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From Lucifer Dignified <vineetdan...@gmail.com>
Subject Re: How to perform queries on Cassandra?
Date Sun, 11 Apr 2010 09:33:16 GMT
Hi Benjamin

I'll try to make it more clear to you.
We have a user table with fields 'id', 'username', and 'password'. Now if
use the ideal way to store key/value, like :
username : vineetdaniel
timestamp
password : <password>
timestamp

second user :

username: <seconduser>
timestamp
password:<password>

and so on, here what i assume is that as we cannot make search on values (as
confirmed by guys on cassandra forums) we are not able to perform robust
'where' queries. Now what i propose is this.

Rather than using a static values for column names use values itself and
unique key as identifier. So, the above example when put in as per me would
be.

vineetdaniel : vineetdaniel
timestamp

<password>:<password>
timestamp

second user
seconduser:seconduser
timestamp

password:password
timestamp

By using above methodology we can simply make search on keys itself rather
than going into using different CF's. But to add further, this cannot be
used for every situation. I am still exploring this, and soon will be
updating the group and my blog with information pertaining to this. As
cassandra is new, I think every idea or experience should be shared with the
community.

I hope I example is clear this time. Should you have any queries feel free
to revert.

On Sun, Apr 11, 2010 at 2:01 PM, Benjamin Black <b@b3k.us> wrote:

> Sorry, I don't understand your example.
>
> On Sun, Apr 11, 2010 at 12:54 AM, Lucifer Dignified
> <vineetdaniel@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 <sikerasakti@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 <paul@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 <sikerasakti@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
> >> >> > <malsmith@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.aktas@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
> >> >> >>
> >> >> >
> >> >> >
> >> >
> >> >
> >
> >
>

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