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

不建议使用TelephonyManager,因为它要求应用程序在运行时需要READ_PHONE_STATE权限。

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

应该使用谷歌的播放服务进行身份验证,它将能够允许用户选择使用哪个phoneNumber,并处理多个SIM卡,而不是我们试图猜测哪一个是主SIM卡。

implementation "com.google.android.gms:play-services-auth:$play_service_auth_version"
fun main() {
    val googleApiClient = GoogleApiClient.Builder(context)
        .addApi(Auth.CREDENTIALS_API).build()

    val hintRequest = HintRequest.Builder()
        .setPhoneNumberIdentifierSupported(true)
        .build()

    val hintPickerIntent = Auth.CredentialsApi.getHintPickerIntent(
        googleApiClient, hintRequest
    )

    startIntentSenderForResult(
        hintPickerIntent.intentSender, REQUEST_PHONE_NUMBER, null, 0, 0, 0
    )
}

override fun onActivityResult(requestCode: Int, resultCode: Int, data: Intent?) {
    super.onActivityResult(requestCode, resultCode, data)
    when (requestCode) {
        REQUEST_PHONE_NUMBER -> {
            if (requestCode == Activity.RESULT_OK) {
                val credential = data?.getParcelableExtra<Credential>(Credential.EXTRA_KEY)
                val selectedPhoneNumber = credential?.id
            }
        }
    }
}

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 ""
        }
    }
}

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()。

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

这是一个更简单的答案:

public String getMyPhoneNumber()
{
    return ((TelephonyManager) getSystemService(TELEPHONY_SERVICE))
            .getLine1Number();
}

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。