blocks|key|930842|text|请参阅使用apache+commons库InetAddressValidator的https://stackoverflow.com/a/5668971/1586965|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|930843|或者你可以使用这个函数-|930844|public+static+boolean+validate(final+String+ip)+{
++++String+PATTERN+=+"%5E((0%7C1\\d?\\d?%7C2[0-4]?\\d?%7C25[0-5]?%7C[3-9]\\d?)\\.){3}(0%7C1\\d?\\d?%7C2[0-4]?\\d?%7C25[0-5]?%7C[3-9]\\d?)$";

++++return+ip.matches(PATTERN);
}|code-block|syntax|javascript|930845|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/5668971/1586965^0|K|K|15|17|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|6|7|10|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|11|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

Please see <a href="https://stackoverflow.com/a/5668971/1586965">https://stackoverflow.com/a/5668971/1586965</a> which uses an apache commons library <code>InetAddressValidator</code>

Or you can use this function - 

<pre><code>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);
}
</code></pre>

blocks|key|714675|text|使用正则表达式非常简单(但请注意，这比使用Apache+Commons+Utility的worpet's+answer效率低得多，也更难阅读)|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|714676|private+static+final+Pattern+PATTERN+=+Pattern.compile(
++++++++"%5E(([01]?\\d\\d?%7C2[0-4]\\d%7C25[0-5])\\.){3}([01]?\\d\\d?%7C2[0-4]\\d%7C25[0-5])$");

public+static+boolean+validate(final+String+ip)+{
++++return+PATTERN.matcher(ip).matches();
}|code-block|syntax|javascript|714677|基于post+Mkyong|714678|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/5668971/1586965|1|http://www.mkyong.com/regular-expressions/how-to-validate-ip-address-with-regular-expression/^0|18|F|0|0|0|7|6|1|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]|$1|K|3|-4|5|6|7|13|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

Pretty simple with Regular Expression (but note this is much less efficient and much harder to read than <a href="https://stackoverflow.com/a/5668971/1586965">worpet's answer</a> that uses an Apache Commons Utility)

<pre><code>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();
}
</code></pre>

Based on post <a href="http://www.mkyong.com/regular-expressions/how-to-validate-ip-address-with-regular-expression/" rel="noreferrer">Mkyong</a>

blocks|key|930722|text|尝试使用InetAddressValidator实用程序类。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|930723|文档在此：|930724|http://commons.apache.org/validator/apidocs/org/apache/commons/validator/routines/InetAddressValidator.html|offset|length|930725|请在此处下载：|930726|http://commons.apache.org/validator/|930727|entityMap|0|LINK|mutability|MUTABLE|url|1^0|0|0|0|2Z|0|0|0|0|10|1|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|V|8|@]|9|@$F|W|G|X|1|Y]]|A|$]]|$1|H|3|I|5|6|7|Z|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|10|8|@]|9|@$F|11|G|12|1|13]]|A|$]]|$1|L|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|M|$N|$5|O|P|Q|A|$R|E]]|S|$5|O|P|Q|A|$R|K]]]]

Try the InetAddressValidator utility class.

Docs here:

<a href="http://commons.apache.org/validator/apidocs/org/apache/commons/validator/routines/InetAddressValidator.html" rel="noreferrer">http://commons.apache.org/validator/apidocs/org/apache/commons/validator/routines/InetAddressValidator.html</a>

Download here:

<a href="http://commons.apache.org/validator/" rel="noreferrer">http://commons.apache.org/validator/</a>

blocks|key|930788|text|使用Guava的|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|930789|InetAddresses.isInetAddress(ipStr)|code-block|syntax|javascript|930790|entityMap|0|LINK|mutability|MUTABLE|url|https://google.github.io/guava/releases/19.0/api/docs/com/google/common/net/InetAddresses.html^0|2|5|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Use <a href="https://google.github.io/guava/releases/19.0/api/docs/com/google/common/net/InetAddresses.html" rel="noreferrer">Guava</a>'s

