簡體   English   中英

Android呼叫過濾器可在我的設備上運行,但不能在客戶端設備上運行

[英]Android Call Filter working on my device, but not client's device

我使用for循環創建了一個簡單的調用過濾器,該循環遍歷數據庫並搜索保存的號碼。 該過濾器在我的設備上很好用,但是當我將其發送給客戶端時,它會自動允許每個呼叫通過。 篩選器顯示“ Deflecktur允許來自...的調用”,因此我知道我們將進入if / else語句。 由於某種原因,它總是與設備上的else語句一起使用。 這是來自廣播接收器的代碼。

if (blocker && isRunning) {
        String appName = context.getString(R.string.app_name);
        ITelephony telephonyService = getTeleService(context);
        if (telephonyService != null) {
            SparseArray<Phone> phones = getPhones(context, false);
            String incomingNumber = null;
            try {
                incomingNumber = intent.getExtras().getString(
                        TelephonyManager.EXTRA_INCOMING_NUMBER);
                if (incomingNumber != null && phones != null && phones.size() > 0) {
                    Log.i(TAG, "incoming call from " + incomingNumber);
                    boolean reject = false;
                    for (int i = 0; i < phones.size(); i++) {
                        Phone phone = phones.valueAt(i);
                        // Remove various characters to ensure a match
                        String matchNumber;
                        matchNumber = phone.getPhoneNumber().replaceAll("[^0-9\\+.,;#\\*N]", "");
                        if (incomingNumber.equals(matchNumber)) {
                            reject = true;
                            break;
                        }
                    }
                    if (reject) {
                        if (VERSION.SDK_INT < VERSION_CODES.GINGERBREAD) {
                            // Old method of silencing the ringer
                            telephonyService.silenceRinger();
                        } else {
                            final AudioManager am = (AudioManager) context
                                    .getSystemService(Context.AUDIO_SERVICE);
                            if (am.getRingerMode() != AudioManager.RINGER_MODE_SILENT) {
                                // Silence the ringer
                                if (mThread == null || !mThread.isAlive()) {
                                    mRingerMode = am.getRingerMode();
                                }
                                am.setRingerMode(AudioManager.RINGER_MODE_SILENT);
                                mThread = new Thread() {

                                    public void run() {
                                        SystemClock.sleep(3000);
                                        am.setRingerMode(mRingerMode);
                                    };
                                };
                                mThread.start();
                            }
                        }
                        telephonyService.endCall();
                        Toast.makeText(context, appName + " rejecting call from " + incomingNumber,
                                Toast.LENGTH_LONG).show();
                        Log.i(TAG, "call ended");
                    } else {
                        Toast.makeText(context, appName + " accepted call from " + incomingNumber,
                                Toast.LENGTH_LONG).show();
                        Log.i(TAG, "call accepted");
                    }
                } else {
                    // A phone state was encountered without incoming number, e.g. an outgoing call; ignore
                }
            } catch (Exception e) {
                if (incomingNumber == null)
                    incomingNumber = "[unknown]";
                e.printStackTrace();
                Log.e(TAG, "failed getting call info", e);
                Toast.makeText(
                        context, appName + " failed rejecting call from " + incomingNumber + "\n"
                                + e.getMessage(),
                        Toast.LENGTH_LONG).show();
            }
        }
    }
}

客戶端正在使用Samsung Galaxy Light,而我正在使用OnePlus One

您是否嘗試過添加日志以查看matchNumber和incomingNumber的值以檢查它們是否確實匹配? 可能是+61 4 ### vs 04 ###的情況嗎?

也許您還需要用正則表達式剝離傳入的值?

暫無
暫無

聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.

 
粵ICP備18138465號  © 2020-2024 STACKOOM.COM