Re: Converting a TimeUUID to a long (timestamp) and vice-versa
It was our original intention on discussing this feature was to have
back-and-forth conversion from timestamps (we were modelling similar
functionality in Pycassa). It's lack of inclusion may have just been
an oversight. We will add this in Hector trunk shortly - thanks for
the complete code sample.
On Tue, Jan 4, 2011 at 10:06 PM, Roshan Dawrani roshandawr...@gmail.com wrote:
Ok, found the solution - finally ! - by applying opposite of what
createTime() does in TimeUUIDUtils. Ideally I would have preferred for this
solution to come from Hector API, so I didn't have to be tied to the private
createTime() implementation.
import java.util.UUID;
import me.prettyprint.cassandra.utils.TimeUUIDUtils;
public class TryHector {
public static void main(String[] args) throws Exception {
final long NUM_100NS_INTERVALS_SINCE_UUID_EPOCH =
0x01b21dd213814000L;
UUID u1 = TimeUUIDUtils.getUniqueTimeUUIDinMillis();
final long t1 = u1.timestamp();
long tmp = (t1 - NUM_100NS_INTERVALS_SINCE_UUID_EPOCH) / 1;
UUID u2 = TimeUUIDUtils.getTimeUUID(tmp);
long t2 = u2.timestamp();
System.out.println(u2.equals(u1));
System.out.println(t2 == t1);
}
}
On Wed, Jan 5, 2011 at 8:15 AM, Roshan Dawrani roshandawr...@gmail.com
wrote:
If I use com.eaio.uuid.UUID directly, then I am able to do what I need
(attached a Java program for the same), but unfortunately I need to deal
with java.util.UUID in my application and I don't have its equivalent
com.eaio.uuid.UUID at the point where I need the timestamp value.
Any suggestion on how I can achieve the equivalent using Hector library's
TimeUUIDUtils?
On Wed, Jan 5, 2011 at 7:21 AM, Roshan Dawrani roshandawr...@gmail.com
wrote:
Hi Victor / Patricio,
I have been using Hector library's TimeUUIDUtils. I also just looked at
TimeUUIDUtilsTest also but didn't find anything similar being tested there.
Here is what I am trying and it's not working - I am creating a Time
UUID, extracting its timestamp value and with that I create another Time
UUID and I am expecting both time UUIDs to have the same timestamp() value -
am I doing / expecting something wrong here?:
===
import java.util.UUID;
import me.prettyprint.cassandra.utils.TimeUUIDUtils;
public class TryHector {
public static void main(String[] args) throws Exception {
UUID someUUID = TimeUUIDUtils.getUniqueTimeUUIDinMillis();
long timestamp1 = someUUID.timestamp();
UUID otherUUID = TimeUUIDUtils.getTimeUUID(timestamp1);
long timestamp2 = otherUUID.timestamp();
System.out.println(timestamp1);
System.out.println(timestamp2);
}
}
===
I have to create the timestamp() equivalent of my time UUIDs so I can
send it to my UI client, for which it will be simpler to compare long
timestamp than comparing UUIDs. Then for the long timestamp chosen by the
client, I need to re-create the equivalent time UUID and go and filter the
data from Cassandra database.
--
Roshan
Blog: http://roshandawrani.wordpress.com/
Twitter: @roshandawrani
Skype: roshandawrani
On Wed, Jan 5, 2011 at 1:32 AM, Victor Kabdebon
victor.kabde...@gmail.com wrote:
Hi Roshan,
Sorry I misunderstood your problem.It is weird that it doesn't work, it
works for me...
As Patricio pointed out use hector standard way of creating TimeUUID
and tell us if it still doesn't work.
Maybe you can paste here some of the code you use to query your columns
too.
Victor K.
http://www.voxnucleus.fr
2011/1/4 Patricio Echagüe patric...@gmail.com
In Hector framework, take a look at TimeUUIDUtils.java
You can create a UUID using TimeUUIDUtils.getTimeUUID(long time); or
TimeUUIDUtils.getTimeUUID(ClockResolution clock)
and later on, TimeUUIDUtils.getTimeFromUUID(..) or just
UUID.timestamp();
There are some example in TimeUUIDUtilsTest.java
Let me know if it helps.
On Tue, Jan 4, 2011 at 10:27 AM, Roshan Dawrani
roshandawr...@gmail.com wrote:
Hello Victor,
It is actually not that I need the 2 UUIDs to be exactly same - they
need to be same timestamp wise.
So, what I need is to extract the timestamp portion from a time UUID
(say, U1) and then later in the cycle, use the same long timestamp value
to
re-create a UUID (say, U2) that is equivalent of the previous one in
terms
of its timestamp portion - i.e., I should be able to give this U2 and
filter
the data from a column family - and it should be same as if I had used
the
original UUID U1.
Does it make any more sense than before? Any way I can do that?
rgds,
Roshan
On Tue, Jan 4, 2011 at 11:46 PM, Victor Kabdebon
victor.kabde...@gmail.com wrote:
Hello Roshan,
Well it
Re: Converting a TimeUUID to a long (timestamp) and vice-versa
Roshan, just a comment in your solution. The time returned is not a simple
long. It also contains some bits indicating the version.
On the other hand, you are assuming that the same machine is processing your
request and recreating a UUID base on a long you provide. The
clockseqAndNode id will vary if another machine takes care of the request
(referring to your use case) .
Is it possible for you to send the UUID to the view? I think that would be
the correct behavior as a simple long does not contain enough information to
recreate the original UUID.
Does it make sense?
On Wed, Jan 5, 2011 at 8:36 AM, Nate McCall n...@riptano.com wrote:
It was our original intention on discussing this feature was to have
back-and-forth conversion from timestamps (we were modelling similar
functionality in Pycassa). It's lack of inclusion may have just been
an oversight. We will add this in Hector trunk shortly - thanks for
the complete code sample.
On Tue, Jan 4, 2011 at 10:06 PM, Roshan Dawrani roshandawr...@gmail.com
wrote:
Ok, found the solution - finally ! - by applying opposite of what
createTime() does in TimeUUIDUtils. Ideally I would have preferred for
this
solution to come from Hector API, so I didn't have to be tied to the
private
createTime() implementation.
import java.util.UUID;
import me.prettyprint.cassandra.utils.TimeUUIDUtils;
public class TryHector {
public static void main(String[] args) throws Exception {
final long NUM_100NS_INTERVALS_SINCE_UUID_EPOCH =
0x01b21dd213814000L;
UUID u1 = TimeUUIDUtils.getUniqueTimeUUIDinMillis();
final long t1 = u1.timestamp();
long tmp = (t1 - NUM_100NS_INTERVALS_SINCE_UUID_EPOCH) / 1;
UUID u2 = TimeUUIDUtils.getTimeUUID(tmp);
long t2 = u2.timestamp();
System.out.println(u2.equals(u1));
System.out.println(t2 == t1);
}
}
On Wed, Jan 5, 2011 at 8:15 AM, Roshan Dawrani roshandawr...@gmail.com
wrote:
If I use com.eaio.uuid.UUID directly, then I am able to do what I need
(attached a Java program for the same), but unfortunately I need to deal
with java.util.UUID in my application and I don't have its equivalent
com.eaio.uuid.UUID at the point where I need the timestamp value.
Any suggestion on how I can achieve the equivalent using Hector
library's
TimeUUIDUtils?
On Wed, Jan 5, 2011 at 7:21 AM, Roshan Dawrani roshandawr...@gmail.com
wrote:
Hi Victor / Patricio,
I have been using Hector library's TimeUUIDUtils. I also just looked at
TimeUUIDUtilsTest also but didn't find anything similar being tested
there.
