获取本机号码

fun getPhoneNumber(): ArrayList<String> {
    val phoneList = arrayListOf<String>()
    try {
        if (Build.VERSION.SDK_INT > Build.VERSION_CODES.LOLLIPOP) {
            val sm =
                getSystemService(Context.TELEPHONY_SUBSCRIPTION_SERVICE) as SubscriptionManager
            val subList = sm.activeSubscriptionInfoList
            subList?.forEach {
                val rawPhone = it.number
                if (!rawPhone.isNullOrEmpty()){
                    val length = rawPhone.length
                    if (length > 11){
                        phoneList.add(rawPhone.substring(length - 11))
                    }else{
                        phoneList.add(rawPhone)
                    }
                }
            }
        } else {
            val tm =
                getSystemService(Context.TELEPHONY_SERVICE) as TelephonyManager
            phoneList.add(tm.line1Number)
        }
    }catch (e: SecurityException){
        Log.e("aa", "**获取手机号码异常:${e.message}")
    }
    return phoneList
}

1、需要权限

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

2、telephonyManager.getLine1Number()

(1)在Q以后版本无法获取到手机号

(2)双卡手机,卡1存在则返回卡1号码,否则返回卡2号码

3、SubscriptionInfo.number()需api22及以上版本

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转载自blog.csdn.net/yufumatou/article/details/105832751