Can do us a query to get the same results of 3 tables

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• Get the same results without line of sight

  Hi gurus,
  
  My version of the database is 10.2.0.4.0.
  
  I have a couple of tables as follows:
  
  

  CREATE TABLE xxgroups(group_number NUMBER, description VARCHAR2(30));
  
  INSERT INTO xxgroups VALUES(1,'Group 1'); 
  INSERT INTO xxgroups VALUES(2,'Group 2');
  INSERT INTO xxgroups VALUES(3,'Group 3');
  INSERT INTO xxgroups VALUES(4,'Group 4');
  INSERT INTO xxgroups VALUES(5,'Group 5');
  
  CREATE TABLE xxgroup_persons(group_number NUMBER, person_number NUMBER, first_name VARCHAR2(30), last_name VARCHAR2(30));
  
  INSERT INTO xxgroup_persons VALUES(1, 1, 'John', 'Smith');
  INSERT INTO xxgroup_persons VALUES(1, 2, 'James', 'Smith');
  INSERT INTO xxgroup_persons VALUES(1, 3, 'Joe', 'William');
  INSERT INTO xxgroup_persons VALUES(2, 4, 'Julien', 'Ford');
  INSERT INTO xxgroup_persons VALUES(2, 5, 'James', 'Kelly');
  INSERT INTO xxgroup_persons VALUES(3, 6, 'Francis', 'Smith');
  INSERT INTO xxgroup_persons VALUES(5, 7, 'Jerry', 'Ford');
  INSERT INTO xxgroup_persons VALUES(5, 8, 'Alfred', 'Willis');
  INSERT INTO xxgroup_persons VALUES(5, 9, 'Jon', 'Preston');
  
  COMMIT;

  I would like to get a query to return for each group, the person with less person_number (if no one assigned to the group, then the query must return again with an empty person group). I know I can do this by using for example:
  

  SELECT group_number, person_number, first_name, last_name 
  FROM (SELECT a.group_number, b.person_number, b.first_name, b.last_name
             , RANK() OVER (PARTITION BY a.group_number ORDER BY person_number) person_order
        FROM xxgroups a, xxgroup_persons b
        WHERE a.group_number = b.group_number (+)
       ) 
  WHERE person_order = 1
  ORDER BY group_number;
  
  GROUP_NUMBER PERSON_NUMBER FIRST_NAME                     LAST_NAME
  ------------ ------------- ------------------------------ ------------------------------
             1             1 John                           Smith
             2             4 Julien                         Ford
             3             6 Francis                        Smith
             4
             5             7 Jerry                          Ford

  but it would be possible to achieve the same results without inline mode?
  
  Thank you.

  You can try this function KEEP:

  scott@orapdt>  SELECT a.group_number,
    2         min(b.person_number) as person_number,
    3         min(b.first_name) keep (dense_rank first order by person_number) as first_name,
    4         min(b.last_name) keep (dense_rank first order by person_number) as last_name
    5  FROM xxgroups a, xxgroup_persons b
    6  WHERE a.group_number = b.group_number (+)
    7  group by a.group_number;
  
  GROUP_NUMBER PERSON_NUMBER FIRST_NAME                     LAST_NAME
  ------------ ------------- ------------------------------ ------------------------------
             1             1 John                           Smith
             2             4 Julien                         Ford
             3             6 Francis                        Smith
             4
             5             7 Jerry                          Ford
  
  5 rows selected.
  

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• SQL query to get the desired result

  I have the array with the following values

  Account	Functional currency	Billing currency	Amount	Header 5
  101	USD	USD	10 h 00
  101	USD	CAD	12 h 00
  102	JPY	USD	9 h 00
  102	JPY	CAD	3.00
  102	JPY	SGD	2.00

  If the account has one different inv_currencypar that the functional currency and then

  I have to summarize the amount in this account by viewing the billing currency

  If the account has more than one other inv_currencypar that the functional currency and then

  I have to summarize the amount in this account showing the functional currency as the currency of invoicing

  I need output like this:

  Account

  Functional currency Billing currency Total amount Header 5101USDCAD10:00 pm102JPYJPY2:00 pm

  Can we get over output in sql without writing pl/sql program?

  Hello

  According to your specific needs, here's a way:

  SELECT account

  functional_currency

  CASE

  WHEN 1 = COUNT (DISTINCT NULLIF (invoice_currency

  functional_currency

  )

  )

  THEN MAX (NULLIF (invoice_currency

  functional_currency

  )

  )

  Of OTHER functional_currency

  END AS invoice_currency

  The SUM of (amount) AS total_amount

  header5

  FROM table_x

  GROUP BY account

  functional_currency

  header5

  ;

  If you would care to post CREATE TABLE and INSERT statements for your sample data, and then I could test this.

