Count distinct users registered

Similar Questions

• Number of distinct users

  I have the following table

  Users	Region
  [email protected]	EMEA
  SSA. [email protected]	APAC
  SSA. [email protected]	APAC
  [email protected]	EMEA
  [email protected]	EMEA
  [email protected]	DVL
  [email protected]	DVL
  [email protected]	EMEA
  [email protected]	EMEA
  [email protected]	EMEA

  I need to know the number of distinct users by region.

  If the result will give me EMEA - 5

  APAC - 1

  LAD - 1

  Can someone help me with the query?

  Hello

  You said.  A way of counting distinct users is COUNT (DISTINCT users):

  SELECT region

  COUNT (DISTINCT users)

  FROM table_x

  GROUP BY region;

  I hope that answers your question.
  If not, post a small example of data (CREATE TABLE and INSERT only relevant columns statements) and the results desired from these data.
  Point where the above query is to produce erroneous results, and explain, using specific examples, how you get the right results from data provided in these places.
  If you change the query at all, post your modified version.
  Always say what version of Oracle you are using (for example, 11.2.0.2.0).

  See the FAQ forum: https://forums.oracle.com/message/9362002

• Count Distinct on a window

  Hello world
  
  I read an article about a measure of viscosity (http://www.seomoz.org/blog/tracking-browse-rate-a-cool-stickiness-metric). It's basically a way to measure how well you attract people to your Web site.
  
  Viscosity for the last 7 days: (County of different users today) / (number of distinct users for the last 7 days)
  Viscosity for the last 30 days: (County of different users today) / (number of distinct users for the last 30 days)
  
  I have a table called WC_WEB_VISITS_F that is in the grain of the user's visit to one of my web sites.
  
  Is there a way I can use OBIEE to calculate values above?
  
  I can get the numerator. I create a logical column called count of separate to everyday users and aggregation set to Distinct Count on USER_ID and then I pivot my report to the Date of visit.
  
  I can't understand how Distinct Count for X days. If there were separate count for this month, I just copy the count of distinct daily users and set the content to the month level, however, it is a floating range, I can't do that.
  
  Does anyone have an idea of Cleaver for this?
  
  Thank you!
  
  -Joe

  11g is the function PERIODROLLING, which is the answer to all these requirements. The PERIODROLLING function allows you to perform an aggregation on a specified set of periods of grain of query rather than a grain of fixed time series. The most common use is to create rolling averages, as "13-week Rolling average." With 10 all we will get is a feature limited according to solutions...

• OVERALL TOTAL incorrect (with COUNT DISTINCT)

  Hello
  
  I get incorrect results in a DISTINCT COUNT measure column GRAND TOTAL.
  
  I have 5 separate in Paris and 10 separate clients in New York, I want the grand total for the sum of the two, it is 15.
  But OBIEE calculates separate customers for all cities, so if there are customers in Paris and New York, the result is false.
  
  
  This is the result I get:
  
  City Number_Distinct_Customers
  ----------------------------------------------------------------------------
  Paris 5
  NEW YORK CITY 10
  GRAND TOTAL 12
  
  
  12 is the number of all separate clients.
  
  The correct GRANT TOTAL is expected to be 5 + 10 = 15
  
  
  
  Thank you
  Concerning

  I guess that COUNT(DISTINCT...) is regarded as the default aggregation in the Oracle replies.
  To come in this, change the State of aggregation of fx (formula column in the criteria of answers tab) to 'SUM '...

  Let me know if that solves your problem.
  -bifacts :-)
  http://www.obinotes.com

• How to count distinct exclusion of a value in the business layer?

  Hi all
  
  I have a column that has a lot of values. I need to do is a measure with aggregator separate count. But I shouldn't count 0 in the column. How can I do that. If I try to use any way to condition the aggregator option is disable. Help, please
  
  Thank you

  Look at this example:

  I did MDB table is DIRTY as:

  Count_Distinct_Prod_Id_Exclude_Prod_Id_144

  I'll count distinct PRODUCTS. PROD_ID, but exclude PROD_ID = 144 when counting.

  Make this measure like this:

  1. new column object/logic
  2. go in tab data type and click EDIT the table logic table source
  3. now, in the general tab add join of a table (in my case of PRODUCTS)
  4. go in the column mapping tab-> see the deleted column mapping

  5. in the new column (in my case Count_Distinct_Prod_Id_Exclude_Prod_Id_144) write similar code:
  CASE WHEN "orcl". » ». "" SH ". "PRODUCTS '." "' PROD_ID ' = 144 THEN ELSE NULL"orcl ". » ». "" SH ". "PRODUCTS '." "' PROD_ID ' END

  6. click OK and close the source logical table
  7. now, in the logical column window go to the tab of the aggregation and choose COUNT DISTINCT.

