我需要确定给定的 IP 地址是否来自某个特殊网络,以便自动进行身份验证。
最佳答案
选项 1:
使用 spring-security-web
的IpAddressMatcher .与 Apache Commons Net 不同,它同时支持 ipv4 和 ipv6。
import org.springframework.security.web.util.matcher.IpAddressMatcher;
...
private void checkIpMatch() {
matches("192.168.2.1", "192.168.2.1"); // true
matches("192.168.2.1", "192.168.2.0/32"); // false
matches("192.168.2.5", "192.168.2.0/24"); // true
matches("92.168.2.1", "fe80:0:0:0:0:0:c0a8:1/120"); // false
matches("fe80:0:0:0:0:0:c0a8:11", "fe80:0:0:0:0:0:c0a8:1/120"); // true
matches("fe80:0:0:0:0:0:c0a8:11", "fe80:0:0:0:0:0:c0a8:1/128"); // false
matches("fe80:0:0:0:0:0:c0a8:11", "192.168.2.0/32"); // false
}
private boolean matches(String ip, String subnet) {
IpAddressMatcher ipAddressMatcher = new IpAddressMatcher(subnet);
return ipAddressMatcher.matches(ip);
}
选项 2(轻量级解决方案!):
上一部分的代码运行良好,但需要包含 spring-security-web
。
如果你不愿意在你的项目中包含 Spring 框架,你可以使用这个类,它是 original class 的略微修改版本。来自 Spring,因此它没有非 JRE 依赖项。
/*
* Copyright 2002-2019 the original author or authors.
*
* Licensed under the Apache License, Version 2.0 (the "License");
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at
*
* https://www.apache.org/licenses/LICENSE-2.0
*
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*/
import java.net.InetAddress;
import java.net.UnknownHostException;
/**
* Matches a request based on IP Address or subnet mask matching against the remote
* address.
* <p>
* Both IPv6 and IPv4 addresses are supported, but a matcher which is configured with an
* IPv4 address will never match a request which returns an IPv6 address, and vice-versa.
*
* @author Luke Taylor
* @since 3.0.2
*
* Slightly modified by omidzk to have zero dependency to any frameworks other than the JRE.
*/
public final class IpAddressMatcher {
private final int nMaskBits;
private final InetAddress requiredAddress;
/**
* Takes a specific IP address or a range specified using the IP/Netmask (e.g.
* 192.168.1.0/24 or 202.24.0.0/14).
*
* @param ipAddress the address or range of addresses from which the request must
* come.
*/
public IpAddressMatcher(String ipAddress) {
if (ipAddress.indexOf('/') > 0) {
String[] addressAndMask = ipAddress.split("/");
ipAddress = addressAndMask[0];
nMaskBits = Integer.parseInt(addressAndMask[1]);
}
else {
nMaskBits = -1;
}
requiredAddress = parseAddress(ipAddress);
assert (requiredAddress.getAddress().length * 8 >= nMaskBits) :
String.format("IP address %s is too short for bitmask of length %d",
ipAddress, nMaskBits);
}
public boolean matches(String address) {
InetAddress remoteAddress = parseAddress(address);
if (!requiredAddress.getClass().equals(remoteAddress.getClass())) {
return false;
}
if (nMaskBits < 0) {
return remoteAddress.equals(requiredAddress);
}
byte[] remAddr = remoteAddress.getAddress();
byte[] reqAddr = requiredAddress.getAddress();
int nMaskFullBytes = nMaskBits / 8;
byte finalByte = (byte) (0xFF00 >> (nMaskBits & 0x07));
// System.out.println("Mask is " + new sun.misc.HexDumpEncoder().encode(mask));
for (int i = 0; i < nMaskFullBytes; i++) {
if (remAddr[i] != reqAddr[i]) {
return false;
}
}
if (finalByte != 0) {
return (remAddr[nMaskFullBytes] & finalByte) == (reqAddr[nMaskFullBytes] & finalByte);
}
return true;
}
private InetAddress parseAddress(String address) {
try {
return InetAddress.getByName(address);
}
catch (UnknownHostException e) {
throw new IllegalArgumentException("Failed to parse address" + address, e);
}
}
}
注意:请注意,使用此选项时,您有责任仔细检查 license为确保使用此代码,您不会违反上述许可规定的任何条款。 (当然,我将这段代码发布到 Stackoverflow.com 并不违法。)
关于java - 如何检查 IP 地址是否来自 Java 中的特定网络/网络掩码?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/577363/