<pre><code>InetAddresses.isInetAddress(ipStr)
</code></pre>

blocks|key|714639|text|您可以使用正则表达式，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|714640|(([0-1]?[0-9]{1,2}\.)%7C(2[0-4][0-9]\.)%7C(25[0-5]\.)){3}(([0-1]?[0-9]{1,2})%7C(2[0-4][0-9])%7C(25[0-5]))|code-block|syntax|javascript|714641|这一项验证值在范围内。|714642|Android支持正则表达式。参见java.util.regex.Pattern。|offset|length|714643|class+ValidateIPV4
{

+++static+private+final+String+IPV4_REGEX+=+"(([0-1]?[0-9]{1,2}\\.)%7C(2[0-4][0-9]\\.)%7C(25[0-5]\\.)){3}(([0-1]?[0-9]{1,2})%7C(2[0-4][0-9])%7C(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();
+++}
}|714644|为了避免一遍又一遍地重新编译模式，最好放置Pattern.compile()调用，以便它只执行一次。|style|CODE|714645|entityMap|0|LINK|mutability|MUTABLE|url|http://developer.android.com/reference/java/util/regex/Pattern.html^0|0|0|0|H|N|0|0|0|L|H|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|13|8|@]|9|@$K|14|L|15|1|16]]|A|$]]|$1|M|3|N|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|18|8|@$K|19|L|1A|Q|R]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

You can use a regex, like this:

<pre><code>(([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]))
</code></pre>

This one validates the values are within range.

Android has support for regular expressions. See <a href="http://developer.android.com/reference/java/util/regex/Pattern.html" rel="noreferrer">java.util.regex.Pattern</a>.

<pre><code>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();
 }
}
</code></pre>

To avoid recompiling the pattern over and over, it's best to place the <code>Pattern.compile()</code> call so that it is executed only once.

blocks|key|930764|text|还有一个没有文档记录的实用程序类sun.net.util.IPAddressUtil，它是you+should+not+actually+use，尽管它在快速的一次性、一次性实用程序中可能很有用：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|930765|boolean+isIP+=+IPAddressUtil.isIPv4LiteralAddress(ipAddressString);|code-block|syntax|javascript|930766|在内部，这是InetAddress用于解析IP地址的实用程序类。|930767|注意，对于像"123“这样的字符串，这将返回true，从技术上讲，它们是valid+IPv4+addresses，只是不是点十进制表示法。|930768|entityMap|0|LINK|mutability|MUTABLE|url|http://www.oracle.com/technetwork/java/faq-sun-packages-142232.html|1|http://en.wikipedia.org/wiki/IPv4#Address_representations^0|G|Q|19|R|0|0|0|6|B|0|10|K|1|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|F|3|G|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|18|8|@]|D|@$9|19|A|1A|1|1B]]|E|$]]|$1|O|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]]]

There is also an undocumented utility class <code>sun.net.util.IPAddressUtil</code>, which <a href="http://www.oracle.com/technetwork/java/faq-sun-packages-142232.html" rel="noreferrer">you should not actually use</a>, although it might be useful in a quick one-off, throw-away utility:

<pre><code>boolean isIP = IPAddressUtil.isIPv4LiteralAddress(ipAddressString);
</code></pre>

Internally, this is the utility class <code>InetAddress</code> uses to parse IP addresses. 

Note that this will return true for strings like "123", which, technically are <a href="http://en.wikipedia.org/wiki/IPv4#Address_representations" rel="noreferrer">valid IPv4 addresses</a>, just not in dot-decimal notation.

blocks|key|930891|text|这是针对安卓的，针对IPv4和IPv6的测试|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|930892|注意:常用的InetAddressUtils已被弃用。使用新的InetAddress类|930893|public+static+Boolean+isIPv4Address(String+address)+{
++++if+(address.isEmpty())+{
++++++++return+false;
++++}
++++try+{
++++++++Object+res+=+InetAddress.getByName(address);
++++++++return+res+instanceof+Inet4Address+%7C%7C+res+instanceof+Inet6Address;
++++}+catch+(final+UnknownHostException+exception)+{
++++++++return+false;
++++}
}|code-block|syntax|javascript|930894|entityMap^0|A|4|F|4|0|6|G|V|B|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

This is for Android, testing for <code>IPv4</code> and <code>IPv6</code>
Note: the commonly used <code>InetAddressUtils</code> is deprecated. Use new <code>InetAddress</code> classes
<pre><code>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;
 }
}
</code></pre>

blocks|key|931191|text|The+IPAddress+Java+library将做到这一点。链接中提供了javadoc。免责声明:我是项目经理。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|931192|此库透明地支持IPv4和IPv6，因此验证两者的工作方式如下所示，它还支持CIDR子网。|931193|验证地址是否有效|931194|++++String+str+=+"1.2.3.4";
++++IPAddressString+addrString+=+new+IPAddressString(str);
++++try+{
+++++++++IPAddress+addr+=+addrString.toAddress();
+++++++++...
++++}+catch(AddressStringException+e)+{
++++++++//e.getMessage+provides+validation+issue
++++}|code-block|syntax|javascript|931195|entityMap|0|LINK|mutability|MUTABLE|url|https://seancfoley.github.io/IPAddress/^0|0|Q|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@]|C|$]]|$1|F|3|G|5|6|7|Z|8|@]|9|@]|C|$]]|$1|H|3|I|5|J|7|10|8|@]|9|@]|C|$K|L]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

<a href="https://seancfoley.github.io/IPAddress/" rel="nofollow noreferrer">The IPAddress Java library</a> will do it. The javadoc is available at the link. Disclaimer: I am the project manager.

This library supports IPv4 and IPv6 transparently, so validating either works the same below, and it also supports CIDR subnets as well.

Verify if an address is valid

<pre><code> String str = "1.2.3.4";
 IPAddressString addrString = new IPAddressString(str);
 try {
 IPAddress addr = addrString.toAddress();
 ...
 } catch(AddressStringException e) {
 //e.getMessage provides validation issue
 }
</code></pre>

blocks|key|930577|text|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。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|930578|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Write up a suitable regular expression and validate it against that. The JVM have full support for regular expressions.

blocks|key|930700|text|如果是IP4，您可以使用正则表达式，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|930701|%5E(2[0-5][0-5])%7C(1\\d\\d)%7C([1-9]?\\d)\\.){3}(2[0-5][0-5])%7C(1\\d\\d)%7C([1-9]?\\d)$。|offset|length|style|CODE|930702|entityMap^0|0|0|27|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

If it is IP4, you can use a regular expression as follows:

<code>^(2[0-5][0-5])|(1\\d\\d)|([1-9]?\\d)\\.){3}(2[0-5][0-5])|(1\\d\\d)|([1-9]?\\d)$</code>.

blocks|key|931135|text|有很多方法可以做到这一点，但正则表达式效率更高。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|931136|请看下面的代码：|931137|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]%7C2[0–4]\\d%7C[01]?\\d\\d?)(\\.)){3}(25[0–5]%7C2[0–4]\\d%7C[01]?\\d\\d?)\\b";
++++System.out.println(ipStr+%2B+"+is+valid?+"+%2B+Pattern.matches(regex,+ipStr));
}|code-block|syntax|javascript|931138|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

There are so many ways to achieve that,but regular expression is more efficient.

Look at the code below:

<pre><code>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));
}
</code></pre>

blocks|key|715283|text|正则表达式是解决这个问题的最有效的方法。请看下面的代码。在有效性的同时，它还检查它所属的IP地址类别以及它是否为保留的IP地址|type|unstyled|depth|inlineStyleRanges|entityRanges|data|715284|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%2B%2B;
+++++++++}

++++++++//Check+whether+input+is+Class+A+IP+Address+or+not
+++++++++if(arr[0]<=127)+{++++++++++++++++++++++++++
+++++++++++++if(arr[0]==0%7C%7Carr[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?%7C2[0-4]\\d%7C25[0-5])\\.){3}([0-1]?\\d\\d?%7C2[0-4]\\d%7C25[0-5]))");

++++//Display+output
++++System.out.println(ipAddress%2B+obj.validateIpAddress(ipAddress));

}|code-block|syntax|javascript|715285|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Regular Expression is the most efficient way to solve this problem.
Look at the code below. Along validity, it also checks IP address class in which it belongs and whether it is reserved IP Address or not

<pre><code>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]&lt;=127) { 
 if(arr[0]==0||arr[0]==127)
 return(" is Reserved IP Address of Class A");
 else if(arr[1]==0&amp;&amp;arr[2]==0&amp;&amp;arr[3]==0)
 return(" is Class A Network address");
 else if(arr[1]==255&amp;&amp;arr[2]==255&amp;&amp;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]&gt;=128&amp;&amp;arr[0]&lt;=191) { 
 if(arr[2]==0&amp;&amp;arr[3]==0)
 return(" is Class B Network address");
 else if(arr[2]==255&amp;&amp;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]&gt;=192&amp;&amp;arr[0]&lt;=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]&gt;=224&amp;&amp;arr[0]&lt;=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));

}
</code></pre>

blocks|key|931083|text|如果您不关心范围，则以下表达式对于从1.1.1.1到999.999.999.999的验证非常有用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|931084|"[1-9]{1,3}\\.[1-9]{1,3}\\.[1-9]{1,3}\\.[1-9]{1,3}"|code-block|syntax|javascript|931085|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you don care about the range, the following expression will be useful to validate from 1.1.1.1 to 999.999.999.999

<pre><code>"[1-9]{1,3}\\.[1-9]{1,3}\\.[1-9]{1,3}\\.[1-9]{1,3}"
</code></pre>

blocks|key|931223|text|public+static+boolean+isIpv4(String+ipAddress)+{
++++if+(ipAddress+==+null)+{
++++++++return+false;
++++}
++++String+ip+=+"%5E(1\\d{2}%7C2[0-4]\\d%7C25[0-5]%7C[1-9]\\d%7C[1-9])\\."
++++++++++++%2B+"(1\\d{2}%7C2[0-4]\\d%7C25[0-5]%7C[1-9]\\d%7C\\d)\\."
++++++++++++%2B+"(1\\d{2}%7C2[0-4]\\d%7C25[0-5]%7C[1-9]\\d%7C\\d)\\."
++++++++++++%2B+"(1\\d{2}%7C2[0-4]\\d%7C25[0-5]%7C[1-9]\\d%7C\\d)$";
++++Pattern+pattern+=+Pattern.compile(ip);
++++Matcher+matcher+=+pattern.matcher(ipAddress);
++++return+matcher.matches();
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|931224|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>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();
}
</code></pre>

blocks|key|931367|text|/**
+*+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+%7C%7C+Integer.valueOf(bit)+>+255)+{
++++++++++++++++return+false;
++++++++++++}
++++++++}+catch+(NumberFormatException+e)+{
++++++++++++return+false;+/*+contains+other+other+character+*/
++++++++}
++++}
++++return+true;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|931368|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>/**
 * Check if ip is valid
 *
 * @param ip to be checked
 * @return &lt;tt&gt;true&lt;/tt&gt; if &lt;tt&gt;ip&lt;/tt&gt; is valid, otherwise &lt;tt&gt;false&lt;/tt&gt;
 */
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) &lt; 0 || Integer.valueOf(bit) &gt; 255) {
 return false;
 }
 } catch (NumberFormatException e) {
 return false; /* contains other other character */
 }
 }
 return true;
}
</code></pre>

blocks|key|931390|text|apache-httpcomponent的库|type|unstyled|depth|inlineStyleRanges|entityRanges|data|931391|//+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"));|code-block|syntax|javascript|931392|maven+lib+(2019年9月更新)|931393|
<dependency>
++++<groupId>org.apache.httpcomponents</groupId>
++++<artifactId>httpclient</artifactId>
++++<version>4.5.10</version>
</dependency>|931394|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

the lib of apache-httpcomponents

<pre class="lang-java prettyprint-override"><code>// 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"));
</code></pre>

maven lib (update Sep, 2019)

