我如何通过编程方式获得正在运行我的android应用程序的设备的电话号码?
当前回答
while working on a security app which needed to get the phone number of who so ever my phone might get into their hands, I had to do this; 1. receive Boot completed and then try getting Line1_Number from telephonyManager which returns a string result. 2. compare the String result with my own phone number and if they don't match or string returns null then, 3. secretly send an SMS containing the string result plus a special sign to my office number. 4. if message sending fails, start a service and keep trying after each hour until sent SMS pending intent returns successful. With this steps I could get the number of the person using my lost phone. it doesn't matter if the person is charged.
其他回答
以下是我找到的解决方案的组合(这里是示例项目,如果你也想检查自动填充):
清单
<uses-permission android:name="android.permission.READ_PHONE_STATE" />
build.gradle
implementation "com.google.android.gms:play-services-auth:17.0.0"
MainActivity.kt
class MainActivity : AppCompatActivity() {
private lateinit var googleApiClient: GoogleApiClient
override fun onCreate(savedInstanceState: Bundle?) {
super.onCreate(savedInstanceState)
setContentView(R.layout.activity_main)
tryGetCurrentUserPhoneNumber(this)
googleApiClient = GoogleApiClient.Builder(this).addApi(Auth.CREDENTIALS_API).build()
if (phoneNumber.isEmpty()) {
val hintRequest = HintRequest.Builder().setPhoneNumberIdentifierSupported(true).build()
val intent = Auth.CredentialsApi.getHintPickerIntent(googleApiClient, hintRequest)
try {
startIntentSenderForResult(intent.intentSender, REQUEST_PHONE_NUMBER, null, 0, 0, 0);
} catch (e: IntentSender.SendIntentException) {
Toast.makeText(this, "failed to show phone picker", Toast.LENGTH_SHORT).show()
}
} else
onGotPhoneNumberToSendTo()
}
override fun onActivityResult(requestCode: Int, resultCode: Int, data: Intent?) {
super.onActivityResult(requestCode, resultCode, data)
if (requestCode == REQUEST_PHONE_NUMBER) {
if (resultCode == Activity.RESULT_OK) {
val cred: Credential? = data?.getParcelableExtra(Credential.EXTRA_KEY)
phoneNumber = cred?.id ?: ""
if (phoneNumber.isEmpty())
Toast.makeText(this, "failed to get phone number", Toast.LENGTH_SHORT).show()
else
onGotPhoneNumberToSendTo()
}
}
}
private fun onGotPhoneNumberToSendTo() {
Toast.makeText(this, "got number:$phoneNumber", Toast.LENGTH_SHORT).show()
}
companion object {
private const val REQUEST_PHONE_NUMBER = 1
private var phoneNumber = ""
@SuppressLint("MissingPermission", "HardwareIds")
private fun tryGetCurrentUserPhoneNumber(context: Context): String {
if (phoneNumber.isNotEmpty())
return phoneNumber
if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.M) {
val subscriptionManager = context.getSystemService(Context.TELEPHONY_SUBSCRIPTION_SERVICE) as SubscriptionManager
try {
subscriptionManager.activeSubscriptionInfoList?.forEach {
val number: String? = it.number
if (!number.isNullOrBlank()) {
phoneNumber = number
return number
}
}
} catch (ignored: Exception) {
}
}
try {
val telephonyManager = context.getSystemService(Context.TELEPHONY_SERVICE) as TelephonyManager
val number = telephonyManager.line1Number ?: ""
if (!number.isBlank()) {
phoneNumber = number
return number
}
} catch (e: Exception) {
}
return ""
}
}
}
对于android版本>= LOLLIPOP_MR1:
增加权限:
叫它:
val subscriptionManager =
getSystemService(Context.TELEPHONY_SUBSCRIPTION_SERVICE) as SubscriptionManager
if (ActivityCompat.checkSelfPermission(this, Manifest.permission.READ_PHONE_STATE) == PackageManager.PERMISSION_GRANTED) {
val list = subscriptionManager.activeSubscriptionInfoList
for (info in list) {
Log.d(TAG, "number " + info.number)
Log.d(TAG, "network name : " + info.carrierName)
Log.d(TAG, "country iso " + info.countryIso)
}
}
更新:这个答案不再可用,因为Whatsapp已经停止将电话号码作为帐户名,请忽略这个答案。
实际上,如果你不能通过电话服务获得它,你可以考虑另一种解决方案。
到今天为止,你可以依靠另一个大型应用程序Whatsapp,使用AccountManager。数以百万计的设备安装了这个应用程序,如果你不能通过TelephonyManager获得电话号码,你可以试试这个。
许可:
<uses-permission android:name="android.permission.GET_ACCOUNTS" />
代码:
AccountManager am = AccountManager.get(this);
Account[] accounts = am.getAccounts();
for (Account ac : accounts) {
String acname = ac.name;
String actype = ac.type;
// Take your time to look at all available accounts
System.out.println("Accounts : " + acname + ", " + actype);
}
检查WhatsApp帐户的actype
if(actype.equals("com.whatsapp")){
String phoneNumber = ac.name;
}
当然,如果用户没有安装WhatsApp,你可能不会得到它,但无论如何都值得一试。 记住,你应该总是询问用户的确认。
private String getMyPhoneNumber(){
TelephonyManager mTelephonyMgr;
mTelephonyMgr = (TelephonyManager)
getSystemService(Context.TELEPHONY_SERVICE);
return mTelephonyMgr.getLine1Number();
}
private String getMy10DigitPhoneNumber(){
String s = getMyPhoneNumber();
return s != null && s.length() > 2 ? s.substring(2) : null;
}
代码摘自http://www.androidsnippets.com/get-my-phone-number
一点小小的贡献。在我的例子中,代码启动了一个错误异常。我需要把一个注释,为代码运行和修复这个问题。这里我让这段代码。
public static String getLineNumberPhone(Context scenario) {
TelephonyManager tMgr = (TelephonyManager) scenario.getSystemService(Context.TELEPHONY_SERVICE);
@SuppressLint("MissingPermission") String mPhoneNumber = tMgr.getLine1Number();
return mPhoneNumber;
}
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