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.

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:

如果我从voiceMailNumer得到数字，那么它工作得很好-

val telephonyManager = getSystemService(TELEPHONY_SERVICE) as TelephonyManager
    if (ActivityCompat.checkSelfPermission(this,
                    Manifest.permission.READ_PHONE_STATE) == PackageManager.PERMISSION_GRANTED
    ) {
        Log.d("number", telephonyManager.voiceMailNumber.toString())
    }

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()。

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。

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