Here is what I am trying and it's not working - I am creating a Time
UUID, extracting its timestamp value and with that I create another
Time
UUID and I am expecting both time UUIDs to have the same timestamp()
value -
am I doing / expecting something wrong here?:
===
import java.util.UUID;
import me.prettyprint.cassandra.utils.TimeUUIDUtils;
public class TryHector {
public static void main(String[] args) throws Exception {
UUID someUUID = TimeUUIDUtils.getUniqueTimeUUIDinMillis();
long timestamp1 = someUUID.timestamp();
UUID otherUUID = TimeUUIDUtils.getTimeUUID(timestamp1);
long timestamp2 = otherUUID.timestamp();
System.out.println(timestamp1);
System.out.println(timestamp2);
}
}
===
I have to create the timestamp() equivalent of my time UUIDs so I can
send it to my UI client, for which it will be simpler to compare long
timestamp than comparing UUIDs. Then for the long timestamp chosen by
the
client, I need to re-create the equivalent time UUID and go and filter
the
data from Cassandra database.
--
Roshan
Blog: http://roshandawrani.wordpress.com/
Twitter: @roshandawrani
Skype: roshandawrani
On Wed, Jan 5, 2011 at 1:32 AM, Victor Kabdebon
victor.kabde...@gmail.com wrote:
Hi Roshan,
Sorry I misunderstood your problem.It is weird that it doesn't work,
it
works for me...
As Patricio pointed out use hector standard way of creating TimeUUID
and tell us if it still doesn't work.
Maybe you can paste here some of the code you use to query your
columns
too.
Victor K.
http://www.voxnucleus.fr
2011/1/4 Patricio Echagüe patric...@gmail.com
In Hector framework, take a look at TimeUUIDUtils.java
You can create a UUID using TimeUUIDUtils.getTimeUUID(long time);
or
TimeUUIDUtils.getTimeUUID(ClockResolution clock)
and later on, TimeUUIDUtils.getTimeFromUUID(..) or just
UUID.timestamp();
There are some example in TimeUUIDUtilsTest.java
Let me know if it helps.
On Tue, Jan 4,
Re: Converting a TimeUUID to a long (timestamp) and vice-versa
Hi Patricio, Thanks for your comment. Replying inline. 2011/1/5 Patricio Echagüe patric...@gmail.com Roshan, just a comment in your solution. The time returned is not a simple long. It also contains some bits indicating the version. I don't think so. The version bits from the most significant 64 bits of the UUID are not used in creating timestamp() value. It uses only time_low, time_mid and time_hi fields of the UUID and not version, as documented here: http://download.oracle.com/javase/1.5.0/docs/api/java/util/UUID.html#timestamp%28%29. When the same timestamp comes back and I call TimeUUIDUtils.getTimeUUID(tmp), it internally puts the version back in it and makes it a Time UUID. On the other hand, you are assuming that the same machine is processing your request and recreating a UUID base on a long you provide. The clockseqAndNode id will vary if another machine takes care of the request (referring to your use case) . When I recreate my UUID using the timestamp() value, my requirement is not to arrive at the exactly same UUID from which timestamp() was derived in the first place. I need a recreated UUID *that should be equivalent in terms of its time value* - so that filtering the time-sorted columns using this time UUID works fine. So, if the lower order 64 bits (clockseq + node) become different, I don't think it is of any concern because the UUID comparison first goes by most significant 64 bits, i.e. the time value and that should settle the time comparison in my use case. Is it possible for you to send the UUID to the view? I think that would be the correct behavior as a simple long does not contain enough information to recreate the original UUID. In my use case, the non-Java clients will be receiving a number of such UUIDs then and they will have to sort them chronologically. I wanted to avoid bits based UUID comparison in these clients. Long timestamp() value is perfect for such ordering of data elements and I send much lesser amount of data over the wire. Does it make sense? Nearly everything makes sense to me :-) -- Roshan Blog: http://roshandawrani.wordpress.com/ Twitter: @roshandawrani http://twitter.com/roshandawrani Skype: roshandawrani
Re: Converting a TimeUUID to a long (timestamp) and vice-versa
Roshan, the first 64 bits does contain the version. The method UUID.timestamp() indeed takes it out before returning. You are right in that point. I based my comment on the UUID spec. What I am not convinced is that the framework should provide support to create an almost identical UUID where only the timestamp is the same (between the two UUIDs). UUID.equals() and UUID.compareTo() does compare the whole bit set to say that two objects are the same. It does compare the first 64 bits to avoid comparing the rest in case the most significant bits already show a difference. But coming to your point, should Hector provide that kind of support or do you feel that the problem you have is specific to your application ? I feel like UUID is as it says an Unique Identifier and creating a sort-of UUID based on a previous timestamp disregarding the least significant bits is not the right support Hector should expose. Thoughts? On Wed, Jan 5, 2011 at 6:30 PM, Roshan Dawrani roshandawr...@gmail.comwrote: Hi Patricio, Thanks for your comment. Replying inline. 2011/1/5 Patricio Echagüe patric...@gmail.com Roshan, just a comment in your solution. The time returned is not a simple long. It also contains some bits indicating the version. I don't think so. The version bits from the most significant 64 bits of the UUID are not used in creating timestamp() value. It uses only time_low, time_mid and time_hi fields of the UUID and not version, as documented here: http://download.oracle.com/javase/1.5.0/docs/api/java/util/UUID.html#timestamp%28%29. When the same timestamp comes back and I call TimeUUIDUtils.getTimeUUID(tmp), it internally puts the version back in it and makes it a Time UUID. On the other hand, you are assuming that the same machine is processing your request and recreating a UUID base on a long you provide. The clockseqAndNode id will vary if another machine takes care of the request (referring to your use case) . When I recreate my UUID using the timestamp() value, my requirement is not to arrive at the exactly same UUID from which timestamp() was derived in the first place. I need a recreated UUID *that should be equivalent in terms of its time value* - so that filtering the time-sorted columns using this time UUID works fine. So, if the lower order 64 bits (clockseq + node) become different, I don't think it is of any concern because the UUID comparison first goes by most significant 64 bits, i.e. the time value and that should settle the time comparison in my use case. Is it possible for you to send the UUID to the view? I think that would be the correct behavior as a simple long does not contain enough information to recreate the original UUID. In my use case, the non-Java clients will be receiving a number of such UUIDs then and they will have to sort them chronologically. I wanted to avoid bits based UUID comparison in these clients. Long timestamp() value is perfect for such ordering of data elements and I send much lesser amount of data over the wire. Does it make sense? Nearly everything makes sense to me :-) -- Roshan Blog: http://roshandawrani.wordpress.com/ Twitter: @roshandawrani http://twitter.com/roshandawrani Skype: roshandawrani -- Patricio.-
Re: Converting a TimeUUID to a long (timestamp) and vice-versa
Hi Patricio, Some thoughts inline. 2011/1/6 Patricio Echagüe patric...@gmail.com Roshan, the first 64 bits does contain the version. The method UUID.timestamp() indeed takes it out before returning. You are right in that point. I based my comment on the UUID spec. I know 64 bits have the version, but timestamp() doesn't and hence it is OK to use it for chronological ordering. Anyway, we agree on it now and this point is out. What I am not convinced is that the framework should provide support to create an almost identical UUID where only the timestamp is the same (between the two UUIDs). Well, I didn't really ask for framework to provide me such an almost identical UUID. What I raised was that since Hector is computing UTC time in 100 nano-sec units as utcTime = msec * 1000 + 0x01B21DD213814000L (NUM_100NS_INTERVALS_SINCE_UUID_EPOCH), it should, at the minimum, give a utility function to do the opposite msec = (utcTime - 0x01B21DD213814000L / 1000), so that if someone has to create an almost identical UUID, where timestamp is same, as I needed, *he shouldn't need to deal with such magic numbers that are linked to Hector's guts.* So, I don't mind creating the UUID myself, but I don't want to do magic calculations that should be done inside Hector to-and-fro, as it is an Hector's internal design thing. UUID.equals() and UUID.compareTo() does compare the whole bit set to say that two objects are the same. It does compare the first 64 bits to avoid comparing the rest in case the most significant bits already show a difference. I know it may need to look at all 128 bits eventually - but it first looks at first 64 bits (time stamp) and then the next 64. That's why I qualified it with for my usecase. It works for me, because the data I am filtering is already within a particular user's data-set - and the possibility of a user having 2 data-points at the same nano-second value (so that clockseq+node bits come into picture) is functionally nil. But coming to your point, should Hector provide that kind of support or do you feel that the problem you have is specific to your application ? As covered above, half of my solution should go inside Hector API, I feel. Other half about re-creating the same-timestamp-UUID and comparison using it is specific to my application. I feel like UUID is as it says an Unique Identifier and creating a sort-of UUID based on a previous timestamp disregarding the least significant bits is not the right support Hector should expose. The support Hector should expose is to keep its magic calculations inside to-and-fro. Does it make any sense? -- Roshan Blog: http://roshandawrani.wordpress.com/ Twitter: @roshandawrani http://twitter.com/roshandawrani Skype: roshandawrani
Re: Converting a TimeUUID to a long (timestamp) and vice-versa
Hello Victor, It is actually not that I need the 2 UUIDs to be exactly same - they need to be same timestamp wise. So, what I need is to extract the timestamp portion from a time UUID (say, U1) and then later in the cycle, use the same long timestamp value to re-create a UUID (say, U2) that is equivalent of the previous one in terms of its timestamp portion - i.e., I should be able to give this U2 and filter the data from a column family - and it should be same as if I had used the original UUID U1. Does it make any more sense than before? Any way I can do that? rgds, Roshan On Tue, Jan 4, 2011 at 11:46 PM, Victor Kabdebon victor.kabde...@gmail.comwrote: Hello Roshan, Well it is normal to do not be able to get the exact same UUID from a timestamp, it is its purpose. When you create an UUID you have in fact two information : random 64 bits number - 64 bits timestamp. You put that together and you have your uuid. . So unless you save your random number two UUID for the same milli( or micro) second are different. Best regards, Victor K. http://www.voxnucleus.fr 2011/1/4 Roshan Dawrani roshandawr...@gmail.com Hi, I am having a little difficulty converting a time UUID to its timestamp equivalent and back. Can someone please help? Here is what I am trying. Is it not the right way to do it? === UUID someUUID = TimeUUIDUtils.getUniqueTimeUUIDinMillis(); long time = someUUID.timestamp(); /* convery from UUID to a long timestamp */ UUID otherUUID = TimeUUIDUtils.getTimeUUID(time); /* do the reverse and get back the UUID from timestamp */ System.out.println(someUUID); /* someUUID and otherUUID should be same, but are different */ System.out.println(otherUUID); === -- Roshan Blog: http://roshandawrani.wordpress.com/ Twitter: @roshandawrani http://twitter.com/roshandawrani Skype: roshandawrani -- Roshan Blog: http://roshandawrani.wordpress.com/ Twitter: @roshandawrani http://twitter.com/roshandawrani Skype: roshandawrani
Re: Converting a TimeUUID to a long (timestamp) and vice-versa
In Hector framework, take a look at TimeUUIDUtils.java You can create a UUID using TimeUUIDUtils.getTimeUUID(long time); or TimeUUIDUtils.getTimeUUID(ClockResolution clock) and later on, TimeUUIDUtils.getTimeFromUUID(..) or just UUID.timestamp(); There are some example in TimeUUIDUtilsTest.java Let me know if it helps. On Tue, Jan 4, 2011 at 10:27 AM, Roshan Dawrani roshandawr...@gmail.comwrote: Hello Victor, It is actually not that I need the 2 UUIDs to be exactly same - they need to be same timestamp wise. So, what I need is to extract the timestamp portion from a time UUID (say, U1) and then later in the cycle, use the same long timestamp value to re-create a UUID (say, U2) that is equivalent of the previous one in terms of its timestamp portion - i.e., I should be able to give this U2 and filter the data from a column family - and it should be same as if I had used the original UUID U1. Does it make any more sense than before? Any way I can do that? rgds, Roshan On Tue, Jan 4, 2011 at 11:46 PM, Victor Kabdebon victor.kabde...@gmail.com wrote: Hello Roshan, Well it is normal to do not be able to get the exact same UUID from a timestamp, it is its purpose. When you create an UUID you have in fact two information : random 64 bits number - 64 bits timestamp. You put that together and you have your uuid. . So unless you save your random number two UUID for the same milli( or micro) second are different. Best regards, Victor K. http://www.voxnucleus.fr 2011/1/4 Roshan Dawrani roshandawr...@gmail.com Hi, I am having a little difficulty converting a time UUID to its timestamp equivalent and back. Can someone please help? Here is what I am trying. Is it not the right way to do it? === UUID someUUID = TimeUUIDUtils.getUniqueTimeUUIDinMillis(); long time = someUUID.timestamp(); /* convery from UUID to a long timestamp */ UUID otherUUID = TimeUUIDUtils.getTimeUUID(time); /* do the reverse and get back the UUID from timestamp */ System.out.println(someUUID); /* someUUID and otherUUID should be same, but are different */ System.out.println(otherUUID); === -- Roshan Blog: http://roshandawrani.wordpress.com/ Twitter: @roshandawrani http://twitter.com/roshandawrani Skype: roshandawrani -- Roshan Blog: http://roshandawrani.wordpress.com/ Twitter: @roshandawrani http://twitter.com/roshandawrani Skype: roshandawrani -- Patricio.-
Re: Converting a TimeUUID to a long (timestamp) and vice-versa
If I use *com.eaio.uuid.UUID* directly, then I am able to do what I need
(attached a Java program for the same), but unfortunately I need to deal
with *java.util.UUID *in my application and I don't have its equivalent
com.eaio.uuid.UUID at the point where I need the timestamp value.
Any suggestion on how I can achieve the equivalent using Hector library's
TimeUUIDUtils?
On Wed, Jan 5, 2011 at 7:21 AM, Roshan Dawrani roshandawr...@gmail.comwrote:
Hi Victor / Patricio,
I have been using Hector library's TimeUUIDUtils. I also just looked at
TimeUUIDUtilsTest also but didn't find anything similar being tested there.
Here is what I am trying and it's not working - I am creating a Time UUID,
extracting its timestamp value and with that I create another Time UUID and
I am expecting both time UUIDs to have the same timestamp() value - am I
doing / expecting something wrong here?:
===
import java.util.UUID;
import me.prettyprint.cassandra.utils.TimeUUIDUtils;
public class TryHector {
public static void main(String[] args) throws Exception {
UUID someUUID = TimeUUIDUtils.getUniqueTimeUUIDinMillis();
long timestamp1 = someUUID.timestamp();
UUID otherUUID = TimeUUIDUtils.getTimeUUID(timestamp1);
long timestamp2 = otherUUID.timestamp();
System.out.println(timestamp1);
System.out.println(timestamp2);
}
}
===
I have to create the timestamp() equivalent of my time UUIDs so I can send
it to my UI client, for which it will be simpler to compare long timestamp
than comparing UUIDs. Then for the long timestamp chosen by the client, I
need to re-create the equivalent time UUID and go and filter the data from
Cassandra database.
--
Roshan
Blog: http://roshandawrani.wordpress.com/
Twitter: @roshandawrani http://twitter.com/roshandawrani
Skype: roshandawrani
On Wed, Jan 5, 2011 at 1:32 AM, Victor Kabdebon victor.kabde...@gmail.com
wrote:
Hi Roshan,
Sorry I misunderstood your problem.It is weird that it doesn't work, it
works for me...
