Helps to find employees in each Department with pay the lowest

Similar Questions

• update of the commission of 6 percent of the employees in each Department, who got less pay

  How to upgrade add 6% commission to the salary of the employee in each Department, which received less pay

  Vinodh

  ora_1978 wrote:

  create table emp (number of eno, ename varchar2 (20), number of salary, number of dno);

  Insert into emp values(1,'john',1000,10);

  Insert into emp values(2,'smith',2000,10);

  Insert into emp values(3,'bill',5000,20);

  Insert into emp values(4,'mary',6000,20);

  Insert into emp values(5,'charles',1500,20);

  Insert into emp values(4,'mary',2000,30);

  Insert into emp values(4,'patrick',4500,30);

  create table dept (number of dno, dname varchar2 (10));

  Insert into dept values(10, 'HR');

  Insert into dept values(20, 'IT');

  Insert into dept values(30, 'ADMIN');

  How to add 6% commission to the salary of the employee table in each Department, which received less pay

  for example in the emp table for the dno 20 salary charles should get updated at 1500 * 1.6

  Vinodh

  Assuming that the two last emp records have 6 and 7 of Brian eno (and not 4 as you have your copy and paste error)...

  You can identify the records with the minimum in each Department as...

  SQL > ed
  A written file afiedt.buf

  1 Select keep (dense_rank of first order of salary) of the min (eno) as min_eno
  2 of t_emp
  3 * Group of dno
  SQL > /.
  MIN_ENO
  ----------
  1
  5
  6

  3 selected lines.

• I need help to find a way to correct or update the drivers for two DVDs that I have on my PC. Device Manager indicates that they are missing or have been corrupted, and they are a product of microsoft.

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  Your CD/DVD drive is missing or is not recognized by Windows or other programs:

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• Query help - to get the number of employees joined on each month regardless of the year

  Hi all

  I wrote the code below for number of employees joined each month regardless of the year of the employee table. But I couldn't get the result. Kindly help me where I'm wrong in my code,

  Select to_char (hiredate, 'my') as join, count (empno) under the number

  WCP

  To_char Group (hiredate, 'my')

  After having count (empno) > 1;

  Your application displays the list of months (regardless of the year) where more than one employee was engaged with number of emplyees hired this month here. If you want months even if nobody was hired months thast, you emp table outer join to the list of every month:

  with t as)

  Select the level m

  of the double

  connect by level<=>

  )

  Select to_char (to_date (TM, 'mm'), 'my') Lun,

  Count (e.empno) cnt

  t

  left join

  E EMP

  on Tahina = to_char (e.hiredate, 'mm')

  Group of Tahina

  order of Tahina

  /

  MY CNT
  --- ----------
  1 jan
  February 2
  Mar 0
  Apr 2
  May 2
  1 Jun
  July 0
  August 0
  2 sep
  Oct 0
  1 nov

  MY CNT
  --- ----------
  Dec 3

  12 selected lines.

  SQL >

  SY.

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  Hello Carlapalsrok,

  Thanks for your post.  You experience this problem on all user accounts?  If you have not tested, try the following:

  Click Start > right click on my computer > select Manage

  Expand local users and groups

  Click on the users folder > in the right pane, right-click and select new users

  Fill in the appropriate data to create a new user account.

  To disconnect from the computer and log on as the new user

  You still see the white background?

  We can't wait to hear back on your part.

  See you soon

• How to choose the Department with employees in the same line?

  Hello world

  I need to make this request,

  Suppose we have 2 departments with n used on each Department

  I need to select the departments with only 2 employees in the same line, example here

  Department	Name of Department	Employee1	Employee2
  10	HUMAN RESOURCES	Marc	Steve
  20	HE	Paul
  30	New Department
  40	Finance	Jean	Michael

  This query below did not work because it only displays the first employee, but I need to display 2 employees (same as the above table)

  select
  deptno,
  (select ename from emp, dept where rownum =1 and emp.deptno = dept.deptno) employee1,
  (select ename from emp, dept where rownum = 2 and emp.deptno = dept.deptno) employee2
  
  from dept;
  
  

  ...

  I did this, but with the function returning the Emp2Dep1 Emp1Dep1

  function get_ename(p_dept number,p_col number)
  return varchar2
  is
      v_ename varchar2(4000);
      v_first boolean := true;
  BEGIN
          select ename
                into v_ename
                from
                    (
                    select ename, rownum as rn
                    from (
                          select ename
                          from emp ,dept
                          where    emp.deptno        = p_dept
                          order by 1 asc
                          )
             
                    )
                where rn = p_col;       
            
  
       return v_ename;
  END;
  
  

  and simply put

  select
  get_ename(deptno ,1 ) employee1,
  get_ename(deptno ,2 ) employee2
  from dept;
  
  

  Is there a better way to do it?

  NB: database 11g

  Thank you

  Select the Department, employee1, employee2

  (select Department, name, cnt

  (name, department, row_number (select) at the NTC (partition by order of the Department by null)

  employees

  )

  where NTC<=>

  )

  Pivot (max (name) of the cnt (1 as employee1, 2 as employee2))

  order by Department

  DEPARTMENT	EMPLOYEE1	EMPLOYEE2
  10	Marc	Steve
  20	Paul	-
  30	-	-
  40	Jean	Michael
  50	Alpha	Beta

  Concerning

  Etbin

• Each time after clicking the icon of user account on windows xp to change a password is crashes with this error message: Microsoft (R) HTML Application host has encountered a problem and needs to close.

  Each time after clicking the icon of user account on windows xp to change a password is crashes with this error message: Microsoft (R) HTML Application host has encountered a problem and needs to close.

  There is a Microsoft Knowledge base article which sets up a few steps to perform in the computer in order to solve the problem you are experiencing, read the article that you find on the link I am mentioning below and see if it helps the measures mentioned:

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• Help to find the best way to interior design (Nested Loops of sliders?)

  Oracle 11G

  First here the table I work with, for example data

  ID	FID	ANNOTATION	DESIGNATION	LOW	HIGH	SEQ
  1	1008755360	FC1053, 81-82	FC1053	81	82	1
  2	1008755360	XD, 3-24	XD	3	24	2
  3	1008756293	FC1053, 81-82	FC1053	81	82	1
  4	1008756293	XD, 3-24	XD	3	24	2

  Each FID is a different line, these two lines are connected physically, that's why I've shown them here on the table. Each line has a 'meter', in this case, each line has 2 lines of account that are the same.

  

  From there, we can say that FC1053, 81 and 82 are connected between the 2 cables. XD DESIGNATION means that cables are connect not really so I can't ignore those.

  We have a different way to show that they connect other than only those data and I need to connect each other too much. For this I need the Info

  FID1	LOW	HIGH	FID2
  1008755360	81	82	1008756293

  So my first design was the hepatitis hase sliders 2 with the data in the table, as shown above. I would then have a nested loop. The first slider to select the first record, then go down to the second loop where I compare all other records the first record and then go back in the outer loop to move to the next record, then return once again to the inner loop to compare the values again.

  LOOP1

  LOOP 2

  If LOOP1. IDF! LOOP2 =. IDF THEN

  IF LOOP1. ANNOTATION = LOOP2. ANNOTATION THEN

  -CONNECT THE DATA

  ON THE OTHER

  IF LOOP1. DESCRIPTION = LOOP2. DESIGNATION THEN

  IF LOOP2. LOW BETWEEN LOOP1. LOW AND LOOP1. THEN HIGH

  IF LOOP2. HIGH BETWEEN LOOP1. LOW AND LOOP1. HGH THEN

  -CONNECT THE DATA

  END IF;

  END IF;

  END IF;

  END IF;

  END IF;

  END LOOP;

  -Open and close done to reset the cursor to the front/top

  CLOSE CURSOR LOOP2;

  OPEN CURSOR LOOP2

  END LOOP;

  The problem with this approach is once the outside loop moves the next value of the IDF, that logic will connect them once again, but this time with the FIDs reverse which is essentially the same thing as it does not matter the direction in which they are connected. I almost feel that once I found the link I need to remove that line from the cursor (which isn't possible).

  FID1	LOW	HIGH	FID2
  1008755360	81	82	1008756293
  1008756293	81	82	1008755360

  I was able to

  1. make a list of all those I have connected and check against that

  2 use some kind of collection

  3. use a temporary table to hold the data

  4 something that I have not yet thought

  I really appreciate in advance for any help I get.

  If your query mapping is correct, this is the result of your query with the addition of the values of sum for the strand/down running.  Manually change the incorrect in the output bit numbers so that we can move on to find another solution.

  with gc_count like)

  Select

  Sum(current_high-current_low+1) on strand_high (g3e_fid seq order partition),

  Sum(current_high-current_low+1) more (partition of g3e_fid order of seq).

  strand_low (current_high-current_low),

  t.* from b$ gc_count t

  )

  Select

  a.g3e_fid, a.current_designation, a.current_low, a.current_high, a.strand_high, a.strand_low,

  b.g3e_fid, b.current_designation, b.current_low, b.current_high, b.strand_high, b.strand_low

  gc_count a, gc_count b

  where a.g3e_fid in (1008757155,1008757159,1009999655)

  and in b.g3e_fid (1008757155,1008757159,1009999655)

  and a.g3e_fid > b.g3e_fid

  and ((a.count_annotation = b.count_annotation)

  or (a.current_designation = b.current_designation

  and a.current_low b.current_low and b.current_high and a.current_high between b.current_low and b.current_high)

  order of b.seq

  /

  G3E_FID CURRENT_DESIGNATION CURRENT_LOW CURRENT_HIGH STRAND_HIGH STRAND_LOW G3E_FID CURRENT_DESIGNATION CURRENT_LOW CURRENT_HIGH STRAND_HIGH STRAND_LOW

  ---------- --------------------- ----------- ------------ ----------- ---------- ---------- --------------------- ----------- ------------ ----------- ----------

  GAGA GAGA 1008757159 1008757155 1 1 2 2 1 2 2 1

  1008757159 3 8 8 3 1008757155 3 8 8 3 F1DM F1DM

  1008757159 9 10 10 9 1008757155 9 10 10 9 F2 F2

  1008757159 11 14 14 11 1008757155 11 14 14 11 F1DM F1DM

  1009999655 17 18 6 5 1008757155 15 18 18 15 F2 F2

  1008757159 15 16 16 15 1008757155 15 18 18 15 F2 F2

  1009999655 21 24 4 1 1008757155 21 24 24 21 F2 F2

  7 selected lines.

  Elapsed time: 00:00:00.06

• Display the name of the Department and employees in this Department

  Hi all
  
  I want to display the name of the Department and employees in this Department in a single line. Can you please help me to do so.
  
  Oracle version : 9.2.0.8
  
  create table emp
  (varchar2 (100) number of ename, empno, dept_id number);
  
  create table dept
  (dept_id number, dept_name varchar2 (100));
  
  Start
  insert into emp values (100, 'Raghu', 10);
  insert into emp values (101, 'Nathalie', 10);
  insert into emp values (102, 'Sai', 10);
  insert into emp values (103, "Riquet", 20);
  insert into emp values (104, 'Radha', 20);
  insert into emp values (105, "Bika", 30);
  insert into emp values (106, 'Satish', 30);
  insert into emp values (107, 'Emma', 30);
  end;
  
  Start
  insert into dept values (10, 'Sales');
  insert into dept values (20, 'Marketing');
  insert into dept values (30, 'accounts');
  end;
  
  Expected results
  *********************
  dept_name employees
  -------------- ----------------------
  Sale of Raghu, Tedla, Sai
  Marketing Ramesh, Radha
  Accounts Bilodeau, Satish, James
  
  Thank you
  Rambeau

  select dept_name,
  max(sys_connect_by_path(ename,',')) employess
  from (select dept_name,ename,
        row_number() over(partition by emp.dept_id
        order by empno) r
        from emp,dept
        where emp.dept_id = dept.dept_id)
  start with r=1
  connect by prior r+1=r
         and prior dept_name=dept_name
  group by dept_name;
  

• How the page break to each Department and calculate totals for Department

  I have a stored procedure that queries our sql database and returns the name of the employee, their Department, and a few other calculations. I use CFDocument to create a table and view the results. Here's the code that does this (see below) what I have to do is find a way to create a page break after that I have show a total for each column in each Department. My questions are so
  
  1. How can I insert a page break after the end of each service
  2. just above the page break, how to calculate the totals for the Department and display them?
  
  < cfdocument format = "pdf" >
  < table width = "650" border = "1" >
  < cfoutput query = "getResults" >
  < b >
  < td > #EmpName # < table >
  < td > #Dept # < table >
  < td > #JobType # < table >
  < td > #NonBill # < table >
  < td > #NonBillMeter # < table >
  < td > #NumberFormat ((NonBillPercent*100),.00) # < table >
  < td > #Bill # < table >
  < td > #BillMeter # < table >
  < td > #NumberFormat ((BillPercent*100),.00) # < table >
  < /tr >
  < / cfoutput >
  < /table >
  < / cfdocument >

  If I remember correctly, the tag you want is cfdocumentitem. Also, you want to check the attribute of cfoutput group.

  The details are in the cfml reference manual. If you do not have one, has the internet.

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  Please help me find my password * address email is removed from the privacy *, you can send the password information to my alternative * address email is removed from the privacy *. Thank you!

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  My questions are: 1) the PUM associated MSSE & the boxes unchecked in MS security Center; (2) should I click on "ignore" in the Malwarebytes scan? 3) was right to check all the boxes in the center of security - "altert me if my computer may be at rist b/c of my xxx software settings?  Thanks in advance for your help.

  PC fan2

  Hello

  (1) is associated with MSSE PUM & the boxes unchecked in MS security Center;

  The following thread might answer this question: http://forums.malwarebytes.org/index.php?showtopic=69859

  (2) should I click on "ignore" in the Malwarebytes scan?

  Yes

  3) was right to check all the boxes in the center of security - "altert me if my computer may be at rist b/c of my xxx software settings?

  Only, run antivirus software at the same time. Firewall must be performed, evil-ware bytes can be run at your pleasure.