some group functions

Similar Questions

• No group functionality

  Since around seven 30 - Oct 1st or so I was unable to use all group functions in any group. 'Old' and those that I just created. I can't send or receive calls or messages in any group. I tried these steps below, and none of them worked. Help?

  (1) sign out and back in

  (2) change of profile information

  (3) update the customer on the later (7.11.0.102)

  (4) to uninstall and reinstall Skype

  (5) clean uninstall and reinstall of Skype

  (6) signature mobile to see if that will fix anything on the end of my PC

  Mobile works very well, however my PC version does not. It worked very well in advance and then just suddenly stopped working for groups. Update the client did nothing. However, I can / message of appeal to people individually. I don't understand. I thought it was the linked server, but my friends have not had any problems...

  Skype is based on some Windows Services work properly, or something?

  I'm under 10 Home Windows (64-bit).

  Welp, after scouring the net I found the problem. It turns out that my situation was the same as this post here.

  http://community.Skype.com/T5/Windows-desktop-client/re-Skype-connection-problems-MSNP-connection-of...

• Help with error - 934 group function is not allowed here

  Hey there will, I'm having problems with a request and just does not know how to do it without error.

  I'm trying to get all the employee emerging infectious diseases that have less than 2 number max of DID (dependants) in the table.

  It's my current query

  SELECT Employee.LName. ' ' || Employee.Fname as Full_Name, Employee.EID

  The left outer JOIN employee depends on Employee.EID = Dependent.EID

  Having Count (DID)--2 > ((select Max (N) as From (SELECT Employee.EID, Count (DID) As N FROM Employee Inner Join Dependent On Employee.EID = Dependent.EID group by Employee.EID, Count (DID))) N)

  Order of Employee.Lname, Employee.Fname

  Which gives me an error on column 4, no matter what I do. If I remove the Count (DID) in the group by clause (which I tried it earlier), it gives me a is not an error of the function of single group...

  The most frustrating thing is that

  Select Max (N) as From (SELECT Employee.EID, Count (DID) As N FROM Employee Inner Join Dependent On Employee.EID = Dependent.EID group by Employee.EID) N

  Works perfectly, but because it's a mission, I have to do in one step (no substeps/views)

  Any help?

  Thank you very much

  Hello

  ac981e5d-D10A-4520-BF42-23a894d04fb7 wrote:

  Ok. I'm taking your code in a view... I get this.

  and there is an orange underscore and a text of the error that says

  Select incoherent list in group by... change the group by clause of e.fname, e.lname, e.eid, count, max

  Which isn't what either the Oracle database would do.  Everything about orange (or any other color) sounds like it is caused by a front-end that could be interacting with Oracle.  In addition, the Oracle error messages always come with a 3-letter-5 code, as ORA-00933.

  under the selection internal (first medium)

  You have deleted the WITH clause.  The parser can recognize the error when it has reached the first left parenthesis.

  Create view AS A10T2

  (

  SELECT e.lname. ' ' || e.fname AS full_name

  e.eid

  (D.) AS this_group_count

  MAX (COUNT (d.)) ON (AS highest_group_count)

  E employee

  LEFT OUTER JOIN dependent d ON e.eid = d.eid

  GROUP BY e.lname, e.fname, e.eid

  )

  SELECT full_name

  eid

  Of aggregate_results

  WHERE this_group_count > = highest_group_count - 2

  ORDER BY full_name

  You need the WITH to define this clause means "AGGREGATE_RESULTS":

  Create view AS A10T2

  WITH aggregate_results AS

  (

  SELECT e.lname. ' ' || e.fname AS full_name

  ...

  Why do you have an ORDER BY clause in a view?   (It is probably not cause of your errors, just make the inefficient view)

  Command line error: column 5: 23

  Error report-

  SQL error: ORA-00933: SQL not correctly completed command

  00933 00000 - "not correctly completed SQL command.

