String deviceName = android.os.Build.MODEL; // returns model name 

String deviceManufacturer = android.os.Build.MANUFACTURER; // returns manufacturer

使用以下方法以编程方式获取模型和制造商

  public String getDeviceName() {
        String manufacturer = Build.MANUFACTURER;
        String model = Build.MODEL;
        if (model.toLowerCase().startsWith(manufacturer.toLowerCase())) {
            return capitalize(model);
        } else {
            return capitalize(manufacturer) + " " + model;
        }
    }


    private String capitalize(String s) {
    if (s == null || s.length() == 0) {
        return "";
    }
    char first = s.charAt(0);
    if (Character.isUpperCase(first)) {
        return s;
    } else {
        return Character.toUpperCase(first) + s.substring(1);
    }

你可以尝试跟随函数，它会以字符串格式返回你的phoneModel名字。

public String phoneModel() {

    return Build.MODEL;
}

在许多流行的设备上，该设备的市场名称是不可用的。例如，在三星Galaxy S6上，Build的值。MODEL可以是“SM-G920F”、“SM-G920I”或“SM-G920W8”。

我创建了一个小型库，用于获取设备的市场(消费者友好)名称。它为超过10,000个设备获得正确的名称，并不断更新。如果你想使用我的图书馆，请点击下面的链接:

Github上的AndroidDeviceNames库

如果你不想使用上面的库，那么这是获得一个消费者友好的设备名称的最佳解决方案:

/** Returns the consumer friendly device name */
public static String getDeviceName() {
  String manufacturer = Build.MANUFACTURER;
  String model = Build.MODEL;
  if (model.startsWith(manufacturer)) {
    return capitalize(model);
  }
  return capitalize(manufacturer) + " " + model;
}

private static String capitalize(String str) {
  if (TextUtils.isEmpty(str)) {
    return str;
  }
  char[] arr = str.toCharArray();
  boolean capitalizeNext = true;

  StringBuilder phrase = new StringBuilder();
  for (char c : arr) {
    if (capitalizeNext && Character.isLetter(c)) {
      phrase.append(Character.toUpperCase(c));
      capitalizeNext = false;
      continue;
    } else if (Character.isWhitespace(c)) {
      capitalizeNext = true;
    }
    phrase.append(c);
  }

  return phrase.toString();
}

以我的Verizon HTC One M8为例:

// using method from above
System.out.println(getDeviceName());
// Using https://github.com/jaredrummler/AndroidDeviceNames
System.out.println(DeviceName.getDeviceName());

结果:

HTC6525LVW HTC One (M8)

我使用下面的代码来获取完整的设备名称。它获得模型和制造商字符串并连接它们，除非模型字符串已经包含制造商名称(在一些手机上):

public String getDeviceName() {
    String manufacturer = Build.MANUFACTURER;
    String model = Build.MODEL;
    if (model.toLowerCase().startsWith(manufacturer.toLowerCase())) {
        return capitalize(model);
    } else {
        return capitalize(manufacturer) + " " + model;
    }
}


private String capitalize(String s) {
    if (s == null || s.length() == 0) {
        return "";
    }
    char first = s.charAt(0);
    if (Character.isUpperCase(first)) {
        return s;
    } else {
        return Character.toUpperCase(first) + s.substring(1);
    }
} 

以下是我从用户那里得到的一些设备名称示例:

三星GT-S5830L 摩托罗拉MB860 索尼爱立信LT18i LGE LG-P500 HTC Desire V HTC野火S A510e .．.

芬兰湾的科特林版本

val Name: String by lazy {
    val manufacturer = Build.MANUFACTURER
    val model = Build.MODEL
    if (model.toLowerCase().startsWith(manufacturer.toLowerCase())) {
        capitalize(model)
    } else {
        capitalize(manufacturer) + " " + model
    }
}

private fun capitalize(s: String?): String {
    if (s == null || s.isEmpty()) {
        return ""
    }
    val first = s[0]
    return if (Character.isUpperCase(first)) {
        s
    } else {
        Character.toUpperCase(first) + s.substring(1)
    }
}

方法获取电话设备名称

BluetoothAdapter

如果手机不支持蓝牙，那么您必须从构造设备名称

android.os.Build类

下面是获取电话设备名称的示例代码。

public String getPhoneDeviceName() {  
        String name=null;
        // Try to take Bluetooth name
        BluetoothAdapter adapter = BluetoothAdapter.getDefaultAdapter();
        if (adapter != null) {
            name = adapter.getName();
        }

        // If not found, use MODEL name
        if (TextUtils.isEmpty(name)) {
            String manufacturer = Build.MANUFACTURER;
            String model = Build.MODEL;
            if (model.startsWith(manufacturer)) {
                name = model;
            } else {
                name = manufacturer + " " + model;
            }
        } 
        return name;
}