403

我想知道是否有办法在 Android 中以编程方式读取电话模型。

我想得到一个像 HTC Dream、Milestone、Sapphire 或其他什么的字符串......

4

16 回答 16

462

我使用以下代码来获取完整的设备名称。它获取模型和制造商字符串并将它们连接起来,除非模型字符串已经包含制造商名称(在某些手机上它包含):

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获得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 Wildfire S A510e
...</p>

于 2012-10-03T11:23:56.203 回答
361

在许多流行的设备上,该设备的市场名称不可用。例如,在三星 Galaxy S6 上, 的值Build.MODEL可能是"SM-G920F""SM-G920I""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台设备获得了正确的名称,并且不断更新。如果您想使用我的图书馆,请单击下面的链接:

Github 上的 AndroidDeviceNames 库


如果您不想使用上面的库,那么这是获得消费者友好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();
}

我的 Verizon HTC One M8 示例:

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

结果:

HTC6525LVW

宏达一 (M8)

于 2015-01-08T09:37:05.747 回答
118

是的:Build.MODEL

于 2010-01-03T16:23:53.880 回答
40

对于寻找Build此处完整属性列表的人来说,这是 Sony Z1 Compact 的一个示例:

Build.BOARD = MSM8974
Build.BOOTLOADER = s1
Build.BRAND = Sony
Build.CPU_ABI = armeabi-v7a
Build.CPU_ABI2 = armeabi
Build.DEVICE = D5503
Build.DISPLAY = 14.6.A.1.236
Build.FINGERPRINT = Sony/D5503/D5503:5.1.1/14.6.A.1.236/2031203XXX:user/release-keys
Build.HARDWARE = qcom
Build.HOST = BuildHost
Build.ID = 14.6.A.1.236
Build.IS_DEBUGGABLE = false
Build.MANUFACTURER = Sony
Build.MODEL = D5503
Build.PRODUCT = D5503
Build.RADIO = unknown
Build.SERIAL = CB5A1YGVMT
Build.SUPPORTED_32_BIT_ABIS = [Ljava.lang.String;@3dd90541
Build.SUPPORTED_64_BIT_ABIS = [Ljava.lang.String;@1da4fc3
Build.SUPPORTED_ABIS = [Ljava.lang.String;@525f635
Build.TAGS = release-keys
Build.TIME = 144792559XXXX
Build.TYPE = user
Build.UNKNOWN = unknown
Build.USER = BuildUser

您可以使用 kotlin 使用“评估表达式”对话框在调试模式下轻松列出设备的这些属性:

android.os.Build::class.java.fields.map { "Build.${it.name} = ${it.get(it.name)}"}.joinToString("\n")
于 2016-08-16T12:28:56.250 回答
35

实际上,这不是 100% 正确的。这可以给你模型(有时是数字)。
将为您提供手机的制造商(您请求的 HTC 部分):

 Build.MANUFACTURER

对于产品名称:

 Build.PRODUCT
于 2012-03-06T18:50:20.807 回答
17
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);
    }
于 2014-09-03T11:09:56.063 回答
9

显然您需要使用来自 Google 的列表https://support.google.com/googleplay/answer/1727131

API 不会返回我期望的任何内容或设置中的任何内容。对于我的摩托罗拉 X,这就是我得到的

   Build.MODEL = "XT1053"
   Build.BRAND = "motorola"
   Build.PRODUCT = "ghost"

转到上面提到的页面“ghost”映射到 Moto X。看起来这可能更简单一点......

于 2014-09-12T17:36:02.863 回答
8

当您要检索制造商、设备名称和/或型号时,可以使用以下字符串:

String manufacturer = Build.MANUFACTURER;
String brand        = Build.BRAND;
String product      = Build.PRODUCT;
String model        = Build.MODEL;
于 2014-05-21T11:02:49.967 回答
7

您可以使用以下代码获取设备的品牌名称和品牌型号。

 String brand = Build.BRAND; // for getting BrandName
 String model = Build.MODEL; // for getting Model of the device
于 2013-01-09T05:29:16.903 回答
5

Kotlin 短版:

import android.os.Build.MANUFACTURER
import android.os.Build.MODEL

fun getDeviceName(): String =
    if (MODEL.startsWith(MANUFACTURER, ignoreCase = true)) {
        MODEL
    } else {
        "$MANUFACTURER $MODEL"
    }.capitalize(Locale.ROOT)
于 2019-12-03T13:42:57.307 回答
5
Build.DEVICE // The name of the industrial design.

Build.DEVICE为某些设备提供人类可读的名称,而不是Build.MODEL

Build.DEVICE = OnePlus6
Build.MANUFACTURER = OnePlus
Build.MODEL = ONEPLUS A6003
于 2020-01-10T04:28:03.210 回答
3

这是我的代码,获取制造商、品牌名称、操作系统版本并支持 API 级别

String manufacturer = Build.MANUFACTURER;

String model = Build.MODEL + " " + android.os.Build.BRAND +" ("
           + android.os.Build.VERSION.RELEASE+")"
           + " API-" + android.os.Build.VERSION.SDK_INT;

if (model.startsWith(manufacturer)) {
    return capitalize(model);
} else {
    return capitalize(manufacturer) + " " + model;
}

输出:

System.out: button press on device name = Lava Alfa L iris(5.0) API-21
于 2017-02-22T11:22:31.000 回答
3

Kotlin 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)
    }
}
于 2018-03-28T13:36:24.203 回答
1

稍微更改了 Idolons 代码。这将在获取设备型号时大写单词。

public static String getDeviceName() {
    final String manufacturer = Build.MANUFACTURER, model = Build.MODEL;
    return model.startsWith(manufacturer) ? capitalizePhrase(model) : capitalizePhrase(manufacturer) + " " + model;
}

private static String capitalizePhrase(String s) {
    if (s == null || s.length() == 0)
        return s;
    else {
        StringBuilder phrase = new StringBuilder();
        boolean next = true;
        for (char c : s.toCharArray()) {
            if (next && Character.isLetter(c) || Character.isWhitespace(c))
                next = Character.isWhitespace(c = Character.toUpperCase(c));
            phrase.append(c);
        }
        return phrase.toString();
    }
}
于 2016-09-10T06:01:58.127 回答
0

您可以从

蓝牙适配器

如果手机不支持蓝牙,那么您必须从

android.os.Build 类

这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;
}
于 2018-01-16T13:47:18.853 回答
-3

您可以尝试以下函数,并以字符串格式返回您的 phoneModel 名称。

public String phoneModel() {

    return Build.MODEL;
}
于 2018-01-18T08:18:46.457 回答