在以编程方式添加联系人时,如何使联系人正确聚合？

Question

我看到了这个问题和答案,但添加电话信息(甚至是电子邮件)仍然不会导致联系信息正确聚合(当我检查People应用程序时,我可以看到同名的多个条目).

这是我用来测试它的代码.

//get the account
Account acct = null;
Account[] accounts = AccountManager.get(getContext()).getAccounts(); 
for (Account acc : accounts){
    acct = acc;
}//assuming there's only one account in there (in my case I know there is)

//loop a few times, creating a new contact each time. In theory, if they have the same name they should aggregate
for(int i=0; i<3; i++){
    ArrayList<ContentProviderOperation> ops = new ArrayList<ContentProviderOperation>();
    ops.add(ContentProviderOperation.newInsert(ContactsContract.RawContacts.CONTENT_URI)
                .withValue(ContactsContract.RawContacts.ACCOUNT_TYPE, acct.type)
                .withValue(ContactsContract.RawContacts.ACCOUNT_NAME, acct.name)
                .withValue(ContactsContract.RawContacts.AGGREGATION_MODE, ContactsContract.RawContacts.AGGREGATION_MODE_DEFAULT) 
                .build());
    ops.add(ContentProviderOperation.newInsert(ContactsContract.Data.CONTENT_URI)
                .withValueBackReference(ContactsContract.Data.RAW_CONTACT_ID, 0)
                .withValue(ContactsContract.Data.MIMETYPE,
                        ContactsContract.CommonDataKinds.StructuredName.CONTENT_ITEM_TYPE)
                .withValue(ContactsContract.CommonDataKinds.StructuredName.DISPLAY_NAME, "ContactName")
                .build());
    ops.add(ContentProviderOperation.newInsert(ContactsContract.Data.CONTENT_URI)
                .withValueBackReference(ContactsContract.Data.RAW_CONTACT_ID, 0)
                .withValue(ContactsContract.Data.MIMETYPE,
                        ContactsContract.CommonDataKinds.Phone.CONTENT_ITEM_TYPE)
                .withValue(ContactsContract.CommonDataKinds.Phone.NUMBER, "1234567890")
                .withValue(ContactsContract.CommonDataKinds.Phone.TYPE, 1)
                .build());
    ops.add(ContentProviderOperation.newInsert(ContactsContract.Data.CONTENT_URI)
                .withValueBackReference(ContactsContract.Data.RAW_CONTACT_ID, 0)
                .withValue(ContactsContract.Data.MIMETYPE,
                        ContactsContract.CommonDataKinds.Email.CONTENT_ITEM_TYPE)
                .withValue(ContactsContract.CommonDataKinds.Email.DATA, "email@address.com")
                .withValue(ContactsContract.CommonDataKinds.Email.TYPE, 1)
                .build());

    try{        
        getContentResolver().applyBatch(ContactsContract.AUTHORITY, ops);
    }
    catch (Exception e) {
        Log.e("Contacts", "Something went wrong during creation! " + e);
        e.printStackTrace();
    }
}

Answer 1

如果它们没有自动聚合,您可以通过向AggregationExceptions表添加一行来手动聚合它们.请确保您在文档中注意到不允许插入.你必须做一个更新.那现在抓到了我两次.以下代码应聚合id为1和2的两个原始联系人:

ContentValues cv = new ContentValues();
cv.put(AggregationExceptions.TYPE, AggregationExceptions.TYPE_KEEP_TOGETHER);
cv.put(AggregationExceptions.RAW_CONTACT_ID1, 1);
cv.put(AggregationExceptions.RAW_CONTACT_ID2, 2);
getContentResolver().update(AggregationExceptions.CONTENT_URI, cv, null, null);