Sorry, I don't understand your example.
On Sun, Apr 11, 2010 at 12:54 AM, Lucifer Dignified
<vineetdan...@gmail.com> wrote:
> Benjamin I quite agree to you, but what in case of duplicate usernames,
> suppose if I am not using unique names as in email id's . If we have
> duplicacy in usernames we cannot use it for key, so what should be the
> solution. I think keeping incremental numeric id as key and keeping the name
> and value same in the column family.
>
> Example :
> User1 has password as 123456
>
> Cassandra structure :
>
> 1 as key
> user1 - column name
> value - user1
> 123456 - column name
> value - 123456
>
> I m thinking of doing it this way for my applicaton, this way i can run
> different sorts of queries too. Any feedback on this is welcome.
>
> On Sun, Apr 11, 2010 at 1:13 PM, Benjamin Black <b...@b3k.us> wrote:
>>
>> You would have a Column Family, not a column for that; let's call it
>> the Users CF. You'd use username as the row key and have a column
>> called 'password'. For your example query, you'd retrieve row key
>> 'usr2', column 'password'. The general pattern is that you create CFs
>> to act as indices for each query you want to perform. There is no
>> equivalent to a relational store to perform arbitrary queries. You
>> must structure things to permit the queries of interest.
>>
>>
>> b
>>
>> On Sat, Apr 10, 2010 at 8:34 PM, dir dir <sikerasa...@gmail.com> wrote:
>> > I have already read the API spesification. Honestly I do not understand
>> > how to use it. Because there are not an examples.
>> >
>> > For example I have a column like this:
>> >
>> > UserName Password
>> > usr1 abc
>> > usr2 xyz
>> > usr3 opm
>> >
>> > suppose I want query the user's password using SQL in RDBMS
>> >
>> > Select Password From Users Where UserName = "usr2";
>> >
>> > Now I want to get the password using OODBMS DB4o Object Query and Java
>> >
>> > ObjectSet QueryResult = db.query(new Predicate()
>> > {
>> > public boolean match(Users Myusers)
>> > {
>> > return Myuser.getUserName() == "usr2";
>> > }
>> > });
>> >
>> > After we get the Users instance in the QueryResult, hence we can get the
>> > usr2's password.
>> >
>> > How we perform this query using Cassandra API and Java??
>> > Would you tell me please?? Thank You.
>> >
>> > Dir.
>> >
>> >
>> > On Sat, Apr 10, 2010 at 11:06 AM, Paul Prescod <p...@prescod.net> wrote:
>> >>
>> >> No. Cassandra has an API.
>> >>
>> >> http://wiki.apache.org/cassandra/API
>> >>
>> >> On Fri, Apr 9, 2010 at 8:00 PM, dir dir <sikerasa...@gmail.com> wrote:
>> >> > Does Cassandra has a default query language such as SQL in RDBMS
>> >> > and Object Query in OODBMS? Thank you.
>> >> >
>> >> > Dir.
>> >> >
>> >> > On Sat, Apr 10, 2010 at 7:01 AM, malsmith
>> >> > <malsm...@treehousesystems.com>
>> >> > wrote:
>> >> >>
>> >> >>
>> >> >> It's sort of an interesting problem - in RDBMS one relatively simple
>> >> >> approach would be calculate a rectangle that is X km by Y km with
>> >> >> User
>> >> >> 1's
>> >> >> location at the center. So the rectangle is UserX - 10KmX ,
>> >> >> UserY-10KmY to
>> >> >> UserX+10KmX , UserY+10KmY
>> >> >>
>> >> >> Then you could query the database for all other users where that
>> >> >> each
>> >> >> user
>> >> >> considered is curUserX > UserX-10Km and curUserX < UserX+10KmX and
>> >> >> curUserY
>> >> >> > UserY-10KmY and curUserY < UserY+10KmY
>> >> >> * Not the 10KmX and 10KmY are really a translation from Kilometers
>> >> >> to
>> >> >> degrees of lat and longitude (that you can find on a google
>> >> >> search)
>> >> >>
>> >> >> With the right indexes this query actually runs pretty well.
>> >> >>
>> >> >> Translating that to Cassandra seems a bit complex at first - but you
>> >> >> could
>> >> >> try something like pre-calculating a grid with the right resolution
>> >> >> (like a
>> >> >> square of 5KM per side) and assign every user to a particular grid
>> >> >> ID.
>> >> >> That
>> >> >> way you just calculate with grid ID User1 is in then do a direct key
>> >> >> lookup
>> >> >> to get a list of the users in that same grid id.
