can do without group by clause report through parameter

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• Group by clause

  Hello

  You can use group by clause with COUNT function...

  In the reports, I have to count the stattus data_set group... If I use COUNT (data_set), it gives me 2 different ranks

  1 for the race and the other to succeed.

  Please help me...

  Thank you

  RS

  I'm able to do. I used the formula below:

  COUNT (CASE WHEN "Status". ("Employment status" = 'Succeed' THEN 1 END)

  even for running status and on the other and it works very well...

• Reg: Having without GROUP OF

  Hi all

  Ask you to clarify for me please on the underside of the statements. How the HAVING works, it's like a WHERE clause. As I studied, HAVING will work only on the GROUP BY.

  SQL > ed

  A written file afiedt.buf

  1 declare

  n number of 2;

  3. start

  4. select count (empno) n of PMC with 1 = 2;

  5 dbms_output.put_line (n);

  exception 6

  7 when no_data_found then

  8 dbmS_output.put_line ("' no problem");

  9 * end;

  SQL > /.

  No problem

  PL/SQL procedure successfully completed.

  SQL > select count (empno) of emp have 1 = 3;   -Why HAVING is different here compared to the Sub declaration.

  no selected line

  SQL > select count (empno) of emp having ename = 'SCOTT';

  Select count (empno) in emp having ename = 'SCOTT '.

  *

  ERROR on line 1:

  ORA-00979: not a GROUP BY expression

  SQL >

  Another way in which these two differ in execution.

  SQL > select count (empno) from emp including 1 = 2;

  COUNT (EMPNO)

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

  0

  SQL > select count (empno) of emp with 1 = 2;

  no selected line

  Thank you

  Hello

  You're right; HAVING is like WHERE.  The difference is that the WHERE clause is applied before the aggregation and HAVING is applied after him.

  The documentation implies that you can use a HAVING clause only if you also have a GROUP BY clause.  As you have noticed, Oracle does not trigger an error when using HAVING without GROUP BY.  It's the kind of behavior without papers who may change from one version to another without warning, so I would avoid that.  If you really want to use a HAVING clause, but you don't need a GROUP BY clause, and then use GROUP BY NULL, just to make sure that your return will be syntactically correct, no matter what are the mistakes is the application of the version or the environment...

• -SQL - GROUP BY clause: fields of nonaggregate mandate

  Hello
  
  I study data (especially data recovery) and found something interesting.
  
  When you use an aggregate function in the SELECT clause, it is mandatory to have all fields that are not aggregated in the SELECT clause to be there in the GROUP BY clause.
  For example,.
  
  SELECT dept_no, Salary
  The EMPLOYEE
  GROUP BY dept_no;
  
  The SQL above works fine.
  But what happens if the user forgets the dept_no in the GROUP BY clause or the clause GROUP BY itself is missing?
  Certainly, it is a mistake.
  
  Why this error is not handled by the database. I mean, the database must be smart/pretty smart to add the GROUP BY clause by itself. So let's assume that, if I miss the GROUP BY clause or miss a field no aggregated from the SELECT clause when I get at least an aggregate function on a field with at least a no aggregated field in the SELECT clause, the database should check the GROUP BY clause at compile time and add mandate missed the fields in the GROUP BY clause.
  
  Example,
  
  SQL1:_
  SELECT dept_no, Salary
  The EMPLOYEE
  GROUP BY dept_no;
  
  SQL2:_
  SELECT dept_no, Salary
  The EMPLOYEE;
  
  Here, the SQL1 and SQL2, both should give me same output without error.
  
  I can't understand why this is handled?

  Hello

  998478 wrote:
  ... If we mix the aggregated and non-aggregated values, then there must be a GROUP BY clause that contains all non-aggregated values. Why this is handled by the database/compiler itself?

  It IS managed by the compiler itself. The compiler manages to trigger an error. The compiler has no way of knowing if you want to remove something from the SELECT clause, or add something to the GROUP BY clause, or not to use the aggregate functions or use several aggregate functions, or a combination of the above. If the compiler re-writes your code and none of these things done automatically, it would be wrong more often that he was right, and you would (rightly) complain about his behavior.

  For example, it is clearly wrong:

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

  What is the right way to fix it?

  1. remove something from the SELECT clause

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

  2. add something to the GROUP BY clause

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

  3. do not use aggregate functions

  SELECT    deptno
  ,       job
  ,       sal
  FROM       scott.emp
  ;
  

  4. use several aggregate functions

  SELECT    deptno
  ,       MIN (job)
  ,       SUM (sal)
  FROM       scott.emp
  GROUP BY  deptno
  ;
  

  What are all the options, either. For example, the correct solution would be to use analytical functions instead of aggregate functions.
  How can anyone tell which of them is right? They all have the right answer for some problem.

  Moreover, by saying that everying in the SELECT clause must be an aggregate or in the GROUP BY clause is a bit oversimplified.
  Fuller, here's the ABC of GROUP BY:
  When you use a GROUP BY clause or in an aggregate function, then all in the SELECT clause must be:
  (A) a ggregate function,
  (B) one of the expressions "group By."
  (C) adding to C, or
  (D) something that Depends on the foregoing. (For example, if you "GROUP BY TRUNC (dt)", you can SELECT "TO_CHAR (TRUNC (dt), 'Mon - DD')").

  Published by: Frank Kulash on April 13, 2013 13:44
  Additional code examples.

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  I would like the white area (which was created using the Brush tool and applying a stroke) to display the transparent background behind the red square - or whatever that fall behind the place.

  
  

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  I'm probably just missing something.
  

  
  

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  Or:

  Stroke (s) select 0) and Ctrl + G for group races if more than one,

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• Concatenation of data with the GROUP BY clause

  Hi again!
  
  Following my previous thread...
  Re: Need help with RANK() on data ZERO
  
  I tried to apply the GROUP BY clause instead of preforming my query with RANK() to manage records NULL... I have a scenario where I also need to concatenate data from several lines.
  
  CREATE TABLE T_EMP (NUMBER OF EMP_NO, NAME VARCHAR2 (20));
  INSERT INTO T_EMP VALUES (1001, 'MARK');
  INSERT INTO T_EMP VALUES (1002, 'DAVID');
  INSERT INTO T_EMP VALUES (1003, "SHAUN");
  INSERT INTO T_EMP VALUES (1004, "JILL");
  
  CREATE TABLE T_EMP_DEPT (NUMBER OF EMP_NO, DEPT_NO NUMBER);
  INSERT INTO T_EMP_DEPT VALUES (1001, 10);
  INSERT INTO T_EMP_DEPT VALUES (1001, 20);
  INSERT INTO T_EMP_DEPT VALUES (1002, 10);
  INSERT INTO T_EMP_DEPT VALUES (1002, 20);
  INSERT INTO T_EMP_DEPT VALUES (1002, 30);
  INSERT INTO T_EMP_DEPT VALUES (1003, 20);
  INSERT INTO T_EMP_DEPT VALUES (1003, 30);
  INSERT INTO T_EMP_DEPT VALUES (1004, 10);
  
  CREATE TABLE T_EMP_VISITS (NUMBER OF EMP_NO, DEPT_NO NUMBER, VISITED DATE);
  INSERT INTO T_EMP_VISITS VALUES (1001, 10, 1 JAN 2009');
  INSERT INTO T_EMP_VISITS VALUES (1002, 10, 1 JAN 2009');
  INSERT INTO T_EMP_VISITS VALUES (1002, 30, 11 APR 2009');
  INSERT INTO T_EMP_VISITS VALUES (1003, 20, 3 MAY 2009');
  INSERT INTO T_EMP_VISITS VALUES (1003, 30: 14 FEB 2009');
  COMMIT;
  
  I have a T_EMP master table that stores the name and number of the emp. Each emp is required to visit some departments. This mapping is stored in the T_EMP_DEPT table. An employee can visit one or more departments. T_EMP_VISITS table stores the dates where the employee visited the services required. I need to view the report which should show when an employee all completed visits, which is the maximum date when it finished to visit all departments. If he did not visit any of the report should display date max, otherwise NULL. I was able to do using GROUP BY such proposed by Salim, but how do I show a list separated by commas of the services required for an employee in the same query.
  
