iMessage is combinging 2 different people in a conversation

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• HI, I can buy an extra App and that it is used by 2 different people on 1 account at once? The apps are PS en ID. So the first person will use ID all day and the other uses ID all day?

  HI, I can buy an extra App and that it is used by 2 different people on 1 account at once? The apps are PS en ID. So the first person will use ID all day and the other uses ID all day? and this account a 1

  Hello

  Please see licenses and terms of use | Adobe

  License Adobe CC can be activated on 2 machines using the same Adobe ID, but can be used alone at the same time.

  Kind regards

  Sheena

• Two different people that have Dreamweaver installed CS6 can be changed to the same Web site?

  AS the title says, I'm trying to understand if it is possible for two different people to make the changes on the site even if CS6 is installed on BOTH computers.

  Sure.

  There is even a feature of DW integrated fund which makes so that they don't accidentally edit the same file at the same time.

  I've never used it so I can't vouch for it, but I know it's there.

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• Different people I have the same address.

  I have a table with columns ADDRESS DISPLAY_NAME, Startdate, Enddate, and ACCOUNT. I need sql to find different people living at the same address. Someone help me please. Thank you.

  Hello

  Welcome to the forum!

  Use the COUNT function to see how many times an ioccurs from the given address and displays only those of a COUNT > 1.
  You can use the COUNT aggregation function or the analytical COUNT function, like this:

  WITH     got_address_cnt     AS
  (
       SELECT     x.*     -- Or list whatever columns you need
       ,     COUNT (*) OVER (PARTITION BY addrress)     AS address_cnt
       FROM     table_x   x
  )
  SELECT       *
  FROM       got_address_cnt
  WHERE       address_cnt     > 1
  ORDER BY  address
  ;
  

  I hope that answers your question.
  If not, post a small example data (CREATE TABLE and only relevant columns, INSERT statements) and also publish outcomes from these data.
  It's a good idea to do so whenever you have a problem. It allows you to specify exactly what the problem is and allows people to test their ideas.

  Always tell what version of Oracle you are using. The above query will work in Oracle 9 (and), but can be changed to run in Oracle 8.1.

• Please help me to assign a new Id for different people

  version 11g
  
  Please help me in new id of person assiging for dirrerent so people it's middle name is different
  I've listed on three scenarios in the example below. Please help me
  
  

  WITH NAMES AS( 
   /* Scenario1 -- Three dirrerent people so assign Three diffrrent ID,
                    keeping 1 id as is and assign two new ids from sequence*/
   SELECT 47540310 ID , 'WO' LAST_NAME , 'ROBERT' FIRST_NAME , 'C' MIDDLE_NAME FROM DUAL UNION ALL  
   SELECT 47540310 ID , 'WO' LAST_NAME , 'ROBERT' FIRST_NAME , 'W' MIDDLE_NAME FROM DUAL  UNION ALL  
   SELECT 47540310 ID , 'WO' LAST_NAME , 'ROBERT' FIRST_NAME , 'X' MIDDLE_NAME FROM DUAL  UNION ALL  
  
    /* Scenario2 NULL can be equal to any value if there is only one value to equate with 
      In the below case we have two values that can equate with null so assign three diffrent ids */
   
   SELECT 47540300 ID , 'AMATUZIO' LAST_NAME , 'ALBERT' FIRST_NAME , 'J' MIDDLE_NAME FROM DUAL UNION ALL  
    SELECT 47540300 ID , 'AMATUZIO' LAST_NAME , 'ALBERT' FIRST_NAME , NULL MIDDLE_NAME FROM DUAL  UNION ALL 
   SELECT 47540300 ID , 'AMATUZIO' LAST_NAME , 'ALBERT' FIRST_NAME , 'L' MIDDLE_NAME FROM DUAL  UNION ALL
  
    /* Scenario3 NULL can be equal to any value if there is only one value to equate with 
      In the below case we have ONE VALUE  that can equate with null so DONT ASSIGN ANY IDS*/
   SELECT 17540300 ID , 'AMARONE' LAST_NAME , 'JOSEPH' FIRST_NAME , 'J' MIDDLE_NAME FROM DUAL UNION ALL   
   SELECT 17540300 ID , 'AMARONE' LAST_NAME , 'JOSEPH' FIRST_NAME , NULL MIDDLE_NAME FROM DUAL
  
   )
   Select * from names
   
   o/P Required
   ID    LAST_NAME    FIRST_NAME    MIDDLE_NAME
  
  47540310    WO      ROBERT          C
  99999990    WO      ROBERT          W                   -- New Sequence Number
  99999991     WO      ROBERT         X                   -- New Sequence Number
  47540300    AMATUZIO    ALBERT    J
  99999992    AMATUZIO    ALBERT                          -- New Sequence Number
  99999993    AMATUZIO    ALBERT    L                     -- New Sequence Number
  17540300    AMARONE    JOSEPH    J
  17540300    AMARONE    JOSEPH    

  Thank you
  
  Published by: new learning June 7, 2012 14:12

  I do not understand what does this line:

  FIRST_VALUE (fn_get_person_id) - sequence to function

  I feel that the only difference with my query is to select distinct values, do not consider duplicates. Correct me if I'm wrong.

  For me, this should give the same result as your query:

  WITH NAMES AS ( /* Scenario1 -- Three dirrerent people so assign Three diffrrent ID, keeping 1 id as is and assign two
  new ids from sequence*/
                 SELECT   47540310 ID,
                          'WO' LAST_NAME,
                          'ROBERT' FIRST_NAME,
                          'C' MIDDLE_NAME
                   FROM   DUAL
                 UNION ALL
                 SELECT   47540310 ID,
                          'WO' LAST_NAME,
                          'ROBERT' FIRST_NAME,
                          'C' MIDDLE_NAME
                   FROM   DUAL
                 UNION ALL
                 SELECT   47540310 ID,
                          'WO' LAST_NAME,
                          'ROBERT' FIRST_NAME,
                          'W' MIDDLE_NAME
                   FROM   DUAL
                 UNION ALL
                 SELECT   47540310 ID,
                          'WO' LAST_NAME,
                          'ROBERT' FIRST_NAME,
                          'X' MIDDLE_NAME
                   FROM   DUAL
                 UNION ALL
                 /* Scenario2 NULL can be equal to any value if there is only one value to equate with
                   In the below case we have two values that can equate with null so assign three diffrent ids */
                 SELECT   47540300 ID,
                          'AMATUZIO' LAST_NAME,
                          'ALBERT' FIRST_NAME,
                          'J' MIDDLE_NAME
                   FROM   DUAL
                 UNION ALL
                 SELECT   47540300 ID,
                          'AMATUZIO' LAST_NAME,
                          'ALBERT' FIRST_NAME,
                          NULL MIDDLE_NAME
                   FROM   DUAL
                 UNION ALL
                 SELECT   47540300 ID,
                          'AMATUZIO' LAST_NAME,
                          'ALBERT' FIRST_NAME,
                          'L' MIDDLE_NAME
                   FROM   DUAL
                 UNION ALL
                 /* Scenario3 NULL can be equal to any value if there is only one value to equate with
                   In the below case we have ONE VALUE  that can equate with null so DONT ASSIGN ANY IDS*/
                 SELECT   17540300 ID,
                          'AMARONE' LAST_NAME,
                          'JOSEPH' FIRST_NAME,
                          'K' MIDDLE_NAME
                   FROM   DUAL
                 UNION ALL
                 SELECT   17540300 ID,
                          'AMARONE' LAST_NAME,
                          'JOSEPH' FIRST_NAME,
                          'Y' MIDDLE_NAME
                   FROM   DUAL
                 UNION ALL
                 SELECT   17540300 ID,
                          'AMARONE' LAST_NAME,
                          'JOSEPH' FIRST_NAME,
                          NULL MIDDLE_NAME
                   FROM   DUAL
                 UNION ALL
                 SELECT   17540300 ID,
                          'AMARONE' LAST_NAME,
                          'JOSEPH' FIRST_NAME,
                          'M' MIDDLE_NAME
                   FROM   DUAL
                 UNION ALL
                 SELECT   17540300 ID,
                          'AMARONE' LAST_NAME,
                          'JOSEPH' FIRST_NAME,
                          'M' MIDDLE_NAME
                   FROM   DUAL
                 UNION ALL
                 SELECT   99999999 ID,
                          'LO' LAST_NAME,
                          'CHRISTY' FIRST_NAME,
                          NULL MIDDLE_NAME
                   FROM   DUAL
                 UNION ALL
                 SELECT   99999999 ID,
                          'LO' LAST_NAME,
                          'CHRISTY' FIRST_NAME,
                          'M' MIDDLE_NAME
                   FROM   DUAL
                 UNION ALL
                 SELECT   99999999 ID,
                          'LO' LAST_NAME,
                          'CHRISTY' FIRST_NAME,
                          'M' MIDDLE_NAME
                   FROM   DUAL),
   sel_names AS (  SELECT id
                            , last_name
                            , first_name
                            , middle_name
                            , COUNT (middle_name) OVER (PARTITION BY id ORDER BY middle_name) cnt
                            , ROW_NUMBER () OVER (PARTITION BY id  ORDER BY middle_name) rown
                         FROM (SELECT DISTINCT * FROM names) -- use SELECT DISTINCT here to remove duplicates
                     ORDER BY 1, 2, 3, 4
                     )
  SELECT 8888888 new_id, id, last_name, first_name
       , middle_name
    FROM sel_names
   WHERE cnt > 1 AND rown > 1;
  

  As you can see I have just replace the table with a view of inline using SELECT DISTINCT *.

                         FROM (SELECT DISTINCT * FROM names) -- use SELECT DISTINCT here to remove duplicates
  

  For the new id I put a dummy value 888888, you can use a function or sequence according to your needs.

  Kind regards.
  Al

• Different people living at the same address, same account

  I Display_name, address, account and a few other columns in a table. I need people (Display_name) at the same address with the same account number and a different name (Display_name). For example, spouses living at the same address and account numbers.
  
