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
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 ""
}
}
}
其他回答
代码:
TelephonyManager tMgr = (TelephonyManager)mAppContext.getSystemService(Context.TELEPHONY_SERVICE);
String mPhoneNumber = tMgr.getLine1Number();
需要许可:
<uses-permission android:name="android.permission.READ_PHONE_STATE"/>
警告:
根据得到高度好评的评论,有一些注意事项需要注意。这可以返回null或“”甚至“???????”,它可以返回一个过时的电话号码,不再有效。如果您想要唯一地标识设备,则应该使用getDeviceId()。
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.
正如我之前回答的那样
使用以下代码:
TelephonyManager tMgr = (TelephonyManager)mAppContext.getSystemService(Context.TELEPHONY_SERVICE);
String mPhoneNumber = tMgr.getLine1Number();
在AndroidManifest.xml中,赋予以下权限:
<uses-permission android:name="android.permission.READ_PHONE_STATE"/>
但请记住,这个代码并不总是有效,因为手机号码取决于SIM卡和网络运营商/手机运营商。
另外,试着在电话->设置->关于->电话身份,如果你能在那里看到号码,从上面的代码获得电话号码的概率更高。如果您不能在设置中查看电话号码,那么您将无法通过此代码获得!
建议解决方案:
获取用户的电话号码作为用户的手动输入。 通过短信将代码发送到用户的手机号码。 请用户输入确认电话号码的代码。 保存在sharedpreference中。
在应用首次发布时一次性执行上述4个步骤。稍后,当需要电话号码时,使用共享首选项中可用的值。
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;
}
这是一个更简单的答案:
public String getMyPhoneNumber()
{
return ((TelephonyManager) getSystemService(TELEPHONY_SERVICE))
.getLine1Number();
}
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