以编程方式获取网关和子网掩码详细信息

Question

如何在 Android 中获取网关和子网掩码详细信息？

Answer 1

我在 android.net 包中找到了一个名为 DhcpInfo 的类。它有一些公共变量存储当前网络参数的值。但问题是它们返回的是从 8 位移位二进制转换而来的整数值。

描述场景的示例图像：

****这里是示例代码：**

**java 文件:****

package com.schogini.dhcp;

import android.app.Activity;
import android.content.Context;
import android.os.Bundle;
import android.widget.TextView;
import android.net.*;
import android.net.wifi.WifiManager;

public class dhcpInfo extends Activity {
    public String   s_dns1 ;
    public String   s_dns2;     
    public String   s_gateway;  
    public String   s_ipAddress;    
    public String   s_leaseDuration;    
    public String   s_netmask;  
    public String   s_serverAddress;
    TextView info;
    DhcpInfo d;
    WifiManager wifii;

    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
        wifii= (WifiManager) getSystemService(Context.WIFI_SERVICE);
        d=wifii.getDhcpInfo();

        s_dns1="DNS 1: "+String.valueOf(d.dns1);
        s_dns2="DNS 2: "+String.valueOf(d.dns2);    
        s_gateway="Default Gateway: "+String.valueOf(d.gateway);    
        s_ipAddress="IP Address: "+String.valueOf(d.ipAddress); 
        s_leaseDuration="Lease Time: "+String.valueOf(d.leaseDuration);     
        s_netmask="Subnet Mask: "+String.valueOf(d.netmask);    
        s_serverAddress="Server IP: "+String.valueOf(d.serverAddress);

        //dispaly them
        info= (TextView) findViewById(R.id.infolbl);
        info.setText("Network Info\n"+s_dns1+"\n"+s_dns2+"\n"+s_gateway+"\n"+s_ipAddress+"\n"+s_leaseDuration+"\n"+s_netmask+"\n"+s_serverAddress);
    }
}

xml 编码：

<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
      package="com.schogini.dhcp"
      android:versionCode="1"
      android:versionName="1.0">
    <uses-sdk android:minSdkVersion="4" />

    <application android:icon="@drawable/icon" android:label="@string/app_name">
        <activity android:name=".dhcpInfo"
                  android:label="@string/app_name">
            <intent-filter>
                <action android:name="android.intent.action.MAIN" />
                <category android:name="android.intent.category.LAUNCHER" />
            </intent-filter>
        </activity>
    </application>
    <uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />
    <uses-permission android:name="android.permission.ACCESS_WIFI_STATE" />  
</manifest>

我尝试将整数值转换为其等效值，但我做不到。如果你这样做，你可以发回来..再见..

更新：一些如何设法将IP从整数形式转换为v4格式
转换为 IPv4 格式：

public String intToIp(int i) {

   return ((i >> 24 ) & 0xFF ) + "." +
               ((i >> 16 ) & 0xFF) + "." +
               ((i >> 8 ) & 0xFF) + "." +
               ( i & 0xFF) ;
}

图片提供：http://www.bennadel.com/blog/1830-Converting-IP-Addresses-To-And-From-Integer-Values-With-ColdFusion.htm

Answer 2

Formatter.formatIpAddress(int) 已被弃用，我们不想使用已弃用的方法，不是吗？

AndroidKid 的版本在某种程度上是相反的，但这应该可以修复它：

public String intToIp(int addr) {
    return  ((addr & 0xFF) + "." + 
            ((addr >>>= 8) & 0xFF) + "." + 
            ((addr >>>= 8) & 0xFF) + "." + 
            ((addr >>>= 8) & 0xFF));
}

来源：

Answer 3

要格式化 ip，请尝试使用：

import android.text.format.Formatter;

public String FormatIP(int IpAddress)
{
    return Formatter.formatIpAddress(IpAddress);
}

Answer 4

这是一个旧线程，但我找到了android API使用的官方函数（包android.net.NetworkUtils）：

/**
 * Convert a IPv4 address from an integer to an InetAddress.
 * @param hostAddress an int corresponding to the IPv4 address in network byte order
 */
public static InetAddress intToInetAddress(int hostAddress) {
    byte[] addressBytes = { (byte)(0xff & hostAddress),
                            (byte)(0xff & (hostAddress >> 8)),
                            (byte)(0xff & (hostAddress >> 16)),
                            (byte)(0xff & (hostAddress >> 24)) };

    try {
       return InetAddress.getByAddress(addressBytes);
    } catch (UnknownHostException e) {
       throw new AssertionError();
    }
}

一旦你有了InetAddress，你就可以获得格式化的字符串这样：

intToInetAddress(d.gateway).getHostAddress()

Answer 5

使用Formatter.formatIpAddress(mask); mask是你的int。

String maskk = Formatter.formatIpAddress(mask);

Answer 6

此版本适用于任何网络（Wifi/Cell）。返回 V4 和 V6。

/**
 * Returns the set of all gateways<V4 & V6> for any Network
 *
 */
fun getGatewaySet(network:Network, context:Context):MutableSet<InetAddress> {
    val connectivityManager = context.getSystemService(Context.CONNECTIVITY_SERVICE) as ConnectivityManager
    connectivityManager.getLinkProperties(network)?.routes?.let {
        //Saw dupes on my device, so use a set
        val set = mutableSetOf<InetAddress>()
        for (routeInfo in it) {
            routeInfo?.gateway?.let { inetAddress ->
                //If comes from AOSP isGateway() - requires API 29
                if (!inetAddress.isAnyLocalAddress) {
                    set.add(inetAddress)
                }
            }
        }
        return set
    }
    return Collections.EMPTY_SET as MutableSet<InetAddress>
}

Answer 7

不要获取 255.255.255.0，只需更改返回的顺序即可；）这样您将能够获得正确的顺序...

public String intToIp(int i) {

   return (i & 0xFF) + "." +
               ((i >> 8 ) & 0xFF) + "." +
               ((i >> 16 ) & 0xFF) + "." +
               ((i >> 24 ) & 0xFF) ;
}

Answer 8

虽然老了，但仍然有魅力。

tvGateway.setText(Formatter.formatIpAddress(dhcpInfo.gateway));