Question

    public void ReadContacts() {
    Cursor people = getContentResolver().query(Phone.CONTENT_URI, null, null, null,Phone.DISPLAY_NAME + " ASC ");
    int indexName = people
            .getColumnIndex(ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME);
    int indexNumber = people
            .getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER);

    people.moveToFirst();
    do {
        name = people.getString(indexName);
        number = people.getString(indexNumber);

        contacts.put(name, number);

    } while (people.moveToNext());

    printHashMap(contacts);

}

public void printHashMap(HashMap<String, String> a) {

    for (Entry<String, String> lists : a.entrySet()) {
        Log.d(lists.getKey(), lists.getValue());
    }

}

尽管使用ASC，联系人仍未排序？你可以帮助我解决它的原因吗？我也使用了upper（）方法

Cursor cur = cr.query（ContactsContract.Contacts.CONTENT_URI，null， null，null，＆＃34; upper（＆＃34; + ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME +＆＃34;）ASC＆＃34;）;

Answer 1

您将（已排序）结果放在HashMap中，而HashMap不保留元素的顺序。使用列表，它将起作用。

修改：使用您在评论中提出的Tuple课程，它应如下所示：

static class Tuple {
    public String name;
    public String number;

    public Tuple(String name, String number) {
        this.name = name;
        this.number = number;
    }
}

public void ReadContacts() {
    Cursor people = getContentResolver().query(Phone.CONTENT_URI, null,  null, null, Phone.DISPLAY_NAME + " ASC ");
    int indexName = people
            .getColumnIndex(ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME);
    int indexNumber = people
            .getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER);

    List<Tuple> contacts = new ArrayList<Tuple>();

    people.moveToFirst();
    do {
        name = people.getString(indexName);
        number = people.getString(indexNumber);

        contacts.add(new Tuple(name, number));

    } while (people.moveToNext());

    printHashMap(contacts);

}

public void printList(List<Tuple> list) {

    for (Tuple tuple : list) {
        Log.d(tuple.name + ", " + tuple.number);
    }
}

但我建议在这种情况下将Tuple类重命名为Contact。有一个名为＆＃39; name＆＃39;的成员的元组课程。和＆＃39;数字＆＃39;很奇怪。

在Android中排序联系人

1 个答案: