Messure of dept

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  Original title: receving calls

  receving calls saying they are windows it dept number is 1-218-859-9558 it is a scam

  Wednesday, October 14, 2015, 20:15:49 + 0000, Mary Janeobrien says:

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• SQL script on emp and dept

  Tables to be used:

  1 scott.emp

  2 scott.dept

  For all s deptno in dept table if there is at least 1 used in the emp table output should be 1 otherwise the output must be 0.

  For example, deptno in dept table 40 have no employees present in the emp table if the output should be 0.

  Desired output:

  REQ

  ----------

  0

  Now, if I add a line in the table emp which deptno is 40 (now for all deptnos in the dept table, there is record emp at least 1), production should be 1

  REQ

  ----------

  1

  I formed query as below but do not know how to do furthur

  SELECT count (empno), deptno from emp by deptno group

  Hello

  2742751 wrote:

  Select dept Dept. d

  where 0! =

  (select count (*) in e emp

  where e.deptno = d.deptno

  );

  This will not work; There is no column dept in the dept table.

  Select deptno from dept d

  where there are

  (select count (*) in e emp

  where e.deptno = d.deptno

  );

  Do not know how to perform tried furthur two queries

  Good start.  It's a good way to use a correlated EXISTS subquery.  (EXISTS subqueries are almost always correlated.  Why?  Discuss among you).

  The query above tells you what are departments have used; He said nothing on which federal departments have not used, and this is what primarily interests us.  There are probably still some departments that have employees (even if you test this in any case); the real question here is: Y at - it all departments that have not used?   So change IS not there and see what you get.

  SELECT DeptNo

  D DEPT

  WHERE DOES NOT EXIST)

  SELECT 1 - keep it simple

  FROM emp e

  WHERE e.deptno = d.deptno

  )

  ORDER BY deptno

  ;

  It's okay put you in the SELECT a clause subquery EXISTS; the subquery does not return actually what you say, the subquery just returns TRUE or FALSE. You need to select for syntax, but you might as well put something that is easy to type at this location.

  Here is the result of the above query:

  DEPTNO

  ----------

  40

  What is the problem with that?  I see two problems:

  It produces the real Department deptno; We want that it displays 0 where there is a Ministry without employee.  It's an easy solution: change the SELECT main clause so that it returns the literal 0 instead of deptno.

  The other problem is trickier.  The above query produces 1 row of output for each Department that satisfies the condition in the WHERE clause.  In the table scott.emp, it happens to be just 1 row, but if there were 2 or 3, or 2300 these departments, then the above query would produce 2 or 3 or 2300 lines of output.  What happens if 0 rows meet the condition in the WHERE clause?  Then the query would produce 0 lines of output.  It's no good: we want exactly 1 row of output, regardless of the number of rows satisfy the WHERE clause.  Do you know of any which is guaranteed to produce 1 row of output, regardless of the number of lines it finds, even if it is 0 rows?  An aggregate function without a GROUP BY doing the exact clause that: she produced exactly 1 row of output, no matter what, if anything, that it finds in the table, we can say:

  SELECT MIN (0) AS req

  D DEPT

  WHERE DOES NOT EXIST)

  SELECT 1

  FROM emp e

  WHERE e.deptno = d.deptno

  )

  ORDER BY deptno

  ;

  Output:

  REQ

  ----------

  0

  0 being the only value which is ever produced, no matter if we use MIN or MAX.  We could even use AVG or SUM if we wanted to be cute.

  However, there is still a problem: If no rows satisfy the WHERE clause, no 0 will be produced, so MIN returns null.  We want the query to display 1 in this case, not NULL, so we can use NVL do display 1 instead of NULL, like this:

  SELECT NVL (MIN (0)

  1

  ) AS req

  D DEPT

  WHERE DOES NOT EXIST)

  SELECT 1

  FROM scott.emp e

  WHERE e.deptno = d.deptno

  )

  ORDER BY deptno

  ;

  Furthermore, I copy the scott schema emp and dept tables in my schema to test this, so I could add or remove data to test the situations as

  • several services without employees
  • no representation without employees
  • no departments at all (emply dept table)
  • no departments with emplpoyees

  You don't want to edit any table provided by Oracle; You must keep the scott paintings exactly as they are, then, make copies you can change anything you want.

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  I think that the last three comments were all correct, but I could only score a.

• Matching EMP and DEPT

  Hi all

  11.2.0.1

  I have departments in the DEPT table which are 10 and 20. Department 10 has 100 employees in EMP while 20 Department hasn't in the EMP, so there is no corresponding line.

  How can I create a query such as all the dept with no EMP will be displayed:

  EMP DEPT

  ====      ====

  10 100

  20 0

  30 20

  Thank you

  pK

  You must learn more about Outer Join

  Select d.deptno

  count (e.empno) emp_count

  d Dept

  Join emp left e

  on d.deptno = e.deptno

  Group

  by d.deptno

• Why I don't get a response from the Dept. serial number already contacted them twice with ZERO response. I have included my product code and proof of my job as requested but nowt back! ??

  Why I don't get a response from the Dept. serial number already contacted them twice with ZERO response. I have included my product code and proof of my job as requested but nowt is back!

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• How to display two dept details at the same time in the form of master-detail

  Hi Experts

  In Forms 6I, using the Scott schema, table DEPT & EMP has created a simple form of master-details relationships.

  Currently when user Dept Block shows all lines dept and EMP block display data based on the selection of deptno.

  It's only a deptno both.

  If the user in the No. 10, block EMP Dept will display all the data related to 10

  If the user moves to 20 Deptno, block EMP will display all the data related to 20.

  And so on.

  But our requirement, what happens if we want to see the 10 and 20 at once (both several deptno)

  Thank you

  Thanks for your information.

  In fact, our requirement is only to display data, he own be no matter what update of data/insertion.

  So below and working for our requirement

  • Remove relationship
  • Additional box on master block.
  • Write code cursor on the box selection - which will fill/clear data block information based on the checkbox selection on block Master.

  Oracle Apps training: how to display two Department employee details at the same time in the master/detail relationship

• I need avg for each lines dept of sage, matched and unmatched wage table

  Hi gurus,
  I doubt in sql, the employee table and department table two tables are there.
  
  1. I need pay for avg to each dept wise.
  
  2. matched and unmatched in the rows of the table
  
  in a request format
  
  Thanks and greetings
  k.Raghavan

  945795 wrote:
  Hi gurus,
  I doubt in sql, the employee table and department table two tables are there.

  Personally, it looks more like a HOME and WORK, only a DOUBT! What are your efforts on your bottom of DOUBTS?

  1. I need pay for avg to each dept wise.

  select department_id, avg(salary)
    from hr.employees
   group by department_id;
  

  2. matched and unmatched in the rows of the table

  The number of matching rows:

  select employee_id, department_name
    from hr.employees natural join hr.departments;
  

  Unpaired lines:

  select department_id, department_name
    from hr.departments d
   where not exists (
                  select 'x'
                    from hr.employees e
                   where d.department_id = e.department_id
                )
  

• Details of employees with salary max theor Dept

  Hi all
  
  I'm trying to find some thedetails of all employees with the maximum salary in their Department.
  
  I find using a few max (salary)
  SELECT max (salary) FROM employee GROUP BY Dept.
  But how to find the empno, ename, salary and other details?
  
  Here are the data of employees

  EMP NO  NAME       INIT  LASTNAME     DEPT  NO       HIRE DATE  JOB       LEVEL   SEX  BIRTH DATE   ARY     BONUS    COMM
  000010  CHRISTINE  I     HAAS         A00   3978     1965-01-01 PRES      18      F    1933-08-24   52750   1000     4220
  000020  MICHAEL    L     THOMPSON     B01   3476     1973-10-10 MANAGER   18      M    1948-02-02   41250   800      3300
  000030  SALLY      A     KWAN         C01   4738     1975-04-05 MANAGER   20      F    1941-05-11   38250   800      3060
  000050  JOHN       B     GEYER        E01   6789     1949-08-17 MANAGER   16      M    1925-09-15   40175   800      3214
  000060  IRVING     F     STERN        D11   6423     1973-09-14 MANAGER   16      M    1945-07-07   32250   500      2580
  000070  EVA        D     PULASKI      D21   7831     1980-09-30 MANAGER   16      F    1953-05-26   36170   700      2893
  000090  EILEEN     W     HENDERSON    E11   5498     1970-08-15 MANAGER   16      F    1941-05-15   29750   600      2380
  000100  THEODORE   Q     SPENSER      E21   0972     1980-06-19 MANAGER   14      M    1956-12-18   26150   500      2092
  000110  VINCENZO   G     LUCCHESSI    A00   3490     1958-05-16 SALESREP  19      M    1929-11-05   46500   900      3720
  000120  SEAN             O'CONNELL    A00   2167     1963-12-05 CLERK     14      M    1942-10-18   29250   600      2340
  000130  DOLORES    M     QUINTANA     C01   4578     1971-07-28 ANALYST   16      F    1925-09-15   23800   500      1904
  000140  HEATHER    A     NICHOLLS     C01   1793     1976-12-15 ANALYST   18      F    1946-01-19   28420   600      2274
  000150  BRUCE            ADAMSON      D11   4510     1972-02-12 DESIGNER  16      M    1947-05-17   25280   500      2022
  000160  ELIZABETH  R     PIANKA       D11   3782     1977-10-11 DESIGNER  17      F    1955-04-12   22250   400      1780
  000170  MASATOSHI  J     YOSHIMURA    D11   2890     1978-09-15 DESIGNER  16      M    1951-01-05   24680   500      1974
  000180  MARILYN    S     SCOUTTEN     D11   1682     1973-07-07 DESIGNER  17      F    1949-02-21   21340   500      1707
  000190  JAMES      H     WALKER       D11   2986     1974-07-26 DESIGNER  16      M    1952-06-25   20450   400      1636
  000200  DAVID            BROWN        D11   4501     1966-03-03 DESIGNER  16      M    1941-05-29   27740   600      2217
  000210  WILLIAM    T     JONES        D11   0942     1979-04-11 DESIGNER  17      M    1953-02-23   18270   400      1462
  000220  JENNIFER   K     LUTZ         D11   0672     1968-08-29 DESIGNER  18      F    1948-03-19   29840   600      2387
  000230  JAMES      J     JEFFERSON    D21   2094     1966-11-21 CLERK     14      M    1935-05-30   22180   400      1774
  000240  SALVATORE  M     MARINO       D21   3780     1979-12-05 CLERK     17      M    1954-03-31   28760   600      2301
  000250  DANIEL     S     SMITH        D21   0961     1969-10-30 CLERK     15      M    1939-11-12   19180   400      1534
  000260  SYBIL      P     JOHNSON      D21   8953     1975-09-11 CLERK     16      F    1936-10-05   17250   300      1380
  000270  MARIA      L     PEREZ        D21   9001     1980-09-30 CLERK     15      F    1953-05-26   27380   500      2190
  000280  ETHEL      R     SCHNEIDER    E11   8997     1967-03-24 OPERATOR  17      F    1936-03-28   26250   500      2100
  000290  JOHN       R     PARKER       E11   4502     1980-05-30 OPERATOR  12      M    1946-07-09   15340   300      1227
  000300  PHILIP     X     SMITH        E11   2095     1972-06-19 OPERATOR  14      M    1936-10-27   17750   400      1420
  000310  MAUDE      F     SETRIGHT     E11   3332     1964-09-12 OPERATOR  12      F    1931-04-21   15900   300      1272
  000320  RAMLAL     V     MEHTA        E21   9990     1965-07-07 FILEREP   16      M    1932-08-11   19950   400      1596
  000330  WING             LEE          E21   2103     1976-02-23 FILEREP   14      M    1941-07-18   25370   500      2030
  000340  JASON      R     GOUNOT       E21   5698     1947-05-05 FILEREP   16      M    1926-05-17   23840   500      1907
  200010  DIAN       J     HEMMINGER    A00   3978     1965-01-01 SALESREP  18      F    1933-08-14   46500  1000      4220
  200120  GREG             ORLANDO      A00   2167     1972-05-05 CLERK     14      M    1942-10-18   29250   600      2340
  200140  KIM        N     NATZ         C01   1793     1976-12-15 ANALYST   18      F    1946-01-19   28420   600      2274
  200170  KIYOSHI          YAMAMOTO     D11   2890     1978-09-15 DESIGNER  16      M    1951-01-05   24680   500      1974
  200220  REBA       K     JOHN         D11   0672     1968-08-29 DESIGNER  18      F    1948-03-19   29840   600      2387
  200240  ROBERT     M     MONTEVERDE   D21   3780     1979-12-05 CLERK     17      M    1954-03-31   28760   600      2301
  200280  EILEEN     R     SCHWARTZ     E11   8997     1967-03-24 OPERATOR  17      F    1936-03-28   26250   500      2100
  200310  MICHELLE   F     SPRINGER     E11   3332     1964-09-12 OPERATOR  12      F    1931-04-21   15900   300      1272
  200330  HELENA           WONG         E21   2103     1976-02-23 FIELDREP  14      F    1941-07-18   25370   500      2030
  200340  ROY        R     ALONZO       E21   5698     1947-05-05 FIELDREP  16      M    1926-05-17   23840   500      1907

  I would like to find (empno, name, job, dept, sal) details of all employees earning a maximum wage in their Department.
  I'm not sue if I have to use funcion analysis here.
  
  THX
  Rod.

  select *
  from
  (
     select e.*,
            dense_rank() over (partition by dept_no order by salary desc) dr
     from   employees e
  )
  where dr = 1;
  

• Time for specific dept record display by adding 10 minutes

  Hello
  

  Oracle Database 11g Enterprise Edition Release 11.2.0.1.0 - 64bit Production
  With the Partitioning, OLAP, Data Mining and Real Application Testing options
  /****************************************************************************/
  
  create table dept(
  dept_id number(14) primary key,
  c_id number(14) constraint dept_fk references centers(c_id) NOT NULL,
  deptName varchar2(50)
  );
  
  create sequence dept_seq
  start with 1 increment by 1;
  
  create or replace trigger dept_trig
  before insert on dept
  referencing new as new
  for each row
  begin
  select dept_seq.nextval into :new.dept_id from dual;
  end;
  /
  
  create table shiftManager(
  s_id number(9) primary key,
  dept_id number(14) constraint shiftManager_fk references dept(dept_id) NOT NULL,
  title varchar2(110), /**e.g. Morning, After noon, Evening **/
  startTime date,
  endTime date
  );
  
  create sequence shiftManager_seq
  start with 1 increment by 1;
  
  create or replace trigger shiftManager_trig
  before insert on shiftManager
  referencing new as new
  for each row
  begin
  select shiftManager_seq.nextval into :new.s_id from dual;
  end;
  /
  
  select to_char(startTime,'HH24:MI'),to_char(endTime,'HH24:MI')
  from shiftManager where dept_id=1 order by startTime asc;

  I want to choose startTime adding 10 minutes to the end for dept_id specific for example time
  
  Dept1
  =====
  startTime = 09:00 and 14:00 = endTime
  
  POWER required:
  ================
  09:10
  09:20
  09:30
  09:40
  09:50
  10:00
  ..
  ..
  ..
  ..
  ..
  13:40
  13:50
  14:00
  
  Thank you in anticipation
  
  Best regards

  Raakh wrote:

  I am not able to understand especially startTime + (level - 1) / 24 / 6<= endtime="" +="" (600="" -="" 1)="" 600="">

  Oracle date arithmetic unit is day. So 1 / 24 represents 1 hour and 24 1/6 represents 10 minutes. You start by startTime and continue to add 10 min and startTime + (level - 1) / 24 / 6 impeccable (level starts from 1). Now we must stop when meet us or exceed the end time for the first time. Therefore, we can add add 9 min 59 s to endDate and use it as the end. And (600 - 1) 3600 / 24 represents 9 min 59 sec.

  In fact, we can simplify and use the following logic: difference between current 10 min. and end must be< 10="">

  connect by startTime + (level - 1) / 24 / 6 - endTime< 1="" 4="">

  SQL> select * from shiftManager
    2  /
  
     DEPT_ID STARTTIME ENDTIME
  ---------- --------- ---------
           1 01-JUL-11 01-JUL-11
           2 01-JUL-11 01-JUL-11
  
  SQL> select  to_char(column_value,'HH24:MI') tm
    2    from  shiftManager,
    3          table(
    4                cast(
    5                     multiset(
    6                              select  least(startTime + (level - 1) / 24 / 6,endTime)
    7                                from  dual
    8                                connect by startTime + (level - 1) / 24 / 6 - endTime < 1 /24 /6
    9                             )
   10                     as sys.OdciDateList
   11                    )
   12               )
   13    where dept_id=1
   14    order by column_value
   15  /
  
  TM
  -----
  09:30
  09:40
  09:50
  10:00
  10:10
  10:20
  10:30
  10:40
  10:50
  11:00
  11:10
  
  TM
  -----
  11:20
  11:30
  11:40
  11:50
  12:00
  12:10
  12:20
  12:30
  12:40
  12:45
  
  21 rows selected.
  
  SQL> 
  

  SY.

• Wise Dept sum sal differece

  Hello
  
  We have sal & deptno columns in the emp table. How can I view the difference in the amount of wages on dept wise
  
  dept20 - dept10 - amount wage difference
  dept20 - dept30 - amount wage difference
  
  I want to display the output in the form of two columns.
  
  Please help me... !!
  
  Concerning
  Evelyne

  Hello

  Please try this... !!

  select d20_sal - d10_sal as d20_10_diff,
              d20_sal - d30_sal as d20_30_diff
         from (
       select sum(case when deptno=10 then sal end) as d10_sal,
              sum(case when deptno=20 then sal end) as d20_sal,
              sum(case when deptno=30 then sal end) as d30_sal
         from emp
              ) totals_by_dept
  

  Concerning
  KPR