android - 如何一次查询到联系人的所有详细信息

Question

编辑:我认为重要的联系方式列表:

1.NAME
2.PHONE NUMBER
3.EMAIL ADDRESS
4.WEBSITE
5.PHYSICAL ADDRESS 

我更愿意使用预取的 contactId 来执行此操作...仅使用一个游标来获取所有指定的数据。我更愿意找到正确的查询来执行此操作:

我想立即获取联系人的所有重要详细信息，我正在使用以下代码来执行此操作:

       public void getAllDataByContactId(int contactId)
{
    Log.d(TAG, "Seriously scared it might not work");
    String phoneNo="Phone disconnected";
    String email="Email could not be delivered";
    String website="Website 404";
    String address="Number 13,Dark Street,Area 51,Bermuda Trianlge";
    String name="Clint Eastwood";
    int hasPhoneNumber;
    String selection=ContactsContract.Data.CONTACT_ID+"=?";
    String[] selectionArgs={String.valueOf(contactId)};
    Cursor c=context.getContentResolver().query(ContactsContract.Data.CONTENT_URI, null,selection, selectionArgs,ContactsContract.Data.TIMES_CONTACTED);

    if(c!=null && c.getCount()>0)
    {

        while(c.moveToNext())
        {

            phoneNo=c.getString(c.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));
            Log.d(TAG, "Phone number: "+phoneNo);
            email=c.getString(c.getColumnIndex(ContactsContract.CommonDataKinds.Email.ADDRESS));
            Log.d(TAG, "Email: "+email);
            website=c.getString(c.getColumnIndex(ContactsContract.CommonDataKinds.Website.URL));
            Log.d(TAG, "Website :"+website);
            address=c.getString(c.getColumnIndex(ContactsContract.CommonDataKinds.StructuredPostal.FORMATTED_ADDRESS));
            name=c.getString(c.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.DISPLAY_NAME));
            Log.d(TAG, "Name :"+name);
        }
    }   
}

但是，尽管这不会引发错误，但它会显示许多行，这些行由一个空字符串组成，中间夹杂着实际值。我该如何编写一个消除干扰的查询？

我试过了，这让我得到了所有的值(value):

             String selection=ContactsContract.Data.CONTACT_ID+"=? AND "+ContactsContract.Data.MIMETYPE+"=? OR "+ContactsContract.Data.MIMETYPE+"=? OR "+ContactsContract.Data.MIMETYPE+"=? OR "+ContactsContract.Data.MIMETYPE+"=? OR "+ContactsContract.Data.MIMETYPE+"=?";
    String[] selectionArgs={String.valueOf(contactId),ContactsContract.CommonDataKinds.StructuredName.CONTENT_ITEM_TYPE,ContactsContract.CommonDataKinds.Email.CONTENT_ITEM_TYPE,ContactsContract.CommonDataKinds.Phone.CONTENT_ITEM_TYPE,ContactsContract.CommonDataKinds.Website.CONTENT_ITEM_TYPE,ContactsContract.CommonDataKinds.StructuredPostal.CONTENT_ITEM_TYPE};

Answer 1

来不及回答，但也许它可以帮助将来的人。我对这个问题的解决方案只有一个 while 循环和查询:

private void fetchContacts(ContentResolver contentResolver) {
        if (contentResolver == null) return;

        Cursor cursor = contentResolver.query(ContactsContract.Data.CONTENT_URI,
                null, null, null, null);

        if (cursor == null || cursor.getCount() <= 0) {
            return;
        }

        String prevId = "";
        String contactId = "";
        PersonContact personContact = null;
        while (cursor.moveToNext()) {
            String company = "";
            String columnName = cursor.getString(cursor.getColumnIndex("mimetype"));
            if (columnName.equals(ContactsContract.CommonDataKinds.Organization.CONTENT_ITEM_TYPE)) {
                company = cursor.getString(cursor.getColumnIndex("data1"));
            }

            String email = "";
            if (columnName.equals(ContactsContract.CommonDataKinds.Email.CONTENT_ITEM_TYPE)) {
                email = cursor.getString(cursor.getColumnIndex(ContactsContract.CommonDataKinds.Email.DATA));
            }
            String phone = "";
            if (columnName.equals(ContactsContract.CommonDataKinds.Phone.CONTENT_ITEM_TYPE)) {
                phone = cursor.getString(cursor.getColumnIndex("data1"));
            }
            String first = "";
            String last = "";
            if (columnName.equals(ContactsContract.CommonDataKinds.StructuredName.CONTENT_ITEM_TYPE)) {
                first = cursor.getString(cursor.getColumnIndex("data2"));
                last = cursor.getString(cursor.getColumnIndex("data3"));
            }

            if (!prevId.equals(contactId)) {
                if (!TextUtils.isEmpty(prevId)) {
                    addFilteredList(personContact);
                    allContacts.put(prevId, personContact);
                }
                prevId = contactId;
                personContact = new PersonContact();
            } else {
                if (personContact != null) {
                    personContact.id = prevId;
                    if (TextUtils.isEmpty(personContact.company)) personContact.company = company;
                    if (TextUtils.isEmpty(personContact.firstName)) personContact.firstName = first;
                    if (TextUtils.isEmpty(personContact.lastName)) personContact.lastName = last;
                    if (!TextUtils.isEmpty(email) && personContact.emails.size() == 0) {
                        personContact.emails.add(email);
                    }
                    if (!TextUtils.isEmpty(phone) && personContact.phoneNumbers.size() == 0) {
                        personContact.phoneNumbers.add(phone);
                    }
                }
            }

            contactId = cursor.getString(cursor.getColumnIndex(ContactsContract.CommonDataKinds.Phone.CONTACT_ID));
        }
        cursor.close();
    }

如您所见，我使用了 prevId 字段，因为 cursor.moveToNext 对一个联系人执行了多次(一次是名字和姓氏，一次是电话等)。每次迭代后，我检查以前的联系人标识符与当前标识符，如果为假，我更新 personContact 模型中的字段。