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
首先在Intent中初始化你的符号
private val signInIntent = registerForActivityResult(ActivityResultContracts.StartIntentSenderForResult()) { result ->
try {
val phoneNumber = Identity.getSignInClient(requireContext()).getPhoneNumberFromIntent(result.data)
// Note phone number will be in country code + phone number format
} catch (e: Exception) {
}
}
要打开谷歌播放意图,并显示与谷歌帐户相关联的电话号码,请使用此
val phoneNumberHintIntentRequest = GetPhoneNumberHintIntentRequest.builder()
.build()
Identity.getSignInClient(requireContext())
.getPhoneNumberHintIntent(phoneNumberHintIntentRequest)
.addOnSuccessListener { pendingIntent ->
signInIntent.launch(IntentSenderRequest.Builder(pendingIntent).build())
}.addOnFailureListener {
it.printStackTrace()
}
注意:
如果用户禁用电话号码共享,此操作将失败。如果是这样,用户必须从设置->谷歌->自动填充->电话号码共享 如果您正在使用播放服务不可用的模拟设备,这将不起作用
其他回答
我注意到有几个回复都发了同样的东西。首先,从2021年开始,onActivityResult已弃用。下面是不弃用的解决方案。
private fun requestHint() {
val hintRequest = HintRequest.Builder()
.setPhoneNumberIdentifierSupported(true)
.build()
val intent = Credentials.getClient(this).getHintPickerIntent(hintRequest)
val intentSender = IntentSenderRequest.Builder(intent.intentSender).build()
val resultLauncher = registerForActivityResult(
ActivityResultContracts.StartIntentSenderForResult()
) { result ->
if (result.resultCode == Activity.RESULT_OK) {
val credential: Credential? = result.data?.getParcelableExtra(Credential.EXTRA_KEY)
// Phone number with country code
Log.i("mTag", "Selected phone No: ${credential?.id}")
}
}
resultLauncher.launch(intentSender)
}
注意:虽然很多人认为这可以让你检索用户的手机号码。通常情况并非如此。谷歌播放服务缓存了几个电话号码,有时对话框显示电话号码,其中不属于用户。
一个重要的导入。com.google.android.gms.auth.api.credentials.Credential
参考文档提供了详细信息,但代码有些不推荐。
所以这就是你如何通过Play服务API请求一个电话号码,而没有许可和黑客。源代码和完整的示例。
在你的构建中。Gradle(版本10.2。X及以上要求):
compile "com.google.android.gms:play-services-auth:$gms_version"
在你的活动中(代码被简化了):
@Override
protected void onCreate(Bundle savedInstanceState) {
// ...
googleApiClient = new GoogleApiClient.Builder(this)
.addApi(Auth.CREDENTIALS_API)
.build();
requestPhoneNumber(result -> {
phoneET.setText(result);
});
}
public void requestPhoneNumber(SimpleCallback<String> callback) {
phoneNumberCallback = callback;
HintRequest hintRequest = new HintRequest.Builder()
.setPhoneNumberIdentifierSupported(true)
.build();
PendingIntent intent = Auth.CredentialsApi.getHintPickerIntent(googleApiClient, hintRequest);
try {
startIntentSenderForResult(intent.getIntentSender(), PHONE_NUMBER_RC, null, 0, 0, 0);
} catch (IntentSender.SendIntentException e) {
Logs.e(TAG, "Could not start hint picker Intent", e);
}
}
@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
super.onActivityResult(requestCode, resultCode, data);
if (requestCode == PHONE_NUMBER_RC) {
if (resultCode == RESULT_OK) {
Credential cred = data.getParcelableExtra(Credential.EXTRA_KEY);
if (phoneNumberCallback != null){
phoneNumberCallback.onSuccess(cred.getId());
}
}
phoneNumberCallback = null;
}
}
这会生成一个这样的对话框:
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)
}
}
有一个新的Android api,允许用户选择他们的电话号码,而不需要权限。来看看: https://android-developers.googleblog.com/2017/10/effective-phone-number-verification.html
// Construct a request for phone numbers and show the picker
private void requestHint() {
HintRequest hintRequest = new HintRequest.Builder()
.setPhoneNumberIdentifierSupported(true)
.build();
PendingIntent intent = Auth.CredentialsApi.getHintPickerIntent(
apiClient, hintRequest);
startIntentSenderForResult(intent.getIntentSender(),
RESOLVE_HINT, null, 0, 0, 0);
}
虽然你可以有多个语音信箱帐号,但当你用自己的号码打电话时,运营商会把你转到语音信箱。因此,telephonymanager。getvoicemailnumber()或telephonymanager。getcompletevoicemailnumber(),这取决于你需要的风格。
希望这能有所帮助。
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