如何从android中选择独特的联系人

概述我想从 Android只选择具有电话号码的联系人的唯一联系人.我正在使用此代码 ContentResolver cr = getContentResolver(); Cursor cur = cr.query(ContactsContract.Contacts.CONTENT_URI, null, null, null, ContactsContra 我想从 Android只选择具有电话号码的联系人的唯一联系人.我正在使用此代码

ContentResolver cr = getContentResolver();        Cursor cur = cr.query(ContactsContract.Contacts.CONTENT_URI,null,ContactsContract.Contacts.disPLAY_name);        // Find the ListVIEw resource.        mainListVIEw = (ListVIEw) findVIEwByID(R.ID.mainListVIEw);        // When item is tapped,toggle checked propertIEs of CheckBox and        // Planet.        mainListVIEw                .setonItemClickListener(new AdapterVIEw.OnItemClickListener()                {                    public voID onItemClick(AdapterVIEw<?> parent,VIEw item,int position,long ID)                    {                        ContactsList planet = listadapter.getItem(position);                        planet.toggleChecked();                        PlanetVIEwHolder vIEwHolder = (PlanetVIEwHolder) item                                .getTag();                        vIEwHolder.getCheckBox().setChecked(planet.isChecked());                    }                });        // Create and populate planets.        planets = (ContactsList[]) getLastNonConfigurationInstance();        // planets = new Planet[10];        // planets.Add("asdf");        ArrayList<ContactsList> planetList = new ArrayList<ContactsList>();        String phoneNumber = null;        String phoneType = null;        count = cur.getCount();        contacts = new ContactsList[count];        if (planets == null)        {            if (cur.getCount() > 0)            {                planets = new ContactsList[cur.getCount()];                int i = 0;                //                while (cur.movetoNext())                {                    String ID = cur.getString(cur                            .getColumnIndex(ContactsContract.Contacts._ID));                    String name = cur                            .getString(cur                                    .getColumnIndex(ContactsContract.Contacts.disPLAY_name));                    if (Integer                            .parseInt(cur.getString(cur                                    .getColumnIndex(ContactsContract.Contacts.HAS_PHONE_NUMBER))) > 0)                    {                        // query phone here. Covered next                        Cursor pCur = cr                                .query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI,ContactsContract.CommonDataKinds.Phone.CONTACT_ID                                                + " = ?",new String[]                                        { ID },null);                        // WHILE WE HAVE CURSOR GET THE PHONE NUMERS                        while (pCur.movetoNext())                        {                            // Do something with phones                            phoneNumber = pCur                                    .getString(pCur                                            .getColumnIndex(ContactsContract.CommonDataKinds.Phone.DATA));                            phoneType = pCur                                    .getString(pCur                                            .getColumnIndex(ContactsContract.CommonDataKinds.Phone.TYPE));                            Log.i("Pratik",name + "'s PHONE :" + phoneNumber);                            Log.i("Pratik","PHONE TYPE :" + phoneType);                        }                        pCur.close();                    }                    planets = new ContactsList[]                    { new ContactsList(name,phoneNumber) };                    contacts[i] = planets[0];                    planetList.addAll(Arrays.asList(planets));                    i++;                }            }

此代码检索所有联系人并将其放入列表中.但我想要独特的联系,只有那些没有电话的人.我怎样才能做到这一点？？有什么方法可以在查询中传递一些参数来只选择唯一的联系人???

解决方法 我想你的意思是你有一些联系人的重复记录.因此,您必须为查询添加条件.关键部分是联系人必须在可见组中并且有电话号码.

String selection = ContactsContract.Contacts.IN_VISIBLE_GROUP + " = '"                + ("1") + "'";        String sortOrder = ContactsContract.Contacts.disPLAY_name                + " ColLATE LOCAliZED ASC";cur = context.getContentResolver().query(                ContactsContract.Contacts.CONTENT_URI,projection,selection                        + " AND " + ContactsContract.Contacts.HAS_PHONE_NUMBER                        + "=1",sortOrder);// this query only return contacts which had phone number and not duplicated

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