  8. move the Count_Distinct_Prod_Id_Exclude_Prod_Id_144 measure for presentation

  9 test answers (report cointains columns as follows)

  PROD_CATEGORY_ID
  Count_Distinct_Prod_Id_Exclude_Prod_Id_144

  And the result in the NQQuery.log is:

  Select T21473. PROD_CATEGORY_ID C1,
  Count (distinct from cases when T21473.) PROD_ID is 144, then NULL else T21473. End PROD_ID) C2
  Of
  PRODUCTS T21473
  Group of T21473. PROD_CATEGORY_ID
  order of c1

  Concerning
  Goran
  http://108obiee.blogspot.com

• COUNT (DISTINCT < nom_de_colonne >)

  I would like to display the count (DISTINCT < company >) at the end of the report. I use a SQL query in the form of data / the whole model. Thanks in advance.
  
  NEW YORK CITY
  JANUARY 1, 2008
  
  Company ID DIFF
  ABC 1-1
  2 2 XYZ
  ABC 3-1
  
  FEBRUARY 1, 2008
  
  Company ID DIFF
  MNO 1-1
  2 2 XYZ
  MNO 3-1
  
  Total number of companies: 3

  Use this

• Vs Count Distinct Count

  What is the difference between the meter and the meter separate in the advanced INTERFACE

  Operations-> all settings-> deployed-> Count?

  This metric is available in the advanced user interface and the custom user interface. Measures are not read in the Admin UI.

  Distinct count is the actual number of virtual machines deployed, while the County is # trend / forecasting of the deployed virtual machines. Usually, you will notice the separate account doesn't have a very clear dynamic threshold, but the County has a dynamic rise. Speaking of which, I would say probably a couple more VMs to meet forecasts of my lab... ha.

• Delete the user registered on the Web site ID

  original title: list of the userids

  When you go to different Web sites that require user id connections, I find that when you click in the neighborhood newspaper, all signs, sign user names appear.  Is there a way to remove this option?

  Hi NJ123,

  • What browser you use on the computer.

  Try the step below only if you are using Internet explorer as your web browser.

  You can delete all store passwords and other information that you type in the fields of web form using the AutoComplete feature in internet explore.

  When Internet Explorer starts, AutoComplete is disabled. To turn it on or off, follow these steps:

  1. Open Internet Explorer by clicking the Start button and then about Internet Explorer.
  2. Click Tools and then click Internet Options.
  3. Click the content tab.
  4. Under AutoComplete, click settings.
  5. Click delete AutoComplete history.
  6. Click OK, and then click OK again.

  Fill in website forms and passwords automatically

  http://Windows.Microsoft.com/en-us/Windows-Vista/fill-in-website-forms-and-passwords-automatically

  Note: Above article also applies to Windows XP.

• Count of users of database in the case of MOHAMED Licensing

  Hello

  I wanted to know how Oracle goes on the number of users of a database.

  What, out of what follows, says to explain the use?

  1. all entries in table DBA_USER

  2. all the OPEN and ASSETS entered in the DBA_USER table

  or

  3 stats SESSION_HIGHWATER

  Thanks in advance,

  T

  user13405592 wrote:

  In my view, that there is a vijayan Hemant K collection & Ajay of confusion.

  I ask with respect to an audit of Oracle's LMS. How did they check my user base?

  They will ask you: how many people (and devices) use the product?

• Two distinct users of Adobe Muse sharing site

  How two users with different Adobe ID, share, and work on the same Web site created by Muse? A user created the site for more than a year and asks to help him maintain. Another user also has an Adobe ID with Muse bought and downloaded separately. They both require access to the same files/site independently (and not at the same time, of course).

  Hello

  Same file Muse can be published on various sites, but I think what you want that both users should be able to manage the site of Muse, as well at the end of the published site.

  If the site is hosted on external server then the two users must have access to the server so that they can download the Muse site update in the root of the site.

  If the site is published to Business Catalyst, then ask the previous user who has created the site to add two new users as an administrator for the users of the portal site or partner and they can also make modifications on the site of Muse and publish on the same site.

  Thank you

  Sanjit

• Maximum, Count Distinct functions in the model.

  Hello
  
  I have a set of data that I tried to regroup for a report.
  I need in the report so that I have product code list in a column and Max (Product Code) in another column product code, product code Sub are strings and not numbers.
  
  
  Suppose that my data set has
  
  
  product code:-product-sub-code
  ____________________________
  
  ABC | 1A
  ABC | 1 b
  ABC | 1 c
  def | 1 B
  def | 2B
  def | 2 c
  ...
  ...
  ....
  ...
  ...
  
  I need to report
  
  ABC | 1A
  def | 2B
  
  ..
  
  
  
  I gathered it and I tried to show the Max (product code) for the current group. But it gave me a number instead of text. I tried to select 'text' to the form field. But it is automatically change the number.
  
  
  Also, can you let me know how to write the number (code of separate by-product). I must also add that column.
  
  I can't add in SQL because there are tons of models on this data set. I have to do it in the model.
  
  Any help is greately appreciated.

  Try using this: -substitute your domain under product code name

  Thank you
  Bipuser

• Several group

  Hello
  
  Could someone tell me how I might have several group from the different count function?
  Here's what I'm trying to do.
  
  

  select x.prev_categ, x.next_categ,
         count(distinct user_id) as countprev2next,
         count(distinct user_id) as countprev2any,
         count(distinct user_id) as countany2next,
         count(distinct user_id) as countany2any
  (
       select user_id, prev_categ,  next_categ,
                 dense_rank() over (order by prev_categ, next_categ) as rankprev2next,
                 dense_rank() over (order by prev_categ) as rankprev2any,
                 dense_rank() over (order by next_categ) as rankany2next,
                 dense_rank() over() as rankany2any
          from next_categ_data
          where x.prev_categ IS NOT NULL and x.next_categ IS NOT NULL
  )x
  group by x.prev_categ, x.next_categ
  ;

  In the group by clause, I would like to have group by in the following terms:
  (1) prev_categ and next_categ as shown in the query
  (2) only prev_categ
  (3) only next_categ
  (4) user_id
  
  By this motion, I am trying to accomplish the following:
  For example, I have a transaction in which category A is passed to category B.
  I want to count distinct users who moved from category:
  (1) A to B (A2B)
  (2) A to any category (A2X)
  (3) any to B (X2B)
  (4) all for the whole (X2X)
  
  This must be done for all possible transactions.
  

  Sample Data
  create table final as
  
  (
  select 1 user_id,2 product_id,A categ_id, to_Date('1/1/2009','MM/DD/YYYY') dt from dual union all
  select 1 user_id,3 product_id,B categ_id, to_Date('1/1/2009','MM/DD/YYYY') dt from dual union all
  select 1 user_id,4 product_id,C categ_id, to_Date('1/3/2009','MM/DD/YYYY') dt from dual union all
  select 1 user_id,5 product_id,D categ_id, to_Date('1/3/2009','MM/DD/YYYY') dt from dual union all
  select 1 user_id,6 product_id,E categ_id, to_Date('1/3/2009','MM/DD/YYYY') dt from dual union all
  select 1 user_id,7 product_id,F categ_id, to_Date('1/10/2009','MM/DD/YYYY') dt from dual union all
  select 1 user_id,8 product_id,G categ_id, to_Date('1/11/2009','MM/DD/YYYY') dt from dual union all
  
  select 2 user_id,2 product_id,A categ_id, to_Date('1/1/2009','MM/DD/YYYY') dt from dual union all
  select 2 user_id,3 product_id,B categ_id, to_Date('1/2/2009','MM/DD/YYYY') dt from dual union all
  select 2 user_id,4 product_id,C categ_id, to_Date('1/4/2009','MM/DD/YYYY') dt from dual union all
  select 2 user_id,5 product_id,F categ_id, to_Date('1/5/2009','MM/DD/YYYY') dt from dual union all
  select 2 user_id,6 product_id,H categ_id, to_Date('1/6/2009','MM/DD/YYYY') dt from dual union all
  select 2 user_id,7 product_id,F categ_id, to_Date('1/12/2009','MM/DD/YYYY') dt from dual union all
  select 2 user_id,8 product_id,G categ_id, to_Date('1/15/2009','MM/DD/YYYY') dt from dual union all
  
  select 3 user_id,2 product_id,A categ_id, to_Date('1/11/2009','MM/DD/YYYY') dt from dual union all
  select 3 user_id,3 product_id,C categ_id, to_Date('1/12/2009','MM/DD/YYYY') dt from dual union all
  select 3 user_id,4 product_id,B categ_id, to_Date('1/13/2009','MM/DD/YYYY') dt from dual union all
  
  ) ;

  Sample output
  Prev_categ | Next_categ | countprev2next | countprev2any | countany2next | countany2any
  ---------------------------------------------------------------------------------------
    A            B              2                 3              3               3
    A            C              1                 -              3               3
    B            C              2                 2              -               3
    C            B              1                 3              -               3
    C            D              1                 -              1               3
    C            F              1                 -              2               3
    D            E              1                 1              1               3
    E            F              1                 1              -               3
    F            G              2                 2              2               3
    F            H              1                 -              1               3
    H            F              1                 1              -               3

  Could you also tell me how I could make the County be repeated? For example, I want to count 3 to print for the two A to B and a-C
  under column of prev2any.
  