As Patricio pointed out use hector standard way of creating TimeUUID and
tell us if it still doesn't work.
Maybe you can paste here some of the code you use to query your columns
too.
Victor K.
http://www.voxnucleus.fr
2011/1/4 Patricio Echagüe patric...@gmail.com
In Hector framework, take a look at TimeUUIDUtils.java
You can create a UUID using TimeUUIDUtils.getTimeUUID(long time); or
TimeUUIDUtils.getTimeUUID(ClockResolution clock)
and later on, TimeUUIDUtils.getTimeFromUUID(..) or just UUID.timestamp();
There are some example in TimeUUIDUtilsTest.java
Let me know if it helps.
On Tue, Jan 4, 2011 at 10:27 AM, Roshan Dawrani roshandawr...@gmail.com
wrote:
Hello Victor,
It is actually not that I need the 2 UUIDs to be exactly same - they
need to be same timestamp wise.
So, what I need is to extract the timestamp portion from a time UUID
(say, U1) and then later in the cycle, use the same long timestamp value to
re-create a UUID (say, U2) that is equivalent of the previous one in terms
of its timestamp portion - i.e., I should be able to give this U2 and
filter
the data from a column family - and it should be same as if I had used the
original UUID U1.
Does it make any more sense than before? Any way I can do that?
rgds,
Roshan
On Tue, Jan 4, 2011 at 11:46 PM, Victor Kabdebon
victor.kabde...@gmail.com wrote:
Hello Roshan,
Well it is normal to do not be able to get the exact same UUID from a
timestamp, it is its purpose.
When you create an UUID you have in fact two information : random 64
bits number - 64 bits timestamp. You put that together and you have your
uuid.
.
So unless you save your random number two UUID for the same milli( or
micro) second are different.
Best regards,
Victor K.
http://www.voxnucleus.fr
2011/1/4 Roshan Dawrani roshandawr...@gmail.com
Hi,
I am having a little difficulty converting a time UUID to its
timestamp equivalent and back. Can someone please help?
Here is what I am trying. Is it not the right way to do it?
===
UUID someUUID = TimeUUIDUtils.getUniqueTimeUUIDinMillis();
long time = someUUID.timestamp(); /* convery from UUID to a
long timestamp */
UUID otherUUID = TimeUUIDUtils.getTimeUUID(time); /* do the
reverse and get back the UUID from timestamp */
System.out.println(someUUID); /* someUUID and otherUUID should
be same, but are different */
System.out.println(otherUUID);
===
--
Roshan
Blog: http://roshandawrani.wordpress.com/
Twitter: @roshandawrani http://twitter.com/roshandawrani
Skype: roshandawrani
--
Roshan
Blog: http://roshandawrani.wordpress.com/
Twitter: @roshandawrani
Re: Converting a TimeUUID to a long (timestamp) and vice-versa
Ok, found the solution - finally ! - by applying opposite of what
createTime() does in TimeUUIDUtils. Ideally I would have preferred for this
solution to come from Hector API, so I didn't have to be tied to the private
createTime() implementation.
import java.util.UUID;
import me.prettyprint.cassandra.utils.TimeUUIDUtils;
public class TryHector {
public static void main(String[] args) throws Exception {
final long NUM_100NS_INTERVALS_SINCE_UUID_EPOCH =
0x01b21dd213814000L;
UUID u1 = TimeUUIDUtils.getUniqueTimeUUIDinMillis();
final long t1 = u1.timestamp();
long tmp = (t1 - NUM_100NS_INTERVALS_SINCE_UUID_EPOCH) / 1;
UUID u2 = TimeUUIDUtils.getTimeUUID(tmp);
long t2 = u2.timestamp();
System.out.println(u2.equals(u1));
System.out.println(t2 == t1);
}
}
On Wed, Jan 5, 2011 at 8:15 AM, Roshan Dawrani roshandawr...@gmail.comwrote:
If I use *com.eaio.uuid.UUID* directly, then I am able to do what I need
(attached a Java program for the same), but unfortunately I need to deal
with *java.util.UUID *in my application and I don't have its equivalent
com.eaio.uuid.UUID at the point where I need the timestamp value.
Any suggestion on how I can achieve the equivalent using Hector library's
TimeUUIDUtils?
On Wed, Jan 5, 2011 at 7:21 AM, Roshan Dawrani roshandawr...@gmail.comwrote:
Hi Victor / Patricio,
I have been using Hector library's TimeUUIDUtils. I also just looked at
TimeUUIDUtilsTest also but didn't find anything similar being tested there.
Here is what I am trying and it's not working - I am creating a Time UUID,
extracting its timestamp value and with that I create another Time UUID and
I am expecting both time UUIDs to have the same timestamp() value - am I
doing / expecting something wrong here?:
===
import java.util.UUID;
import me.prettyprint.cassandra.utils.TimeUUIDUtils;
public class TryHector {
public static void main(String[] args) throws Exception {
UUID someUUID = TimeUUIDUtils.getUniqueTimeUUIDinMillis();
long timestamp1 = someUUID.timestamp();
UUID otherUUID = TimeUUIDUtils.getTimeUUID(timestamp1);
long timestamp2 = otherUUID.timestamp();
System.out.println(timestamp1);
System.out.println(timestamp2);
}
}
===
I have to create the timestamp() equivalent of my time UUIDs so I can send
it to my UI client, for which it will be simpler to compare long timestamp
than comparing UUIDs. Then for the long timestamp chosen by the client, I
need to re-create the equivalent time UUID and go and filter the data from
Cassandra database.
--
Roshan
Blog: http://roshandawrani.wordpress.com/
Twitter: @roshandawrani http://twitter.com/roshandawrani
Skype: roshandawrani
On Wed, Jan 5, 2011 at 1:32 AM, Victor Kabdebon
victor.kabde...@gmail.com wrote:
Hi Roshan,
Sorry I misunderstood your problem.It is weird that it doesn't work, it
works for me...
As Patricio pointed out use hector standard way of creating TimeUUID
and tell us if it still doesn't work.
Maybe you can paste here some of the code you use to query your columns
too.
Victor K.
http://www.voxnucleus.fr
2011/1/4 Patricio Echagüe patric...@gmail.com
In Hector framework, take a look at TimeUUIDUtils.java
You can create a UUID using TimeUUIDUtils.getTimeUUID(long time); or
TimeUUIDUtils.getTimeUUID(ClockResolution clock)
and later on, TimeUUIDUtils.getTimeFromUUID(..) or just
UUID.timestamp();
There are some example in TimeUUIDUtilsTest.java
Let me know if it helps.
On Tue, Jan 4, 2011 at 10:27 AM, Roshan Dawrani
roshandawr...@gmail.com wrote:
Hello Victor,
It is actually not that I need the 2 UUIDs to be exactly same - they
need to be same timestamp wise.
So, what I need is to extract the timestamp portion from a time UUID
(say, U1) and then later in the cycle, use the same long timestamp value
to
re-create a UUID (say, U2) that is equivalent of the previous one in terms
of its timestamp portion - i.e., I should be able to give this U2 and
filter
the data from a column family - and it should be same as if I had used the
original UUID U1.
Does it make any more sense than before? Any way I can do that?
rgds,
Roshan
On Tue, Jan 4, 2011 at 11:46 PM, Victor Kabdebon
victor.kabde...@gmail.com wrote:
Hello Roshan,
Well it is normal to do not be able to get the exact same UUID from a
timestamp, it is its purpose.
When you create an UUID you have in fact two information : random 64
bits number - 64 bits timestamp. You put that together and you have your
uuid.
.
So unless you save your random number two UUID for the same milli( or
micro) second are different.
Best regards,
