Question

用户从他们的联系人中选择一个联系人，这是回调函数（我正在跳过onActivityResult(...)代码，它工作正常）：

private User addUserFromContactData(Intent data) {
    Uri contactData = data.getData();

    // Cursor loader to query contact information
    CursorLoader cl = new CursorLoader(this);
    String[] projection = { ContactsContract.Contacts.DISPLAY_NAME,
        ContactsContract.CommonDataKinds.Email.DATA };
    cl.setProjection(projection);
    cl.setUri(ContactsContract.CommonDataKinds.Email.CONTENT_URI);
    cl.setSelection(ContactsContract.CommonDataKinds.Email.CONTACT_ID + " = ?");
    cl.setSelectionArgs(new String[] { contactData.getLastPathSegment() });

    // Execute query and get results from cursor
    Cursor c = cl.loadInBackground();
    final int nameIndex = c.getColumnIndex(projection[0]);
    final int emailIndex = c.getColumnIndex(projection[1]);

    User user = null;
    if (c.moveToFirst()) {
        user = new User();
        user.setName(c.getString(nameIndex));
        user.setEmail(c.getString(emailIndex));
    }

    c.close();
    return user;
}

此方法适用于具有电子邮件地址的联系人。但是当我选择没有它的联系人时，没有找到结果。电子邮件对我来说是可选的，所以如果没有电子邮件，我只会使用联系人姓名。好像我只选择带有电子邮件地址的联系人。

任何人都可以帮助我吗？也许我需要查询两次，第一次获取名称，第二次获得可选的电子邮件？

谢谢你们。

Answer 1

解决。要做到这一点，需要两个查询（如果我只能用一个查询，请纠正我）：

private User addUserFromContactData(Intent data) {
    Uri contactData = data.getData();

    // Cursor loader to query contact name
    CursorLoader clName = new CursorLoader(this);
    clName.setProjection(new String[] { ContactsContract.Contacts.DISPLAY_NAME });
    clName.setUri(ContactsContract.Contacts.CONTENT_URI);
    clName.setSelection(ContactsContract.Contacts._ID + " = ?");
    clName.setSelectionArgs(new String[] { contactData.getLastPathSegment() });

    // Cursor loader to query optional contact email
    CursorLoader clEmail = new CursorLoader(this);
    clEmail.setProjection(new String[] { ContactsContract.CommonDataKinds.Email.DATA });
    clEmail.setUri(ContactsContract.CommonDataKinds.Email.CONTENT_URI);
    clEmail.setSelection(ContactsContract.CommonDataKinds.Email.CONTACT_ID + " = ?");
    clEmail.setSelectionArgs(new String[] { contactData.getLastPathSegment() });

    // Execute queries and get results from cursors
    Cursor cName = clName.loadInBackground();
    Cursor cEmail = clEmail.loadInBackground();
    final int nameIndex = cName.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME);
    final int emailIndex = cEmail.getColumnIndex(ContactsContract.CommonDataKinds.Email.DATA);

    User user = null;
    if (cName.moveToFirst()) {
        user = new User();
        user.setName(cName.getString(nameIndex));
    }

    // This only could be true if selected contact has at least one email
    if (cEmail.moveToFirst() && user != null) {
        user.setEmail(cEmail.getString(emailIndex));
    }

    cName.close();
    cEmail.close();
    return user;
}

感谢线索@Luksprog！在之前的代码中，我只是通过电子邮件查询联系人，因为CursorLoader的{{1}}是Uri。

选择联系人姓名和电子邮件（如果存

1 个答案: