Looks like the trick was to use [] around the udt value literal.
Any way to do this using the java driver?
On Sat, Nov 12, 2016 at 7:58 PM, Ali Akhtar <ali.rac...@gmail.com> wrote:
> Changing the double quotes to single quotes gives:
>
> UPDATE my_table SET labels = labels + {id: 'foo'} where id = '';
>
> InvalidRequest: Error from server: code=2200 [Invalid query]
> message="Invalid user type literal for labels of type list<frozen<label>>"
>
>
> On Sat, Nov 12, 2016 at 7:50 PM, Ali Akhtar <ali.rac...@gmail.com> wrote:
>
>> The question is about appending to a set of frozen<udt> and how to do
>> that while avoiding the race condition.
>>
>> If I run:
>>
>> UPDATE my_table SET labels = labels + {id: "foo"} where id = 'xx';
>>
>> I get:
>>
>> SyntaxException: line 1:57 no viable alternative at input '}' (...=
>> labels + {id: ["fo]o"}...)
>>
>> Here labels is set<frozen<label>>
>>
>> On Sat, Nov 12, 2016 at 7:40 PM, Vladimir Yudovin <vla...@winguzone.com>
>> wrote:
>>
>>> If I used consistency = ALL both when getting the record, and when
>>> saving the record, will that avoid the race condition?
>>> If I use consistency level = all, will that cause it to end up with
>>> [1,2]?
>>> No. Even if you have only one host it's possible that two threads first
>>> both read data and than overwrite existing value one by one.
>>>
>>> The list is actually of a list<frozen<my_udt>> and not a text (I used
>>> text for simplification, apologies).
>>> In that case, will updates still merge the list values instead of
>>> overwriting them?
>>> Do you mean UPDATE cql operation? Yes, it adds new values to list,
>>> allowing duplicates.
>>>
>>> When setting a new value to a list, C* will do a read-delete-write
>>> internally e.g. read the current list, remove all its value (by a range
>>> tombstone) and then write the new list.
>>> As I mentioned duplicates are allowed in LIST, and as DOC says:
>>>
>>> These update operations are implemented internally without any
>>> read-before-write. Appending and prepending a new element to the list
>>> writes only the new element.
>>>
>>> Only when using index
>>>
>>> When you add an element at a particular position, Cassandra reads the
>>> entire list, and then writes only the updated element. Consequently, adding
>>> an element at a particular position results in greater latency than
>>> appending or prefixing an element to a list.
>>>
>>>
>>> Best regards, Vladimir Yudovin,
>>>
>>> *Winguzone <https://winguzone.com?from=list> - Hosted Cloud
>>> CassandraLaunch your cluster in minutes.*
>>>
>>>
>>> ---- On Sat, 12 Nov 2016 07:57:36 -0500*Ali Akhtar
>>> <ali.rac...@gmail.com <ali.rac...@gmail.com>>* wrote ----
>>>
>>> The labels collection is of the type set<frozen<label>> , where label is
>>> a udt containing: id, name, description , all text fields.
>>>
>>> On Sat, Nov 12, 2016 at 5:54 PM, Ali Akhtar <ali.rac...@gmail.com>
>>> wrote:
>>>
>>> The problem isn't just the update / insert though, right? Don't frozen
>>> entities get overwritten completely? So if I had [1] [2] being written as
>>> updates, won't each update overwrite the set completely, so i'll end up
>>> with either one of them instead of [1,2]?
>>>
>>> On Sat, Nov 12, 2016 at 5:50 PM, DuyHai Doan <doanduy...@gmail.com>
>>> wrote:
>>>
>>> Maybe you should use my Achilles mapper, which does generates UPDATE
>>> statements on collections and not only INSERT
>>> Le 12 nov. 2016 13:08, "Ali Akhtar" <ali.rac...@gmail.com> a écrit :
>>>
>>> I am using the Java Cassandra mapper for all of these cases, so my code
>>> looks like this:
>>>
>>> Item myItem = myaccessor.get( itemId );
>>> Mapper<Item> mapper = mappingManager.create( Item.class );
>>>
>>> myItem.labels.add( newLabel );
>>> mapper.save( myItem );
>>>
>>> On Sat, Nov 12, 2016 at 5:06 PM, Ali Akhtar <ali.rac...@gmail.com>
>>> wrote:
>>>
>>> Thanks DuyHai, I will switch to using a set.
>>>
>>> But I'm still not sure how to resolve the original question.
>>>
>>> - Original labels = []
>>> - Request 1 arrives with label = 1, and request 2 arrives with label = 2
>>> - Updates are sent to c* with labels = [1] and labels = [2]
>>> simultaneously.
>>>
>>> What will happen in the above case? Will it cause the labels to end up
>>> as [1,2] (what I want) or either [1] or [2]?
>>>
>>> If I use consistency level = all, will that cause it to end up with
>>> [1,2]?
>>>
>>> On Sat, Nov 12, 2016 at 4:59 PM, DuyHai Doan <doanduy...@gmail.com>
>>> wrote:
>>>
>>> Don't use list, use set instead. If you need ordering of insertion, use
>>> a map<timeuuid,text> where timeuuid is generated by the client to guarantee
>>> insertion order
>>>
>>> When setting a new value to a list, C* will do a read-delete-write
>>> internally e.g. read the current list, remove all its value (by a range
>>> tombstone) and then write the new list. Please note that prepend & append
>>> operations on list do not require this read-delete-write and thus performs
>>> slightly better
>>>
>>> On Sat, Nov 12, 2016 at 11:34 AM, Ali Akhtar <ali.rac...@gmail.com>
>>> wrote:
>>>
>>> I have a table where each record contains a list<string> of labels.
>>>
>>> I have an endpoint which responds to new labels being added to a record
>>> by the user.
>>>
>>> Consider the following scenario:
>>>
>>> - Record X, labels = []
>>> - User selects 2 labels, clicks a button, and 2 http requests are
>>> generated.
>>> - The server receives request for Label 1 and Label 2 at the same time.
>>> - Both requests see the labels as empty, add 1 label to the collection,
>>> and send it.
>>> - Record state as label 1 request sees it: [1], as label 2 sees it: [2]
>>>
>>> How will the above conflict be resolved? What can I do so I end up with
>>> [1, 2] instead of either [1] or [2] after both requests have been processed?
>>>
>>>
>>>
>>
>
