How to handle same usernames. Otherwise seems fine to me. On Sun, Apr 11, 2010 at 6:17 PM, Dop Sun <su...@dopsun.com> wrote:
> Hi, > > > > As far as I can see it, the Cassandra API currently supports criterias on: > > Token – Key – Super Column Name (if applicable) - Column Names > > > > I guess Token is not usually used for the day to day queries, so, Key and > Column Names are normally used for querying. For the user name and password > case, I guess it can be done like this: > > > > Define a CF as UserAuth with type as Super, and Key is user name, while > password can be the SuperKeyName. So, while you receive the user name and > password from the UI (or any other methods), it can be queried via: > multiget_slice or get_range_slices, if there are anything returned, means > that the user name and password matches. > > > > If not using the super column name, and put the password as the column > name, the column name usually not used for these kind of discretionary > values (actually, I don’t see any definitive documents on how to use the > column Names and Super Columns, flexibility is the good of Cassandra, or is > it bad if abused? :P) > > > > Not sure whether this is the best way, but I guess it will work. > > > > Regards, > > Dop > > > > *From:* Lucifer Dignified [mailto:vineetdan...@gmail.com] > *Sent:* Sunday, April 11, 2010 5:33 PM > *To:* user@cassandra.apache.org > *Subject:* Re: How to perform queries on Cassandra? > > > > 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 > <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 > >> >> >> > >> >> > > >> >> > > >> > > >> > > > > > > > >