>> >> >>
>> >> >> A second approach would be to have to column families -- one that
>> >> >> maps
>> >> >> a
>> >> >> Latitude to a list of users who are at that latitude and a second
>> >> >> that
>> >> >> maps
>> >> >> users who are at a particular longitude. You could do the same
>> >> >> rectange
>> >> >> calculation above then do a get_slice range lookup to get a list of
>> >> >> users
>> >> >> from range of latitude and a second list from the range of
>> >> >> longitudes.
>> >> >> You would then need to do a in-memory nested loop to find the list
>> >> >> of
>> >> >> users
>> >> >> that are in both lists. This second approach could cause some
>> >> >> trouble
>> >> >> depending on where you search and how many users you really have --
>> >> >> some
>> >> >> latitudes and longitudes have many many people in them
>> >> >>
>> >> >> So, it seems some version of a chunking / grid id thing would be the
>> >> >> better approach. If you let people zoom in or zoom out - you could
>> >> >> just
>> >> >> have different column families for each level of zoom.
>> >> >>
>> >> >>
>> >> >> I'm stuck on a stopped train so -- here is even more code:
>> >> >>
>> >> >> static Decimal GetLatitudeMiles(Decimal lat)
>> >> >> {
>> >> >> Decimal f = 0.0M;
>> >> >> lat = Math.Abs(lat);
>> >> >> f = 68.99M;
>> >> >> if (lat >= 0.0M && lat < 10.0M) { f = 68.71M; }
>> >> >> else if (lat >= 10.0M && lat < 20.0M) { f = 68.73M; }
>> >> >> else if (lat >= 20.0M && lat < 30.0M) { f = 68.79M; }
>> >> >> else if (lat >= 30.0M && lat < 40.0M) { f = 68.88M; }
>> >> >> else if (lat >= 40.0M && lat < 50.0M) { f = 68.99M; }
>> >> >> else if (lat >= 50.0M && lat < 60.0M) { f = 69.12M; }
>> >> >> else if (lat >= 60.0M && lat < 70.0M) { f = 69.23M; }
>> >> >> else if (lat >= 70.0M && lat < 80.0M) { f = 69.32M; }
>> >> >> else if (lat >= 80.0M) { f = 69.38M; }
>> >> >>
>> >> >> return f;
>> >> >> }
>> >> >>
>> >> >>
>> >> >> Decimal MilesPerDegreeLatitude =
>> >> >> GetLatitudeMiles(zList[0].Latitude);
>> >> >> Decimal MilesPerDegreeLongitude = ((Decimal)
>> >> >> Math.Abs(Math.Cos((Double)
>> >> >> zList[0].Latitude))) * 24900.0M / 360.0M;
>> >> >> dRadius = 10.0M // ten miles
>> >> >> Decimal deltaLat = dRadius / MilesPerDegreeLatitude;
>> >> >> Decimal deltaLong = dRadius / MilesPerDegreeLongitude;
>> >> >>
>> >> >> ps.TopLatitude = zList[0].Latitude - deltaLat;
>> >> >> ps.TopLongitude = zList[0].Longitude - deltaLong;
>> >> >> ps.BottomLatitude = zList[0].Latitude + deltaLat;
>> >> >> ps.BottomLongitude = zList[0].Longitude + deltaLong;
>> >> >>
>> >> >>
>> >> >>
>> >> >> On Fri, 2010-04-09 at 16:30 -0700, Paul Prescod wrote:
>> >> >>
>> >> >> 2010/4/9 Onur AKTAS <onur.ak...@live.com>:
>> >> >> > ...
>> >> >> > I'm trying to find out how do you perform queries with
>> >> >> > calculations
>> >> >> > on
>> >> >> > the
>> >> >> > fly without inserting the data as calculated from the beginning.
>> >> >> > Lets say we have latitude and longitude coordinates of all users
>> >> >> > and
>> >> >> > we
>> >> >> > have
>> >> >> > Distance(from_lat, from_long, to_lat, to_long) function which
>> >> >> > gives distance between lat/longs pairs in kilometers.
>> >> >>
>> >> >> I'm not an expert, but I think that it boils down to "MapReduce" and
>> >> >> "Hadoop".
>> >> >>
>> >> >> I don't think that there's any top-down tutorial on those two words,
>> >> >> you'll have to research yourself starting here:
>> >> >>
>> >> >> * http://en.wikipedia.org/wiki/MapReduce
>> >> >>
>> >> >> * http://hadoop.apache.org/
>> >> >>
>> >> >> * http://wiki.apache.org/cassandra/HadoopSupport
>> >> >>
>> >> >> I don't think it is all documented in any one place yet...
>> >> >>
>> >> >> Paul Prescod
>> >> >>
>> >> >
>> >> >
>> >
>> >
>
>