  * Cause:

  * Action:

  This is another indication that some front is getting involved.  Looks like your front-end reports the exact Oracle error message, "0RA-00933" and then builing it's own error code, "00933. 00000 ", on this basis.  ORA-00933 is a reasonable mistake to wait if you omit the line ' WITH the aggregated results AS.  Once again, until I can actually run your code, I can't test it, and I can't run your code until you post CREATE TABLE and INSERT statements for some examples of data, or change the problem to use commonly available tables, such as those in the scott schema.

  and when I try my code

  CREATE VIEW A10T2 AS

  SELECT Employee.LName. ' ' || Employee.Fname as Full_Name, Employee.EID

  The left outer JOIN employee depends on Employee.EID = Dependent.EID

  Seen (Count (DID)) + 2 > (select Max (N) From (SELECT Employee.EID, Count (DID) As "N" FROM Employee Inner Join Dependent On Employee.EID = Dependent.EID group by Employee.EID))

  Order of Employee.Lname, Employee.Fname

  I get

  Command line error: column 2: 8

  Error report-

  SQL error: ORA-00937: not a function of simple-group

  00937 00000 - 'not a single-group function.

  * Cause:

  * Action:

  Then the orange underscore even under my inner ("select employee. EID, Count (DID) as "N" ") says to change the Group of Employee.eid, Count (DID)

  I just don't understand why he tells me to group them by Count (DID)?

  Isn't that what you did in your original post, and I have explained in answer #2?  If you continue to repeat the same mistake, you can expect continue to get the same error.  Given that you have a code, you know causes an error, do you think really that what makes a vision will cause the error to disappear?

  The inner query works fine on its own...

  Right; It's the outer query where you are missing the GROUP BY clause.

• After you download the security update 2015-008 I lost the function of the function keys and some other functions. Have already played with the keyboard f nothing helps. Are also past system on and outside...

  After you download the security update 2015-008 I lost the function of the function keys and some other functions. Have already played with the keyboard f nothing helps. Are also past system on and outside... need help pls

  PS noticed the update downloaded twice

  disregard my previous question... I found the answer: F1, F2 has been ticked on.

• Some of function boxes are elongated and may not reach the OK box

  original title: Windows Vista - Control Panel customization to enlarge the font size.  Some of function boxes are elongated and may not reach the OK box

  Using Windows Vista. I used the customization Panel control to enlarge the font size.  I discovered that some boxes, like Adobe Reader Print and function, in effect, the Personaization (to reduce the font size to original size) require a click OK. But the box frame with large font size, is elongated until I can't get the OK.  How to get to the OK or reduce the size of the police without the OK?

  Simply press enter instead of OK.

• ISE is associated with the AD, but some groups are missing

  People,

  In my ISE I already joined my announcement on a Windows 2008 Server. But when I retrieve groups with *, some groups are missing. I mean there is a band like XXX.COM\COMPANY\IDG\HR in my 2008 Server, but I can't get back in my ISE, the group is Global. Is this a bug of ISE or are there special limits in import of my groups in ISE?

  Kind regards

  ISE can retrieve the list of groups max 100 and as you have said that the number did not reach 100, check if the configuration of Active Directory in the user node Administration ISE
  the interface is correct or you can add that the lack of groups directly use that check the following to add groups manually

  http://www.Cisco.com/en/us/docs/security/ISE/1.0/user_guide/ise10_man_id_stores.html#wp1059262

• Not a "simple-group function.

  SQL > select max(THREAD#), SEQUENCE #, NAME from v$ archived_log;

  Select max(THREAD#), SEQUENCE #, NAME from v$ archived_log

  *

  ERROR on line 1:

  ORA-00937: not a single group group function

  Often, I face "not a single-group function. I tried with the order by clause, also,

  -Please clarify why this error occurs?

  Hello

  You must use Group by clause in your query when you use the Group feature as count, sum, max in query

  Select max(THREAD#), SEQUENCE #, NAME of Group v$ archived_log by SEQUENCE #, NAME;

  Thank you

  Su.GI

• not a single group function

  Hello
  
  
  I have this request with me

  select  
  mst.ENTRY_DATE    ordered_date , mst.FIELD, 
  V.ADDRESS1        Address   , 
  OPN_JOB_DESC      job_number,
  ra.CUSTOMER_NAME    customer_NAME, 
  DECODE ( GROSS_ITEM_REF , NULL , SUBSTR (PRODUCT_ATTR_VAL_DISP , INSTR( PRODUCT_ATTR_VAL_DISP, '.', -1, 1) + 1  ) , GROSS_ITEM_REF )      item_number, 
  dECODE (GROSS_ITEM_NAME , NULL,itm.attribute2,GROSS_ITEM_NAME  )        item_description, 
  subdtlest.ITEM_UOM                unit_of_measurement, 
  SUM ( subdtlest.QTY_ISSUE  )            Quantity_Issue,
  sum(nvl (subdtlest.CALC_CEMENT_SK ,subdtlest.CALC_AMOUNT ) )    quantity ,
  mst.DN_DESCRIPTION   Description
  --decode(MST.DESCRIPTION ,NULL ,mst.DN_DESCRIPTION,MST.DESCRIPTION) Description
   from 
     XXNP_OPN_JOBLOG_001          mst , 
     XXNP_OPN_JOBLOG_STAGE_002   dtl , 
     XXNP_OPN_JOBLOG_SLURRY_003  subdtl ,
     XXNP_OPN_JOBLOG_EST_002     subdtlest,
     qp_list_lines_v            line  ,
     QP_LIST_HEADERS_TL          head  ,
     QP_LIST_HEADERS_B           curr,   
     ra_customers ra ,
     AR_ADDRESSES_V v ,
     MTL_SYSTEM_ITEMS itm  
     where   
  dtl.OPN_JOBLOG_001_ID                        = mst.OPN_JOBLOG_001_ID         AND 
  subdtl.OPN_JOBLOG_001_ID                = dtl.OPN_JOBLOG_001_ID          AND
  subdtl.OPN_JOBLOG_006_ID                  = dtl.OPN_JOBLOG_006_ID         and  
  subdtlest.OPN_JOBLOG_001_ID           = subdtl.OPN_JOBLOG_001_ID       AND
  subdtlest.OPN_JOBLOG_006_ID         = subdtl.OPN_JOBLOG_006_ID      AND
  subdtlest.OPN_JOBLOG_007_ID         = subdtl.OPN_JOBLOG_007_ID      AND
  line.LIST_HEADER_ID          =       head.LIST_HEADER_ID                          and  
  curr.LIST_HEADER_ID              =   head.LIST_HEADER_ID                        and  
  substr ( PRODUCT_ATTR_VAL_DISP , 1, instr ( PRODUCT_ATTR_VAL_DISP ,'.' )-1 )  = subdtlest.ITEM_NUMBER
  and   itm.SEGMENT3 = substr ( PRODUCT_ATTR_VAL_DISP , INSTR(PRODUCT_ATTR_VAL_DISP, '.', -1, 1) + 1  ) 
  and   mst.CONTRACT                       =  head.name   
  and   RA.CUSTOMER_ID = v.CUSTOMER_ID
  and   ra.customer_id = mst.customer_id
  and   V.ADDRESS1 =  mst.CONTRACT      
  and mst.OPN_JOB_DESC='K/D/UP24/UN006/KOP/PL/0112/1'
  and itm.ORGANIZATION_ID  =103
  and itm.INVENTORY_ITEM_STATUS_CODE = 'Active'
  group by 
  mst.ENTRY_DATE ,  
  V.ADDRESS1     , 
  OPN_JOB_DESC   ,mst.FIELD,
  ra.CUSTOMER_NAME , 
  substr ( PRODUCT_ATTR_VAL_DISP , INSTR(PRODUCT_ATTR_VAL_DISP, '.', -1, 1) + 1  ), 
  itm.ATTRIBUTE2        , 
  subdtlest.ITEM_UOM ,
  mst.DN_DESCRIPTION  ,GROSS_ITEM_REF ,GROSS_ITEM_NAME
  order by  substr ( PRODUCT_ATTR_VAL_DISP , INSTR(PRODUCT_ATTR_VAL_DISP, '.', -1, 1) + 1  )   Asc
  
  i require  a field called balance which is sum of Quantity_issue and Quantity fields

  i tried this
  
  select  
  mst.ENTRY_DATE    ordered_date , mst.FIELD, 
  V.ADDRESS1        Address   , 
  OPN_JOB_DESC      job_number,
  ra.CUSTOMER_NAME    customer_NAME, 
  DECODE ( GROSS_ITEM_REF , NULL , SUBSTR (PRODUCT_ATTR_VAL_DISP , INSTR( PRODUCT_ATTR_VAL_DISP, '.', -1, 1) + 1  ) , GROSS_ITEM_REF )      item_number, 
  dECODE (GROSS_ITEM_NAME , NULL,itm.attribute2,GROSS_ITEM_NAME  )        item_description, 
  subdtlest.ITEM_UOM                unit_of_measurement, 
  SUM ( subdtlest.QTY_ISSUE  )            Quantity_Issue,
  sum(nvl (subdtlest.CALC_CEMENT_SK ,subdtlest.CALC_AMOUNT ) )    quantity ,
  sum(sum(nvl (subdtlest.CALC_CEMENT_SK ,subdtlest.CALC_AMOUNT ) ) + SUM ( subdtlest.QTY_ISSUE  )  )balance,
  mst.DN_DESCRIPTION   Description
  --decode(MST.DESCRIPTION ,NULL ,mst.DN_DESCRIPTION,MST.DESCRIPTION) Description
   from 
     XXNP_OPN_JOBLOG_001          mst , 
     XXNP_OPN_JOBLOG_STAGE_002   dtl , 
     XXNP_OPN_JOBLOG_SLURRY_003  subdtl ,
     XXNP_OPN_JOBLOG_EST_002     subdtlest,
     qp_list_lines_v            line  ,
     QP_LIST_HEADERS_TL          head  ,
     QP_LIST_HEADERS_B           curr,   
     ra_customers ra ,
     AR_ADDRESSES_V v ,
     MTL_SYSTEM_ITEMS itm  
     where   
  dtl.OPN_JOBLOG_001_ID                        = mst.OPN_JOBLOG_001_ID         AND 
  subdtl.OPN_JOBLOG_001_ID                = dtl.OPN_JOBLOG_001_ID          AND
  subdtl.OPN_JOBLOG_006_ID                  = dtl.OPN_JOBLOG_006_ID         and  
  subdtlest.OPN_JOBLOG_001_ID           = subdtl.OPN_JOBLOG_001_ID       AND
  subdtlest.OPN_JOBLOG_006_ID         = subdtl.OPN_JOBLOG_006_ID      AND
  subdtlest.OPN_JOBLOG_007_ID         = subdtl.OPN_JOBLOG_007_ID      AND
  line.LIST_HEADER_ID          =       head.LIST_HEADER_ID                          and  
  curr.LIST_HEADER_ID              =   head.LIST_HEADER_ID                        and  
  substr ( PRODUCT_ATTR_VAL_DISP , 1, instr ( PRODUCT_ATTR_VAL_DISP ,'.' )-1 )  = subdtlest.ITEM_NUMBER
  and   itm.SEGMENT3 = substr ( PRODUCT_ATTR_VAL_DISP , INSTR(PRODUCT_ATTR_VAL_DISP, '.', -1, 1) + 1  ) 
  and   mst.CONTRACT                       =  head.name   
  and   RA.CUSTOMER_ID = v.CUSTOMER_ID
  and   ra.customer_id = mst.customer_id
  and   V.ADDRESS1 =  mst.CONTRACT      
  --and   OPN_JOB_DESC = :P_JOB_NUMBER
  and itm.ORGANIZATION_ID  =103
  and itm.INVENTORY_ITEM_STATUS_CODE = 'Active'
  and mst.OPN_JOB_DESC='K/D/UP24/UN006/KOP/PL/0112/1'
  group by 
  mst.ENTRY_DATE ,  
  V.ADDRESS1     , 
  OPN_JOB_DESC   ,mst.FIELD,
  ra.CUSTOMER_NAME , 
  substr ( PRODUCT_ATTR_VAL_DISP , INSTR(PRODUCT_ATTR_VAL_DISP, '.', -1, 1) + 1  ), 
  itm.ATTRIBUTE2        , 
  subdtlest.ITEM_UOM ,
  subdtlest.CALC_CEMENT_SK ,subdtlest.CALC_AMOUNT, subdtlest.QTY_ISSUE ,
  mst.DN_DESCRIPTION  ,GROSS_ITEM_REF ,GROSS_ITEM_NAME
  order by  substr ( PRODUCT_ATTR_VAL_DISP , INSTR(PRODUCT_ATTR_VAL_DISP, '.', -1, 1) + 1  )   Asc
  
  but i get the cursor pointing to mst.entry_date and the message "not a single group function"
  
  kindly guide me

  thanking in advance
  

• Group functions

  Hello
  
  Is it possible to aggregate group functions based on a field in responses, my query is below:
  
  SELECT Absence_Days,
  Absence_Reason,
  Absence_Type,
  Employee_name,
  End_date,
  First name,
  Last_name,
  Person_Id,
  Start_date,
  DEPARTMENT,
  DIVISION,
  INTERNAL_LOCATION,
  CASE
  WHERE LAST_NAME = 'Gordon '.
  THEN COUNT (person_id) * COUNT (Person_Id) * SUM (Absence_Days)
  0 OTHERWISE
  END
  FROM EMP
  WHERE LAST_NAME = 'Gordon '.
  GROUP OF ABSENCE_DAYS, ABSENCE_REASON, ABSENCE_TYPE, EMPLOYEE_NAME, END_DATE, FIRST_NAME, LAST_NAME, PERSON_ID, START_DATE, DEPARTMENT, DIVISION, INTERNAL_LOCATION
  
  Count (person_id) * COUNT (Person_Id) * SUM (Absence_Days) returns a value by each line and aggregate in the column of fx. I want to this group by person_id, is this possible?
  
  Thank you

  Hello

  Try this in Fx

  SUM (Absence_Days by person_id)

  Thank you
  Saichand.v

• The local users and 'Add Member to group' groups function displays the password in clear text

  Hello

  I saw something strange that happens when you add the connected locally to the domain user to a local security group via lusrmgr.msc. This seems to be repeated in our Organization, and I was wondering if it's something to do with our generation, or if it is more widespread? After you have selected the locally connected user (domain) to be added to a local security group, but before confirming with OK / apply, the user name appears in the ACL in the format:

  Domain\username (password@fulldomain)

  The password appears when the user name should be, which is probably just a minor bug where an incorrect value is selected on the screen, but the fact that it is displayed in plain text rather than a hash of password is a little more disconcerting. There seems to be some caveats: it affects only the users in the domain (local users are displayed differently); It is not affected by the local administrator rights (although /are/ to add the user to the local administrator rights); It doesn't affect that the user logged on locally (I can't reproduce as a runas user).

  The steps to reproduce this are:
  (1) start | Run | lusrmgr.msc
  (2) go to the groups. [Any group] | Right-click | Add to group
  3) click on add
  (4) enter the username of the user locally. Ok

  Anyone else see this behavior?

  See you soon,.

  Dines keV

  Hello

  The question you posted would be better suited to the TechNet community. Please visit the link below to find a community that will provide the support you want.

  http://social.technet.Microsoft.com/forums/en-us/category/w7itpro

  Hope this information is useful.

• Analytical and group functions of

  Oracle 11g Release 2 Server

  SQL, SQLPLUS not

  I need to 'break' when a column value changes and insert a line with only a dash.

  CREATE TABLE tab
  (   id            number
    , part_num      number
    , part_type_id  number
    , part_type_txt varchar2(50)
    , CONSTRAINT tab_pk PRIMARY KEY(part_num,part_type_id)
  )
  /
  
  insert into tab values(1,10,100,'hose') ;
  insert into tab values(1,10,110,'hose clamp') ;
  insert into tab values(1,20,200,'plastic value') ;
  insert into tab values(1,20,210,'brass value') ;
  insert into tab values(1,30,300,'headlamp') ;
  insert into tab values(1,30,310,'lamp misc') ;
  commit ;
  
  select part_num,part_type_id,part_type_txt, count(part_num) CNT
  from   tab
  where  id = 1
  group by part_num,part_type_id,part_type_txt
  order by part_num,part_type_id ;
  
    PART_NUM PART_TYPE_ID PART_TYPE_TXT       CNT
  ---------- ------------ ------------------- ----------
          10          100 hose                         1
          10          110 hose clamp                   1
          20          200 plastic value                1
          20          210 brass value                  1
          30          300 headlamp                     1
          30          310 lamp misc                    1
  

  My query:

  WITH   data
  AS
  (
     select   case when nvl( lag( part_num ) over( order by rownum ), ' ' ) != part_num
                   then part_num
              end part_num
           ,  part_type_id, part_type_txt, count(part_num) CNT
     from
              tab
     where    id = 1 
     group by    case when nvl( lag( part_num ) over( order by rownum ), ' ' ) != part_num
                      then part_num
                 end 
               , part_type_id
               , part_type_txt
     order by part_num,part_type_id 
  )
  SELECT rownum, d.part_num,d.part_type_id, d.part_type_txt
  FROM   data   d ;
  
  ERROR at line 11:
  ORA-30483: window  functions are not allowed here
  
  
  

  I guess that analytical functions are not allowed in a group by clause.

  Output desired, he had to 'break' on when PART_NUM changes and insertion of a line with a dash (-):

    PART_NUM PART_TYPE_ID PART_TYPE_TXT       CNT
  ---------- ------------ ------------------- ----------
          10          100 hose                         1
          10          110 hose clamp                   1
          -
          20          200 plastic value                1
          20          210 brass value                  1
          -
          30          300 headlamp                     1
          30          310 lamp misc                    1
  

  Any help appreciated.

  Hello

  I see: XYZ123 is linked only to a part_type_id, 7777, so you do not want any output to XYZ123.

  Similarly, XYZ456 is only linked to a part_type_id, 8888, so you do not want XYZ456.

  One thing you can do is to start with the query in response to #4 above.  The results of this query, you can count part_type_ids h: different lots, each part_num is bound to and then only display the part_nums who have more than 1 part_type_id.  For example:

  WITH got_aggregates AS

  (

  SELECT THE CHECK BOX

  WHEN you GROUP (part_type_id) = 0

  THEN TO_CHAR (part_num)

  ELSE '-'

  END                               AS txt

  part_type_id, part_type_txt

  CASE

  WHEN you GROUP (part_type_id) = 0

  THEN COUNT (part_num)

  END                               AS cnt

  part_num,

  COUNT (DISTINCT part_type_id)

  COURSES (PARTITION BY part_num) AS part_type_id_cnt

  TAB

  ID WHERE = 1

  GROUP BY part_num, ROLLUP (part_type_id, part_type_txt))

  )

  SELECT txt, part_type_id, part_type_txt, cnt

  OF got_aggregates

  WHERE part_type_id_cnt > 1

  ORDER BY part_num, part_type_id

  ;

  Note that the subquery got_aggregates is almost identical to the response query #4.  The only differences are that the subquery is not an ORDERBY clause (ORDER BY is almost always useless to subqueries) and includes two additional columns, which will be needed in the WHERE and ORDER BY the main query clauses.

• LR buy on Amazon or Adobe? questions... It sounds as if you're going on the road to Adobe and just decide to stay on for 1 year, you then to lose the ability to "develop" and some other functions? What is buy the disc or download also

  Just trying to figure my best options, looks like your hook with Adobe.

  There are actually two routes from Adobe, you can take. One of them is to subscribe to the creative cloud and pay a monthly subscription of $9.99. For this tax, you get the latest Photoshop and the latest Lightroom and all improvements that come along. It is true that if you decide that you do not want to continue your subscription, you will lose functionality in the programs. The other method is to acquire directly from Adobe Lightroom. It can be purchased from this site:

  Products

  You can buy right now the 5 Lightroom and download it. If there are several updates Lightroom 5, you will get free updates. You will be able to use Lightroom 5 "forever". You have the license to use 5 Lightroom, and it does not expire. Don't forget that when Lightroom 6 is out you will have more no support from Adobe, nor will there be additional updates at your disposal. It's a bit like buying a used car. You can use it for as long as it works for you, but there is no support from society.

  The choice is yours, of course. Personally, I think that plan creative cloud works really well. It is economical, and he assures you that you will always have the latest versions of Photoshop and Lightroom. But if this plan doesn't appeal to you, then you can buy 5 Lightroom from the link I provided above. 5 Lightroom is an excellent program. But he reaches the end of its life cycle. You can now buy it and install updates that may come. And it is possible that there will be another update or two who might support some new cameras. But when Lightroom 6 came out, if you want the new features and support for new cameras, then you will need to purchase a version update of Lightroom 6. One that suits you best, it is something only you can decide.

• Group function

  Function group can be nested?

  Hello

  MU peleipua wrote:
  Function group can be nested?

  One of the aggregate functions can be nested inside of each other. The GROUP BY clause will be aplly to domestic.
  For example:

  SELECT    AVG (SUM (sal))    AS avg_dept_sal
  FROM      scott.emp
  GROUP BY  deptno
  ;
  

  First of all, the SUM is calculated for each Department, then the average of these numbers is taken.
  The result set of a query with a nested aggregate functions always contains 1 row.

  AVG_DEPT_SAL
  ------------
          9675
  

  To understand what is happening, you may find it useful to run a similar query without the external function:

  SELECT       deptno
  ,       SUM (sal)     AS dept_sal
  FROM       scott.emp
  GROUP BY  deptno
  ;
  

  Output:

  .   DEPTNO   DEPT_SAL
  ---------- ----------
          30       9400
          20      10875
          10       8750
  

  You cannot nest deeper than that of aggregation functions.

• Need help with Group functions

  I'm a total novice with SQL, so please forgive me if the answer to my question seems to be too obvious
  I work with diagrams of the sample (in particular with the employees table):
  
  DESC employees;
  result
  
  What I have to do is select all the managers, including the number of subordinates is higher than the average number of subordinates of managers who work in the same Department. What I've done so far is as follows:
  
  SELECT mgr.employee_id manager_id, Director of mgr.last_name, mgr.department_id, COUNT (emp.employee_id)
  Employees emp employees JOIN Bishop
  ON emp.manager_id = mgr.employee_id
  GROUP OF mgr.employee_id, mgr.last_name, mgr.department_id
  ORDER BY mgr.department_id;
  result
  
  As you can see, I'm almost done. Now, I need only to calculate the average of the result of the COUNT function for each Department. But I'm totally stuck at this point.
  All advice?

  Hello

  Welcome to the forum!

  user12107811 wrote:
  I'm a total novice with SQL, so please forgive me if the answer to my question seems to be too obvious

  Just the opposite! Looks like a very difficult mission.

  I work with diagrams of the sample (in particular with the employees table):

  DESC employees;
  result

  What I have to do is select all the managers, including the number of subordinates is higher than the average number of subordinates of managers who work in the same Department. What I've done so far is as follows:

  SELECT mgr.employee_id manager_id, Director of mgr.last_name, mgr.department_id, COUNT (emp.employee_id)
  Employees emp employees JOIN Bishop
  ON emp.manager_id = mgr.employee_id
  GROUP OF mgr.employee_id, mgr.last_name, mgr.department_id
  ORDER BY mgr.department_id;
  result

  As you can see, I'm almost done. Now, I need only to calculate the average of the result of the COUNT function for each Department. But I'm totally stuck at this point.
  All advice?

  Yes, you're almost done. You just need to add one more condition. You have to calculate the average value of total_cnt (the COUNT (*) you already do) of a Department and compare that to total_cnt.

  There are several ways to do this, including
  a scalar subquery (in a HAVING clause)
  (b) make a result set with one line per Department, containing the average_cnt and reach than your current result set
  (c) analytical functions. Analytical functions are calculated after the GROUP BY clause is applied and aggregate functions are calculated, it is legitimate to say "AVG (COUNT (*)) MORE (...)").

  If thinking (c) is the simplest. It involves the use of a query of Tahina, but (a) and (b) also require subqueries.

  This sounds like homework, so I'll do it for you.
  Instead, here is a very similar problem with the hr.employees table.
  Let's say that we are interested in total wages given each type of work in each Department.

  SELECT        department_id
  ,        job_id
  ,        SUM (salary)     AS sum_sal
  FROM        hr.employees
  GROUP BY   department_id
  ,             job_id
  ORDER BY   department_id
  ,             job_id
  ;
  

  Results:

  DEPARTMENT_ID JOB_ID        SUM_SAL
  ------------- ---------- ----------
             10 AD_ASST          4400
             20 MK_MAN          13000
             20 MK_REP           6000
             30 PU_CLERK        13900
             30 PU_MAN          11000
             40 HR_REP           6500
             50 SH_CLERK        64300
             50 ST_CLERK        55700
             50 ST_MAN          36400
             60 IT_PROG         28800
             70 PR_REP          10000
             80 SA_MAN          61000
             80 SA_REP         243500
             90 AD_PRES         24000
             90 AD_VP           34000
            100 FI_ACCOUNT      39600
            100 FI_MGR          12000
            110 AC_ACCOUNT       8300
            110 AC_MGR          12000
                SA_REP           7000
  

  Now suppose we want to find out which of these sum_sals is higher than the average sum_sal of his Department.
  For example, in detriment 110 (near the end OIF the list) there two types of work (AC_ACCOUND and AC_MGR) that have sum_sals of 8300 and 12000. The average of these two numbers is 10150, so we selected AC_MGR (because its sum_sal, 12000, is superior to 10150, and we do not want to include AC_ACCOUNT, because its sum_sal, 8300, is less than or equal to the average of the Department.
  In departments where there is only one job type (for example, Department 70, or null "Department" at the end of the list above) the only sum_sal will be the average; and because the sum_sal is not greater than the average, we want to exclude this line.

  Let's start with the calculation of the avg_sum_sal using the analytical function AVG:

  SELECT        department_id
  ,        job_id
  ,        SUM (salary)                              AS sum_sal
  ,        AVG (SUM (salary)) OVER (PARTITION BY department_id)     AS avg_sum_sal
  FROM        hr.employees
  GROUP BY   department_id
  ,             job_id
  ORDER BY   department_id
  ,             job_id
  ;
  

  Output:

  DEPARTMENT_ID JOB_ID        SUM_SAL AVG_SUM_SAL
  ------------- ---------- ---------- -----------
             10 AD_ASST          4400        4400
             20 MK_MAN          13000        9500
             20 MK_REP           6000        9500
             30 PU_CLERK        13900       12450
             30 PU_MAN          11000       12450
             40 HR_REP           6500        6500
             50 SH_CLERK        64300  52133.3333
             50 ST_CLERK        55700  52133.3333
             50 ST_MAN          36400  52133.3333
             60 IT_PROG         28800       28800
             70 PR_REP          10000       10000
             80 SA_MAN          61000      152250
             80 SA_REP         243500      152250
             90 AD_PRES         24000       29000
             90 AD_VP           34000       29000
            100 FI_ACCOUNT      39600       25800
            100 FI_MGR          12000       25800
            110 AC_ACCOUNT       8300       10150
            110 AC_MGR          12000       10150
                SA_REP           7000        7000
  

  Now all we have to do is to compare the sum_sal and avg_sum_sal columns.
  Given that the analytic functions are calculated after the WHERE clause is applied, we cannot use avg_sum_sal in the WHERE clause of the query, even where it has been calculated. But we can do that in a subquery; Then, we can use avg_sum_sal in any way that we love in the Super-requete:

  WITH     got_avg_sum_sal       AS
  (
       SELECT        department_id
       ,        job_id
       ,        SUM (salary)                              AS sum_sal
       ,        AVG (SUM (salary)) OVER (PARTITION BY department_id)     AS avg_sum_sal
       FROM        hr.employees
       GROUP BY   department_id
       ,             job_id
  )
  SELECT        department_id
  ,        job_id
  ,        sum_sal
  FROM        got_avg_sum_sal
  WHERE        sum_sal     > avg_sum_sal
  ORDER BY   department_id
  ,             job_id
  ;
  

  Results:

  DEPARTMENT_ID JOB_ID        SUM_SAL
  ------------- ---------- ----------
             20 MK_MAN          13000
             30 PU_CLERK        13900
             50 SH_CLERK        64300
             50 ST_CLERK        55700
             80 SA_REP         243500
             90 AD_VP           34000
            100 FI_ACCOUNT      39600
            110 AC_MGR          12000
  

• I followed some groups and contributors

  But did not get any updates by email all weekend.

  Change something?

  Thank you!

  Holly

  Make sure that you always follow these groups and that it contributes.  Over the years, I found that from time to time the forum destroyed my following, and I need to re - follow the things that I'm interested.

  I don't know if that's a problem, or if there is a way to ensure that the user is always really interested in getting these emails, and if they aren't, it will reduce the number of servers forum need to send emails.  TOTAL speculation on my part, but it happened to me several times since 2005 when I began to participate in the forums.