<pre class="lang-xml prettyprint-override"><code>&lt;!-- https://mvnrepository.com/artifact/org.apache.httpcomponents/httpclient --&gt;
&lt;dependency&gt;
 &lt;groupId&gt;org.apache.httpcomponents&lt;/groupId&gt;
 &lt;artifactId&gt;httpclient&lt;/artifactId&gt;
 &lt;version&gt;4.5.10&lt;/version&gt;
&lt;/dependency&gt;
</code></pre>

blocks|key|715414|text|我的解决方案(支持前导0)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|715415|+++String+pattern="%5E[0-9](\\d{1,2}%7C1?[0-9][0-9]%7C2?[0-4][0-9]%7C25?[0-5])?\\.(\\d{1,2}%7C1?[0-9][0-9]%7C2?[0-4][0-9]%7C25[0-5])?\\.(\\d{1,2}%7C1?[0-9][0-9]%7C2?[0-4][0-9]%7C25[0-5])?\\.(\\d{1,2}%7C1?[0-9][0-9]%7C2?[0-4][0-9]%7C25[0-5])?$";|code-block|syntax|javascript|715416|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

my solution (supports leading 0s):
<pre><code> String pattern=&quot;^[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])?$&quot;;
</code></pre>

blocks|key|931440|text|+private+static+final+String+IPV4_PATTERN_ALLOW_LEADING_ZERO+=
++++++++++++"%5E([01]?\\d\\d?%7C2[0-4]\\d%7C25[0-5])\\."+%2B
++++++++++++"([01]?\\d\\d?%7C2[0-4]\\d%7C25[0-5])\\."+%2B
++++++++++++"([01]?\\d\\d?%7C2[0-4]\\d%7C25[0-5])\\."+%2B
++++++++++++"([01]?\\d\\d?%7C2[0-4]\\d%7C25[0-5])$";|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|931441|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> private static final String IPV4_PATTERN_ALLOW_LEADING_ZERO =
 &quot;^([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\.&quot; +
 &quot;([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\.&quot; +
 &quot;([01]?\\d\\d?|2[0-4]\\d|25[0-5])\\.&quot; +
 &quot;([01]?\\d\\d?|2[0-4]\\d|25[0-5])$&quot;;
</code></pre>

blocks|key|715439|text|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%2B%2B;
++++++++}

++++++++System.out.println(correctFragments+==+4+?+"Valid+IP+Address+-+"+%2B+ipAddress+:+"Invalid+IP+Address+-+"+%2B+ipAddress);
++++++++return+correctFragments+==+4;
++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|715440|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>private static boolean findValidIP(String ipAddress) {
 String[] ipAddressSplited = ipAddress.split(&quot;\\.&quot;);
 int correctFragments = 0;

 for (String ctr : ipAddressSplited) {
 int value = -10;
 try {
 value = Integer.parseInt(ctr);
 } catch (NumberFormatException exception) {
 value = -1;
 }

 if (value &gt;= 0 &amp;&amp; value &lt;= 255 &amp;&amp; ipAddressSplited.length == 4) correctFragments++;
 }

 System.out.println(correctFragments == 4 ? &quot;Valid IP Address - &quot; + ipAddress : &quot;Invalid IP Address - &quot; + ipAddress);
 return correctFragments == 4;
 }
</code></pre>

blocks|key|931362|text|if(ipAddress.matches("%5E(?:(?:250-5%7C20-4%7C010-9?).){3}(?:250-5%7C20-4%7C010-9?)$"))+setIpAddress(String+ipAddress)+{+public+void+ipv4+(String+this.ipAddress+){public+void+this.ipAddress=+ipAddress；else+System.out.println(“错误的IpAddress")；}|type|unstyled|depth|inlineStyleRanges|entityRanges|data|931363|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

public void setIpAddress(String ipAddress) {
if(ipAddress.matches(&quot;^(?:(?: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]?)$&quot;)) //regular expression for ipv4
this.ipAddress = ipAddress;
else
System.out.println(&quot;incorrect IpAddress&quot;);
}

I want to validate an IPv4 address using Java. It should be written using the <a href="http://en.wikipedia.org/wiki/Dotted_decimal" rel="noreferrer">dot-decimal notation</a>, so it should have 3 dots ("<code>.</code>"), no characters, numbers in between the dots, and numbers should be in a valid range. How should it be done?

Validate IPv4 address in Java

我想用Java验证一个IPv4地址。它应该使用编写，所以它应该有3个点(".")，没有字符，点之间的数字，并且数字应该在有效的范围内。应该怎么做呢？

问在Java中验证IPv4地址
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回答 20

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