显示联系人号码在Android 2.1的我自己的ListView [英] Show contact number in my own listview in android 2.1

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本文介绍了显示联系人号码在Android 2.1的我自己的ListView的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我是新来的Andr​​oid和我想在我自己的列表视图显示从电话簿中的联系人号码。我已经实现显示的名字,但是当我试图显示数字,则返回布尔数为1或0。

I am new to android and I want to display a contact number from phone book in my own listview. I had achieved to display the name but when I am trying to display the number then it returns the Boolean number that is 1 or 0.

我的code是

public class Test extends ListActivity {
    static final String[] CONTACTS_SUMMARY_PROJECTION = new String[] {
            Contacts._ID,  
            Contacts.DISPLAY_NAME, 
            Contacts.STARRED,  
            Contacts.TIMES_CONTACTED,  
            Contacts.CONTACT_PRESENCE,  
            Contacts.PHOTO_ID,  
            Contacts.LOOKUP_KEY,  
            Contacts.HAS_PHONE_NUMBER                  
    };

    static final int SUMMARY_ID_COLUMN_INDEX = 0;
    static final int SUMMARY_NAME_COLUMN_INDEX = 1;
    static final int SUMMARY_STARRED_COLUMN_INDEX = 2;
    static final int SUMMARY_TIMES_CONTACTED_COLUMN_INDEX = 3;
    static final int SUMMARY_PRESENCE_STATUS_COLUMN_INDEX = 4;
    static final int SUMMARY_PHOTO_ID_COLUMN_INDEX = 5;
    static final int SUMMARY_LOOKUP_KEY = 6;
    static final int SUMMARY_HAS_PHONE_COLUMN_INDEX = 7;


    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        String select = "((" + Contacts.DISPLAY_NAME + " NOTNULL) AND ("
                + Contacts.HAS_PHONE_NUMBER + "=1) AND ("
                + Contacts.DISPLAY_NAME + " != '' ))";
        Cursor c =
                getContentResolver().query(Contacts.CONTENT_URI, CONTACTS_SUMMARY_PROJECTION, select,
                null, Contacts.DISPLAY_NAME + " COLLATE LOCALIZED ASC");
        startManagingCursor(c);



        ContactListItemAdapter adapter = new ContactListItemAdapter(this, R.layout.main, c);
        setListAdapter(adapter);

    }

    private final class ContactListItemAdapter extends ResourceCursorAdapter {
        public ContactListItemAdapter(Context context, int layout, Cursor c) {
            super(context, layout, c);
        }


        @Override
        public void bindView(View view, Context context, Cursor cursor) {
            final ContactListItemCache cache = (ContactListItemCache) view.getTag();

            cursor.copyStringToBuffer(SUMMARY_NAME_COLUMN_INDEX, cache.nameBuffer);
            int size = cache.nameBuffer.sizeCopied;

            cursor.copyStringToBuffer(SUMMARY_HAS_PHONE_COLUMN_INDEX, cache.numberView);
            int number = cache.numberView.sizeCopied;


            cache.nameView.setText(cache.nameBuffer.data, 0, size);
            cache.numView.setText(cache.numberView.data, 0, number);
            final long contactId = cursor.getLong(SUMMARY_ID_COLUMN_INDEX);
            final String lookupKey = cursor.getString(SUMMARY_LOOKUP_KEY);
            cache.photoView.assignContactUri(Contacts.getLookupUri(contactId, lookupKey));

        }

        @Override
        public View newView(Context context, Cursor cursor, ViewGroup parent) {
            View view = super.newView(context, cursor, parent);
            ContactListItemCache cache = new ContactListItemCache();
            cache.nameView = (TextView) view.findViewById(R.id.name);
            cache.numView = (TextView) view.findViewById(R.id.number);
            cache.photoView = (QuickContactBadge) view.findViewById(R.id.badge);
            view.setTag(cache);

            return view;
        }
    }

    final static class ContactListItemCache {
        public TextView nameView;
        public TextView numView;
        public CharArrayBuffer numberView = new CharArrayBuffer(128);
        public QuickContactBadge photoView;
        public CharArrayBuffer nameBuffer = new CharArrayBuffer(128);
    }

在此先感谢。

Thanks in advance.

请参阅我的更新code - >

See my updated code ->

public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);


    ContentResolver cr = new ContentResolver(this.getApplicationContext()) {
    };

    Cursor phones = cr.query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, null,
            ContactsContract.CommonDataKinds.Phone.CONTACT_ID + " = ?", null, null);

    phones.getString(phones.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));

    ContactListItemAdapter adapter = new ContactListItemAdapter(this, R.layout.main, phones);
    setListAdapter(adapter);

}

但它显示错误。帮我解决这个问题!

But it shows error. Help me to solve this!

感谢....

推荐答案

这看起来不正确对我说:

This does not look right to me:

ContactsContract.CommonDataKinds.Phone.CONTACT_ID +" = "+ number, 

如何涉及到 CONTACT_ID

这篇关于显示联系人号码在Android 2.1的我自己的ListView的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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