  Please give me an idea with the sql. Thank you.

  Hello

  This sounds like a job for the COUNT function:

  WITH     got_cnt       AS
  (
       SELECT  display_name          -- or whatever columns you want
       ,     COUNT (DISTINCT display_name) OVER ( PARTITION BY  address
                                                       ,                    account
                                          ) AS display_name_cnt
       FROM    a_table
  )
  SELECT  display_name          -- or whatever columns you want
  FROM     got_cnt
  WHERE     display_name_cnt     > 1
  ;
  

  Analytical functions are calculated after the clause WHERE is applied. Use the results of an analytic function in a WHERE clause, you must calculate the function in a subquery, then we can use this value in the WHERE clause (or elsewhere) of the request of the mother.

  I hope that answers your question.
  If not, post a small example data (CREATE TABLE and only relevant columns, INSERT statements) and also publish outcomes from these data.
  Explain, using specific examples, how you get these results from these data.
  Always tell what version of Oracle you are using.

  Published by: Frank Kulash, February 1, 2011 10:54

• Fastest way to SELECT different people, same date of birth, same ZIP (postal code)

  I have a pretty large table of people over 25 M records. There NAME, ADDRESS, DOB, ZIP, OTHER, etc.

  Ago total of about 9 M people separate for one person it might be more then a recording (that is, change of address, name, etc.)

  I want to choose the DISTINCT people born on the same date and in the same ZIP (postal) code. You are looking for the most effective way to do it.

  Any idea is welcome

  Hello

  Here's one way:

  WITH got_cnt AS

  (

  SELECT person_id, person_name, address, date of birth, zip - or other columns of your choice

  , COUNT (DISTINCT person_id) over (PARTITION BY dob, zip) AS cnt

  FROM table_x

  )

  SELECT *.

  OF got_cnt

  WHERE cnt > 1

  ORDER BY dob, zip

  ;

  It will be faster than other methods because it requires only 1 pass through the table.

• I have a PDF file that contains 5 different documents, I want to send each document for different people.

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  Hello
  
  I install my email address to receive grid to backup emails, alerts, etc...
  
  
  I would like to add a guy (my backup). I added it in directors and implemented his email in preferences of Grid Control. But not at all, it never receives emails.
  
  If anyone can help me?
  
  Thanks in advance.
  
  oldschool

  I mean, exactly, you should see in the list and will be able to subscribe to them.

  You have publicly them accessible when you created them?

  Of course, no one sees here how to set up your work of OS, then you will need to provide detailed information, step by step lists the actions that you perform, and what you see. Details on how they were defined, etc., if you are not able to understand these tips form.

  Can subscribe to the new account for default alerts? Is it only the custom OS jobs you can't subscribe to?, etc.

  Here's a complete documentation on how administrators can subscribe to user defined Metrics created by a different administrator:

  http://docs.huihoo.com/Oracle/docs/B16240_01/doc/EM.102/b16242/UDM2.htm

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  As with regular measurements, you can receive notifications by email when User-Defined metric critical or warning alert severities are reached. Assuming that you have already defined your e-mail addresses and timing of the notification, the remaining task is to set up a notification rule for the metric defined by the user.

  To configure notification rules:

  1.

  Click on Preferences.
  2.

  In the vertical navigation bar, click rules, if you are a Super Administrator or my rules if you are a regular Enterprise Manager administrator.
  3.

  Click on create to define a new notification rule. The pages create a Notification rule appear.
  4.

  On the general page, enter the necessary definition rule and choose host Type target for BONE at base of the parameters set by the user or database for basic SQL Instance defined by the user settings.
  5.

  On the settings page, click Add. A list of available parameters. To display all parameters at the same time, choose Show all in the menu drop-down.
  6.

  Select Digital metric defined by the user or User - Defined String metric based on the type of the value returned by your metric defined by the user.
  7.

  In the objects column, choose whether you want to receive a notification for all parameters defined by the user (all objects) or specific Metrics (Select) user-defined.

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  You can search for available measures defined by the user using the search function (flashlight icon). However, search results will not show parameters defined by the user who has at least one point of data collection. For measures which have not yet perceived at least a given point, as can be the case for a newly created user-defined measure, you must specify the in the Select text entry field.
  8.

  Select the severity or the State of corrective action for which you want to receive the notification and then click on continue.
  9.

  If you wish to receive e-mail for the metric defined by the specified user, go to the Notification rule and check the option "send me a message.
  10.

  Click OK to create the new notification rule. If you have done the public notification rule, administrators can subscribe to the same rule.

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  Hello

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  21:05 Monday; March 28, 2016

   iMac 2.5 Ghz i5 2011 (El Capitan)
   G4/1GhzDual MDD (Leopard 10.5.8)
   MacBookPro (Snow Leopard 10.6.8) 2 GB
   Mac OS X (10.6.8).
   a few iPhones and an iPad

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  Hi all

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  Concerning

  Steve

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