Kotlin极简版

fun getIpv4HostAddress(): String {
    NetworkInterface.getNetworkInterfaces()?.toList()?.map { networkInterface ->
        networkInterface.inetAddresses?.toList()?.find {
            !it.isLoopbackAddress && it is Inet4Address
        }?.let { return it.hostAddress }
    }
    return ""
}

编译一些想法，以更好的kotlin解决方案从WifiManager获得wifi ip:

private fun getWifiIp(context: Context): String? {
  return context.getSystemService<WifiManager>().let {
     when {
      it == null -> "No wifi available"
      !it.isWifiEnabled -> "Wifi is disabled"
      it.connectionInfo == null -> "Wifi not connected"
      else -> {
        val ip = it.connectionInfo.ipAddress
        ((ip and 0xFF).toString() + "." + (ip shr 8 and 0xFF) + "." + (ip shr 16 and 0xFF) + "." + (ip shr 24 and 0xFF))
      }
    }
  }
}

或者，您可以通过NetworkInterface获取ip4环回设备的ip地址:

fun getNetworkIp4LoopbackIps(): Map<String, String> = try {
  NetworkInterface.getNetworkInterfaces()
    .asSequence()
    .associate { it.displayName to it.ip4LoopbackIps() }
    .filterValues { it.isNotEmpty() }
} catch (ex: Exception) {
  emptyMap()
}

private fun NetworkInterface.ip4LoopbackIps() =
  inetAddresses.asSequence()
    .filter { !it.isLoopbackAddress && it is Inet4Address }
    .map { it.hostAddress }
    .filter { it.isNotEmpty() }
    .joinToString()

您不需要像目前提供的解决方案那样添加权限。以字符串形式下载此网站:

http://www.ip-api.com/json

or

http://www.telize.com/geoip

下载一个网站作为字符串可以用java代码完成:

http://www.itcuties.com/java/read-url-to-string/

像这样解析JSON对象:

https://stackoverflow.com/a/18998203/1987258

json属性“query”或“ip”包含ip地址。

在AndroidManifest.xml中声明ACCESS_WIFI_STATE权限:

<uses-permission
    android:name="android.permission.ACCESS_WIFI_STATE"/>

可以通过WifiManager获取IP地址:

Context context = requireContext().getApplicationContext();
WifiManager wm = (WifiManager) context.getSystemService(Context.WIFI_SERVICE);
String ip = Formatter.formatIpAddress(wm.getConnectionInfo().getIpAddress());

我不使用Android，但我会用完全不同的方式来解决这个问题。

发送一个查询到谷歌，像这样: https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=my%20ip

并引用发布响应的HTML字段。您也可以直接查询到源。

谷歌最可能比你的应用程序存在的时间长。

只要记住，这可能是你的用户在这个时候没有互联网，你希望发生什么!

祝你好运

这里是@Nilesh和@anargund的kotlin版本

  fun getIpAddress(): String {
    var ip = ""
    try {
        val wm = applicationContext.getSystemService(WIFI_SERVICE) as WifiManager
        ip = Formatter.formatIpAddress(wm.connectionInfo.ipAddress)
    } catch (e: java.lang.Exception) {

    }

    if (ip.isEmpty()) {
        try {
            val en = NetworkInterface.getNetworkInterfaces()
            while (en.hasMoreElements()) {
                val networkInterface = en.nextElement()
                val enumIpAddr = networkInterface.inetAddresses
                while (enumIpAddr.hasMoreElements()) {
                    val inetAddress = enumIpAddr.nextElement()
                    if (!inetAddress.isLoopbackAddress && inetAddress is Inet4Address) {
                        val host = inetAddress.getHostAddress()
                        if (host.isNotEmpty()) {
                            ip =  host
                            break;
                        }
                    }
                }

            }
        } catch (e: java.lang.Exception) {

        }
    }

   if (ip.isEmpty())
      ip = "127.0.0.1"
    return ip
}