  Is it possible to have 2 (or more) of different functional_currencies in the same account?

• How to rewrite this query to get the correct results?

  Friends,
  
  DB: 9iR2
  
  I need to get the name of the employee and the employee number that are not in the table of presence.
  but this query is not the right answer.
  

  select e.eno,e.ename from empl e
  where e.eno not in (select a.eno from attendance a)

  Thank you

  Depending on your data

  SQL> create table attendance(
    2  ENO   VARCHAR2(5),
    3   TDATE VARCHAR2(10),
    4   IN_TIME  VARCHAR2(6),
    5   OUT_TIME VARCHAR2(6),
    6   SHIFT_NO  NUMBER(1));
  
  Table created.
  
  SQL>  create table empl(
    2   ENO VARCHAR2(5),
    3   ENAME  VARCHAR2(75));
  
  Table created.
  
  SQL> insert into empl values('11','AA');
  
  1 row created.
  
  SQL>   insert into empl values('12','AB');
  
  1 row created.
  
  SQL>    insert into empl values('13','AC');
  
  1 row created.
  
  SQL>     insert into empl values('14','AD');
  
  1 row created.
  
  SQL>   insert into empl values('15','AF');
  
  1 row created.
  
  SQL>  insert into attendance values('11','23-3-2009','9.00','6.00',1);
  
  1 row created.
  
  SQL>  insert into attendance values('14','24-3-2009','9.00','6.00',1);
  
  1 row created.
  
  SQL>  insert into attendance values('11','25-3-2009','9.00','6.00',1);
  
  1 row created.
  
  SQL>  insert into attendance values('13','23-3-2009','9.00','6.00',1);
  
  1 row created.
  
  SQL>  insert into attendance values('15','23-3-2009','9.00','6.00',1);
  
  1 row created.
  
  SQL> commit;
  
  Commit complete.
  
  select e.eno,e.ename
  from empl e
  where not exists(select 1 from attendance a where a.eno=e.eno);
  
  ENO     ENAME
  
  12     AB
  

  Twinkle

• How to get the same result on different characterset?

  Hello experts,

  I use

  Oracle Database 11g Enterprise Edition Release 11.2.0.3.0 - 64bit Production
  PL/SQL Release 11.2.0.3.0 - Production
  "CORE 11.2.0.3.0 Production"
  TNS for 64-bit Windows: Version 11.2.0.3.0 - Production
  NLSRTL Version 11.2.0.3.0 - Production
  
  

  Needing a result even on my example.

  SELECT ASCII('ˆ') -- when characterset is -NLS_CHARACTERSET WE8MSWIN1252
  FROM DUAL;
  --result 136
  
  

  SELECT ASCII('ˆ') -- when characterset is -NLS_CHARACTERSET AL32UTF8
  FROM DUAL;
  --result 52102 but need 136
  
  

  I need same result.

  Thanks for your time...

  EDITED RESULT.

  Ask2Learn

  You test something that you call "encrypted" data, which are not at all encryption.

  For example speaking of letter t.

  The CHR (ASCII('t') + 20) operation is pretty useless. You try to find the character that has the code binary and longer than 20 that the letter 't'. This can work with single-byte as WE8MSWIN1252 character sets, but it does not work with variable width as AL32UTF8 character sets. The letter 't' a code 0 x 74 in WE8MSWIN1252 and AL32UTF8. If you add 20 (decimal), you get 0 x 88. Byte 0 x 88 is a valid code in the WE8MSWIN1252, but is not a valid code in AL32UTF8. Bytes in this range may bytes of continuation in AL32UTF8, i.e. a second, third or the fourth byte of a multi-byte sequence.

  The "circumflex accent" character code 0 x 88 (in WE8MSWIN1252) has the code of two bytes 0xCB 0 x 86 in AL32UTF8.

  So, why not take a look at DBMS_CRYPTO integrated

  
  

  Hope this helps

  
  

  Hamid

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  Why?

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  Hi friends
  
  My data in the table are given below
  
  user transaction_date transaction_type
  1 1st August 2011
  August 2, 2011 1 c
  1 3 August 2011
  1 b 3 August 2011
  1 August 4, 2011 a
  1 b 4 August 2011
  2 3 August 2011
  2 b 3 August 2011
  August 4, 2011 2 c
  2 b 4 August 2011
  2 August 5, 2011 a
  2 b August 5, 2011
  2 August 7, 2011 a
  2 b 7 August 2011
  
  I want the count for each user as, how many times he made a transaction type transaction immediately after 'b' type?
  
  As the output of above data should be
  
  number of users
  1 2
  2 3
  
  Thanks in advance :)

  Maybe (NOT TESTED).

  select "user",
         count(criteria) "count"
    from (select "user",
                 transaction_date,
                 transaction_type,
                 case when transaction_type = 'b'
                       and lag(transaction_type,1) over (partition by "user",transaction_date
                                                             order by transaction_date,transaction_type) = 'a'
                      then 1
                 end criteria
            from the_table
         )
   group by "user"
   order by "user"
  

  Concerning

  Etbin

• I need assistance of query SQL, a query to get the same without union clause

  EmpNo, ename sum (sal) sum (deptno)
  200-2000-10 Tom
  201 3000 10 smith
  Alfred 202 20 3000
  40 8000 total

  select NVL(TO_CHAR(employee_id), 'Total'),
         max(last_name),
         sum(salary),
         sum(department_id)
    from employees
   group by rollup(employee_id), ()
  

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  Hello
  
  I have 3 related tables and you must combine customers table1 and table3. My Control Panel is constantly evolving based schedule changes or order_completed qty. I have two problems that I can't solve.
  
  Table 1: Manager has fields {managerId, Mangager_name, customerId & customer}
  Table2: client includes the fields {customerId, orderId, product manager, managerId and}
  Table 3: order has fields {orderId, customerId, customer, date, amount_order, qty, order_completed}
  
  1. I need check what are customers or away from table Manager / order based on customerId and managerId
  
  Here is the example of union that I would like to convert a join because I need to include aggregated order table fields and the customer at table eating based on the field of the client:
  
  Select m.client
  Manager m
  where m.managerId = 22
  Union select m.client in order o where o.customerId = 65
  
  2. create a view or a temporary table (called State) to the report with an additional field that determines if the client has been found in the roll of the order and not in the table Manager.
  
  Example of a view of the State or the temporary table select output
  rownum. customer | status
  1. Mary | The two Tables
  2. Tomas | The two Tables
  3. John | Customer table
  4. April | Customer table
  
  Thanks for the help

  Hello

  SQL> create table manager (m_id number, m_name char(10), customer_id number, client char(10));
  
  Table created.
  
  SQL>
  SQL> insert into manager values (10,'M_1',111,'AAA');
  
  1 row created.
  
  SQL> insert into manager values (20,'M_2',222,'BBB');
  
  1 row created.
  
  SQL>
  SQL> create table customer (C_id number, m_id number, order_id number);
  
  Table created.
  
  SQL>
  SQL> insert into customer values (111,10,5000);
  
  1 row created.
  
  SQL> insert into customer values (222,20,5001);
  
  1 row created.
  
  SQL> insert into customer values (333,NULL,5002);
  
  1 row created.
  
  SQL> create table order_dt (Order_id number, c_id number, Client char(10));
  
  Table created.
  
  SQL> insert into order_dt values (5000,111,'AAA');
  
  1 row created.
  
  SQL> insert into order_dt  values (5002,333,'CCC');
  
  1 row created.
  
  SQL> select  rownum, a.c2, CASE WHEN ((a.ct2+a.ct1) = 2) THEN 'Both Tables'
    2         WHEN (a.ct2 = 1) THEN 'Client table'
    3         WHEN (a.ct1 = 1) THEN 'Manager table'
    4         ELSE 'NULL'
    5    END AS status
    6  from (
    7  select NVL(m.Client,o.Client) c2,count(m.Client) Ct1, count(O.Client) ct2
    8  from   manager m,
    9  customer c,
   10  order_dt O
   11  where m.m_id(+) = c.m_id
   12    and c.c_id = O.c_id(+)
   13  group by m.Client,O.Client
   14  ) A
   15  ;
  
      ROWNUM C2         STATUS
  ---------- ---------- -------------
           1 AAA        Both Tables
           2 CCC        Client table
           3 BBB        Manager table
  

  -Pavan Kumar N

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  I have a Dell computer with a 32-bit version Vista Ultimate factory loaded. Service Pack 1 is loaded.  I tried to load Vista Service Pack 2 several times by using the following procedure and it does not load.
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  I repeated this process several times and always get the same result.
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  ========================

  Free unlimited installation and compatibility support is available for Windows Vista, but only for Service Pack 2 (SP2). This support for Windows Vista Service Pack 2 (SP2) is valid until November 26, 2009. Availability of support chat or messaging differs depending on your location. Go to http://support.microsoft.com/oas/default.aspx?prid=13014&gprid=582034 & select appropriate category (i.e., download problem;) Installation problem; Problems after installing a service pack). ~ Robear Dyer (PA Bear) ~ MS MVP (that is to say, mail, security, Windows & Update Services) since 2002 ~ WARNING: MS MVPs represent or work for Microsoft