为了唯一地识别每个设备,我想使用IMEI(或CDMA设备的ESN号)。如何以编程方式访问它?


当前回答

API等级11或以上:

case TelephonyManager.PHONE_TYPE_SIP: 
return "SIP";

TelephonyManager tm= (TelephonyManager)getSystemService(Context.TELEPHONY_SERVICE);
textDeviceID.setText(getDeviceID(tm));

其他回答

获取IMEI(国际移动设备标识符)

public String getIMEI(Activity activity) {
    TelephonyManager telephonyManager = (TelephonyManager) activity
            .getSystemService(Context.TELEPHONY_SERVICE);
    return telephonyManager.getDeviceId();
}

获取设备唯一id

public String getDeviceUniqueID(Activity activity){
    String device_unique_id = Secure.getString(activity.getContentResolver(),
            Secure.ANDROID_ID);
    return device_unique_id;
}

TelephonyManager的getDeviceId()方法返回唯一的设备ID,例如GSM电话的IMEI, CDMA电话的MEID或ESN。如果设备ID不可用,则返回null。

Java代码

package com.AndroidTelephonyManager;

import android.app.Activity;
import android.content.Context;
import android.os.Bundle;
import android.telephony.TelephonyManager;
import android.widget.TextView;

public class AndroidTelephonyManager extends Activity {
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    TextView textDeviceID = (TextView)findViewById(R.id.deviceid);

    //retrieve a reference to an instance of TelephonyManager
    TelephonyManager telephonyManager = (TelephonyManager)getSystemService(Context.TELEPHONY_SERVICE);

    textDeviceID.setText(getDeviceID(telephonyManager));

}

String getDeviceID(TelephonyManager phonyManager){

 String id = phonyManager.getDeviceId();
 if (id == null){
  id = "not available";
 }

 int phoneType = phonyManager.getPhoneType();
 switch(phoneType){
 case TelephonyManager.PHONE_TYPE_NONE:
  return "NONE: " + id;

 case TelephonyManager.PHONE_TYPE_GSM:
  return "GSM: IMEI=" + id;

 case TelephonyManager.PHONE_TYPE_CDMA:
  return "CDMA: MEID/ESN=" + id;

 /*
  *  for API Level 11 or above
  *  case TelephonyManager.PHONE_TYPE_SIP:
  *   return "SIP";
  */

 default:
  return "UNKNOWN: ID=" + id;
 }

}
}

XML

<linearlayout android:layout_height="fill_parent" android:layout_width="fill_parent" android:orientation="vertical" xmlns:android="http://schemas.android.com/apk/res/android">
<textview android:layout_height="wrap_content" android:layout_width="fill_parent" android:text="@string/hello">
<textview android:id="@+id/deviceid" android:layout_height="wrap_content" android:layout_width="fill_parent">
</textview></textview></linearlayout> 

许可要求 manifest文件中的READ_PHONE_STATE。

方法getDeviceId()已弃用。 这里有一个getImei(int)的新方法

检查在这里

以下代码是为我工作。

val uid: String = Settings.Secure.getString(ctx.applicationContext.contentResolver, Settings.Secure.ANDROID_ID)
if (ContextCompat.checkSelfPermission(ctx, Manifest.permission.READ_PHONE_STATE) == PackageManager.PERMISSION_GRANTED) {
            imei = when {
                Build.VERSION.SDK_INT >= Build.VERSION_CODES.Q -> {
                    uid
                }
                Build.VERSION.SDK_INT >= Build.VERSION_CODES.O -> {
                    telephonyManager.imei
                }
                else -> {
                    telephonyManager.deviceId
                }
            }
        }

对于那些寻找Kotlin版本的人,您可以使用这样的东西;

private fun telephonyService() {
    val telephonyManager = getSystemService(TELEPHONY_SERVICE) as TelephonyManager
    val imei = if (android.os.Build.VERSION.SDK_INT >= 26) {
        Timber.i("Phone >= 26 IMEI")
        telephonyManager.imei
    } else {
        Timber.i("Phone IMEI < 26")
        telephonyManager.deviceId
    }

    Timber.i("Phone IMEI $imei")
}

注意:您必须使用checkSelfPermission或任何您使用的方法将上面的telephonyService()与权限检查一起包装。

还可以在清单文件中添加此权限;

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