Question

我在Contacts表中进行迭代，每次迭代都在Phone表上进行迭代，所以它需要appx。 4秒为243个联系人做联系人列表（没有迭代它立即加载的电话号码）。我确信有办法让它更快。

已经进行了预测，只能获取所需的列，但还没有那么多改进。也许我应该改进SQLite查询或以其他方式迭代：

Contact realmContact = new Contact();

        Uri uri = Contacts.CONTENT_URI;

        String[] projection = {
                Contacts.LOOKUP_KEY,
                Contacts.DISPLAY_NAME_PRIMARY,
                Contacts.LAST_TIME_CONTACTED,
                Contacts.HAS_PHONE_NUMBER
        };

        String selection = "((" + CommonDataKinds.Phone.DISPLAY_NAME_PRIMARY + " NOTNULL) AND ("
                + Contacts.HAS_PHONE_NUMBER + "=1) AND ("
                + CommonDataKinds.Phone.DISPLAY_NAME_PRIMARY + " != '' ))";

        Cursor phones = getActivity()
                .getContentResolver()
                .query(uri, projection, selection, null, null);

        while (phones.moveToNext()) {
            String id = phones.getString(phones.getColumnIndex(Contacts.LOOKUP_KEY));

            String name = phones.getString(phones.getColumnIndex(Contacts.DISPLAY_NAME_PRIMARY));

            String lastTimeContacted = phones.getString(phones.getColumnIndex(Contacts.LAST_TIME_CONTACTED));
            long data = Long.parseLong(lastTimeContacted);
            String date = new SimpleDateFormat("dd/MM/yyyy | HH:mm").format(new Date(data));

            if (Integer.parseInt(phones.getString(phones.getColumnIndex(Contacts.HAS_PHONE_NUMBER))) > 0) {
                Cursor pCur = getActivity().getContentResolver().query(
                        CommonDataKinds.Phone.CONTENT_URI,
                        new String[]{CommonDataKinds.Phone.NUMBER},
                        CommonDataKinds.Phone.LOOKUP_KEY + " = ?",
                        new String[]{id}, null);

                int iterationCounter = 0;
                String phoneNumber = "";
                while (pCur.moveToNext()) {
                    phoneNumber += iterationCounter == 0 ?
                            pCur.getString(pCur.getColumnIndex(CommonDataKinds.Phone.NUMBER))
                             : "," + pCur.getString(pCur.getColumnIndex(CommonDataKinds.Phone.NUMBER));

                    iterationCounter += 1;
                }
                realmContact.setNumber(phoneNumber);
                Log.i("asd-number", phoneNumber);
                pCur.close();
            }
            realmContact.setId(id);
            realmContact.setName(name);
            realmContact.setLastTimeContacted(
                    getActivity().getResources()
                            .getString(R.string.contacts_list_fragment_last_call)
                            + date);
            realmContact.setIsBeingSaved(true);
            _realm.insertOrUpdate(realmContact);

Answer 1

正如评论中提到的@pskink，您需要查询Phone表。要通过联系人获取手机列表，您可以保留HashMap从contactId到List手机。

这里有一个可能有用的小片段（你可能想稍微调整一下这个集合，每个联系人只能存储一次显示名称）：

HashMap<Long, ArrayList<String>> phones = new HashMap<Long, ArrayList<String>>();
String[] projection = new String[] { Phone.CONTACT_ID, Phone.DISPLAY_NAME, Phone.NUMBER };
Cursor c = cr.query(Phone.CONTENT_URI, projection, null, null, null);
while (c != null && c.moveToNext()) {
    long contactId = c.getLong(0);
    ArrayList<String> list = phones.get(contactId);
    if (list == null) {
        ArrayList<String> list = new ArrayList<String>();
        phones.put(contactId, list);
    }
    list.add(c.getString(1) + " " + c.getString(2));
}

如何使用CommonDataKinds.Phone内部查询更快地进行ContactsContract.Contact查询？

1 个答案: