访问 Android 联系人组名称

Question

您能否告诉我如何以编程方式获取联系人组 存储在我们的 Android 手机中？

Answer 1

final String[] GROUP_PROJECTION = new String[] {
            ContactsContract.Groups._ID, ContactsContract.Groups.TITLE };
    cursor = getContentResolver().query(
    ContactsContract.Groups.CONTENT_URI, GROUP_PROJECTION, null,
            null, ContactsContract.Groups.TITLE);

            GlobalConfig.groupList.clear();
    Group g = new Group();
    g.GroupIdList += "0";
    g.setGroupTitle("ALL");
    GlobalConfig.groupList.add(g);
    while (cursor.moveToNext()) {

        String id = cursor.getString(cursor
                .getColumnIndex(ContactsContract.Groups._ID));

        String gTitle = (cursor.getString(cursor
                .getColumnIndex(ContactsContract.Groups.TITLE)));

        if (gTitle.contains("Group:")) {
            gTitle = gTitle.substring(gTitle.indexOf("Group:") + 6).trim();

        }
        if (gTitle.contains("Favorite_")) {
            gTitle = "Favorites";
        }
        if (gTitle.contains("Starred in Android")
                || gTitle.contains("My Contacts")) {
            continue;
        }

        Group gObj = new Group();

        int pos = GlobalConfig.GroupContainsTitle(gTitle);
        if (pos != -1) {
            gObj = GlobalConfig.groupList.get(pos);
            gObj.GroupIdList += "," + id;
            GlobalConfig.groupList.set(pos, gObj);

        } else {
            gObj.GroupIdList += id;
            gObj.setGroupTitle(gTitle);
            GlobalConfig.groupList.add(gObj);

        }

        // Log.d("GrpId  Title", gObj.getGroupIdList() +
        // gObj.getGroupTitle());
    }

Answer 2

@Abhi 的答案是好的，但有一些限制：

• 将列出已删除的联系人
• 将列出不可见的组
• 将列出“幽灵”组（即应该已删除但仍处于不确定状态的组）

-

private class GroupInfo {
    String id;
    String title;

    @Override
    public String toString() {
        return title+ " ("+id+")";
    }

    public String getId() {
        return id;
    }
}

List<GroupInfo> groups = new ArrayList<GroupInfo>();

public void loadGroups() {
   final String[] GROUP_PROJECTION = new String[] {
            ContactsContract.Groups._ID, 
            ContactsContract.Groups.TITLE,
            ContactsContract.Groups.SUMMARY_WITH_PHONES
            };

    Cursor c = getContentResolver().query(
            ContactsContract.Groups.CONTENT_SUMMARY_URI,
            GROUP_PROJECTION,
            ContactsContract.Groups.DELETED+"!='1' AND "+
            ContactsContract.Groups.GROUP_VISIBLE+"!='0' "
            ,
            null,
            null);
    final int IDX_ID = c.getColumnIndex(ContactsContract.Groups._ID);
    final int IDX_TITLE = c.getColumnIndex(ContactsContract.Groups.TITLE);

    Map<String,GroupInfo> m = new HashMap<String, GroupInfo>();

    while (c.moveToNext()) {
        GroupInfo g = new GroupInfo();
        g.id = c.getString(IDX_ID);
        g.title = c.getString(IDX_TITLE);
        int users = c.getInt(c.getColumnIndex(ContactsContract.Groups.SUMMARY_WITH_PHONES));
        if (users>0) {
            // group with duplicate name?
            GroupInfo g2 = m.get(g.title);
            if (g2==null) {
                m.put(g.title, g);
                groups.add(g);
            } else {
                g2.id+=","+g.id;
            }
        }
    }
    c.close();
  }

Answer 3

不需要旧的过分的答案。这里的解决方案要简单得多。

final String[] GROUP_PROJECTION = new String[] {
            ContactsContract.Groups._ID, ContactsContract.Groups.TITLE };
Cursor gC = getContentResolver().query(
            ContactsContract.Groups.CONTENT_URI, GROUP_PROJECTION,null,null,null);
gC.moveToFirst();
while (!gC.isAfterLast()) {
        int idcolumn = gC.getColumnIndex(ContactsContract.Groups.TITLE);
        String Id = gC.getString(idcolumn);
        ArrayL.add(Id);
        gC.moveToNext();
}
        LinkedHashSet<String> s = new LinkedHashSet<String>();
        s.addAll(ArrayL);
        ArrayL.clear();
        ArrayL.addAll(s);