  I appreciate all help.
  
  Thanks again,

  Hello

  You can do it with the analytical COUNT function:

  SELECT DISTINCT
  ,       prev_categ
  ,       next_categ
  ,       COUNT (DISTINCT user_id) OVER (PARTITION BY prev_id
                            ,           next_id
                           )               AS countprev2next
  ,       COUNT (DISTINCT user_id) OVER (PARTITION BY prev_id)     AS countprev2any
  ,       COUNT (DISTINCT user_id) OVER (PARTITION BY next_id)     AS countany2next
  ,       COUNT (DISTINCT user_id) OVER ()               AS countany2any
  FROM       next_categ_data
  WHERE       next_categ     IS NOT NULL
  ORDER BY  prev_categ
  ,       next_categ
  ;
  

  Sorry, I'm not a database now, so I can't test it for 12 hours.

  Looking at the code you posted, it seems as if you were on the right track with the partitions, only you were trying the wrong analytical function.

  You really have a table like next_categ_data? Most people would use a view, if this isn't a subquery for this, unless the query speed was very important.

• meter registered user

  Is it possible to add a counter users registered on my site of muse?

  What catalyst for business package is the most suitable for this option?

  I need of the saving options and password only for customers.

  Ciao

  Geoff

  If you are looking to use the CRM features as well as the ability to create secure areas to ensure certain pages through a system of registration/login, webMarketing BC plan or above would be the way to go, see this thread - http://forums.adobe.com/message/5552081. However, please note that there is no native integration of such features within the Muse CMS and put all this is more of a manual process much easier to follow.

  Unfortunately, it is not a module in British Colombia which would be a counter users registered to the output interface.

  Thank you

  Vinayak

• Distinct count every time a month goes to the rear

  Hello
  
  I'm trying to find a good way to turn a repetitive motion to something a little more streamlined. Basically, I need generate the output of distinct users all the time per month, so if I ran the query as I would like to:
  
  Select (separate cd. INVESTIGATION period), "Jan," such as 'Month', cd.state, cd.zip
  from the cd from mytable
  where cd.date < to_date('20120101090001','yyyymmddhh24miss')
  
  can I change the date back a month and the name of 'Months' and the same union query.
  
  Any ideas?

  918455 wrote:
  The only problem I see is that when I want to group by zip code, etc.

  Then add the code postal etc to the PARTITION BY clause:

  with mytable as (
                   select 55 user_id,to_date('3/1/2012','mm/dd/yyyy') DateOfActivity from dual union all
                   select 45,to_date('2/1/2012','mm/dd/yyyy') from dual union all
                   select 45,to_date('2/3/2012','mm/dd/yyyy') from dual union all
                   select 45,to_date('1/1/2012','mm/dd/yyyy') from dual union all
                   select 25,to_date('2/1/2012','mm/dd/yyyy') from dual union all
                   select 25,to_date('2/3/2012','mm/dd/yyyy') from dual union all
                   select 35,to_date('1/1/2012','mm/dd/yyyy') from dual union all
                   select 35,to_date('12/1/2011','mm/dd/yyyy') from dual
                  ),
             t as (
                   select  first_value(DateOfActivity) over(partition by user_id,zipcode order by DateOfActivity) FirstOccurrence,
                           zipcode,
                           DateOfActivity
                     from  mytable
                  )
  select  distinct trunc(DateOfActivity,'MM') MonthOfActivity,
                   zipcode,
                   count(case DateOfActivity when FirstOccurrence then 1 end) over(partition by zipcode order by DateOfActivity) NumUsers
    from  t
    order by MonthOfActivity
  /
  

  SY.

• How oracle to count the number of users of Hyperion

  Hello

  Can I find out how Oracle count of users for Hyperion Planning? I disabled some dummy accounts in the Production environment, but they exist the planning app security group. Will be what they considered authorized users?

  Thank you

  As I see it is provisioned to users who could potentially access the system, so I assumed that disabled accounts could not access to the system so don't are not counted, just my opinion, I do not speak for Oracle

  As suggested, it always better to speak to the representative of account for any clarification.