在您的投影中，您将查询限制为 MIMETYPE CommonDataKinds.Event.CONTENT_ITEM_TYPE 行，因此您只会得到生日.

In your projection you're limiting your query to rows of MIMETYPE CommonDataKinds.Event.CONTENT_ITEM_TYPE only, so you'll only get birthdays.

您需要询问电子邮件和电话 mimetypes，但请注意，这些附加信息将出现在同一联系人的不同行中.例如，对于拥有 2 部电话、3 封电子邮件和生日的联系人 A，您将在光标中看到 6 个结果.因此，您需要使用 CONTACT_ID 字段将它们组合在一起.

You need to ask for emails and phones mimetypes, but note that these additional information will come in separate rows for the same contact. For example, for contact A that has 2 phones, 3 emails and a birthday, you'll get 6 results in your cursor. So you need to group them all together using the CONTACT_ID field.

这是让您入门的简单代码，打印生成的 HashMap，您将获得每个联系人的所有姓名、电子邮件、电话和生日:

Here's simple code to get you started, print the resulting HashMap and you'll get for each contact all his/hers name, emails, phones and birthday:

Map<Long, List<String>> contacts = new HashMap<Long, List<String>>();

String[] projection = {Data.CONTACT_ID, Data.DISPLAY_NAME, Data.MIMETYPE, Data.DATA1, Data.DATA2, Data.DATA3};

// query only emails/phones/events
String selection = Data.MIMETYPE + " IN ('" + Phone.CONTENT_ITEM_TYPE + "', '" + Event.CONTENT_ITEM_TYPE"', '" + Email.CONTENT_ITEM_TYPE + "')";
ContentResolver cr = getContentResolver();
Cursor cur = cr.query(Data.CONTENT_URI, projection, selection, null, null);

while (cur != null && cur.moveToNext()) {
    long id = cur.getLong(0);
    String name = cur.getString(1); // full name
    String mime = cur.getString(2); // type of data (phone / birthday / email)
    String data = cur.getString(3); // the actual info, e.g. +1-212-555-1234

    String kind = "unknown";

    switch (mime) {
        case Phone.CONTENT_ITEM_TYPE: 
            kind = "phone"; 
            break;
        case Event.CONTENT_ITEM_TYPE: 
            kind = "birthday";
            break;
        case Email.CONTENT_ITEM_TYPE: 
            kind = "email";
            break;
    }
    Log.d(TAG, "got " + id + ", " + name + ", " + kind + " - " + data);

    // add info to existing list if this contact-id was already found, or create a new list in case it's new
    List<String> infos;
    if (contacts.containsKey(id)) {
        infos = contacts.get(id);
    } else {
        infos = new ArrayList<String>();
        infos.add("name = " + name);
        contacts.put(id, infos);
    }
    infos.add(kind + " = " + data);
}