,3}\\.[1-9]{1,3}"
有很多方法可以实现这一点，但正则表达式更有效 请看下面的代码：
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));
}
String str = "1.2.3.4";
IPAddressString addrString = new IPAddressString(str);
try {
IPAddress addr = addrString.toAddress();
...
} catch(AddressStringException e) {
//e.getMessage provides validation issue
}
正则表达式是解决这个问题最有效的方法。 请看下面的代码。除了有效性，它还检查它所属的IP地址类，以及它是否为保留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));
}
模式i模式；
int[]arr=新的int[4]；
int i=0；
//检验有效性的方法
私有字符串验证ipAddress（字符串ipAddress）{
Matcher ipMatcher=ipPattern.Matcher（ipAddress）；
//检查输入IP格式的条件
如果（ipMatcher.matches（））{
//基于的拆分输入IP地址。
字符串[]octate=ipAddress.split（“[.]”）；
对于（字符串x:octate）{
//将字符串数转换为整数
arr[i]=整数.parseInt（x）；
i++；
}
//检查输入是否为A类IP地址
如果（arr[0]=128&&arr[0]=192&&arr[0]=224&&arr[0]public void setIpAddress（字符串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；
其他的
System.out.println（“不正确的IP地址”）；
}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;;
}
/**
*检查ip是否有效
*
*@param ip待检查
*@如果ip有效，则返回true，否则返回false
*/
私有静态布尔值有效（字符串ip）{
字符串[]位=ip.split（“\\”）；
如果（位长度！=4）{
返回false；
}
for（字符串位：位）{
试一试{
if（整型.valueOf（位）<0 | |整型.valueOf（位）>255）{
返回false；
}
}捕获（数字格式）{
返回false；/*包含其他字符*/
}
}
返回true；
}
apache httpcomponents的库
//ipv4是真的
资产真实（InetAddressUtils.isIPv4Address（“127.0.0.1”）；
//未检测到ipv6
assertFalse（InetAddressUtils.isIPv4Address（“2001:0db8:85a3:0000:0000:8a2e:0370:7334”）；
maven lib（2019年9月更新）
org.apache.httpcomponents
httpclient
4.5.10
我的解决方案（支持领先的0）：
DVM也支持它吗？我正在使用AndroidOk，我会再次检查它…！！它将匹配“001.xxx.xxx.xxx.xxx”。我不确定，所以发布了一个表达式，明确限制0xx，0x
。静态私有最终字符串IPV4\u REGEX=“（[0-1]？[0-9]{1,2}）（2[0-4][0-9]\）（25[0-5]\）{3}（[0-1]？[0-9]{1,2}[0-5]）这一行给了我错误：（@khachik有点晚了，但是：那是一个。事实上，RFC1166在第5页给出了128.009.000.032
作为一个例子。当有4个零时它有效吗？哦，它在android上不起作用。当应用程序启动时，它给出了强制关闭的错误。那么奇怪格式的IP地址呢，比如127.1（这相当于127.0.0.1）？IPv6呢？您只需编写：return ip！=null&&ip.matches（模式）正则表达式的一个潜在问题是，它似乎与010.020.030.040之类的内容相匹配，但这可能会有问题，因为以0开头的数字通常会被解释为八进制数。没错！这忽略了前面的0大小写。@Akarshit Wal的解决方案效果更好。最好为请注意，并非所有技术上有效的IP地址符号都具有t
String str = "1.2.3.4";
IPAddressString addrString = new IPAddressString(str);
try {
IPAddress addr = addrString.toAddress();
...
} catch(AddressStringException e) {
//e.getMessage provides validation issue
}
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();
}
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));
}
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;;
}
/**
* 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;
}
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])?$";
private static final String IPV4_PATTERN_ALLOW_LEADING_ZERO =
"^([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." +
"([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." +
"([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\." +
"([01]?\\d\\d?|2[0-4]\\d|25[0-5])$";