  SELECT
  EMP_NO,
  NAME,
  MAX (DEPT_NO) KEEP (DENSE_RANK LAST ORDER BY VISITED) MAX_DEPT_NO,.
  MAX (VISITED) KEEP (DENSE_RANK LAST ORDER PER VISIT) VISITS_COMP
  DE)
  SELECT
  T_EMP. EMP_NO,
  NAME,
  T_EMP_DEPT. DEPT_NO,
  VISITED
  OF T_EMP
  LEFT OUTER JOIN T_EMP_DEPT
  ON T_EMP. EMP_NO = T_EMP_DEPT. EMP_NO
  LEFT OUTER JOIN T_EMP_VISITS
  ON T_EMP_DEPT. EMP_NO = T_EMP_VISITS. EMP_NO
  AND T_EMP_DEPT. DEPT_NO = T_EMP_VISITS. DEPT_NO)
  GROUP EMP_NO, NAME;
  
  Output
  EMP_NO NAME MAX_DEPT_NO VISITS_COMP
  1001 MARK 20
  1002 DAVID 20
  1003 SHAUN 20 3 MAY 09
  JILL 1004
  
  Power required
  EMP_NO NAME REQ_DEPTS MAX_DEPT_NO VISITS_COMP
  1001 MARC 20 10.20
  1002 DAVID 10,20,30 20
  1003 SHAUN 20,30 20 3 MAY 09
  JILL 10 1004
  
  Can we do this in a single query?

  Hello

  user512647 wrote:
  ... Sanjay
  The query you provided that stragg() use seems to work but my requirement is not in the result set. I don't know how to use stragg with
  MAX (DEPT_NO) KEEP (DENSE_RANK LAST ORDER BY VISITED) MAX_DEPT_NO,.
  MAX (VISITED) KEEP (DENSE_RANK LAST ORDER PER VISIT) VISITS_COMP
  I need more, these two columns these gives me the date when they have completed all visits. If they missed any Department then the result must be NULL in the VISITS_COMP field.

  Just add them to the SELECT clause:

  SELECT    t_emp.emp_no,
         name,
         STRAGG (t_emp_dept.dept_no)     AS deptno,
         MAX (t_emp_dept.dept_no) KEEP (DENSE_RANK LAST ORDER BY visited)
                                    AS max_dept_no,
         MAX (visited)                      AS visits_comp
  FROM             t_emp
  LEFT OUTER JOIN      t_emp_dept     ON   t_emp.emp_no     = t_emp_dept.emp_no
  LEFT OUTER JOIN      t_emp_visits     ON   t_emp_dept.emp_no     = t_emp_visits.emp_no
                               AND  t_emp_dept.dept_no = t_emp_visits.dept_no
  GROUP BY  t_emp.emp_no
  ,            name
  ;
  

  The column called visit_comp is simply the last visited, regardless of how the employee visited departments.
  If you want to have the NULL value if the employee has not yet visited all 3 departments:

  ...       CASE
            WHEN  COUNT (DISTINCT t_emp_dept.dept_no) = 3
            THEN  MAX (visited)
         END                    AS visits_comp
  

  The 'magic number' 3 is the total number of departments.
  If you want to understand the correct value of that at the time of the execution of the query, replace the code literal 3 hard with a scalar subquery.

  Note that 'KEEP MAX (x) (DENSE_RANK OVER LAST SERVICE BY x)' (where the exact same column is used as an argument and that the ORDER BY column) is just "MAX (x)".

• A question about the analytical function used with the GROUP BY clause in SHORT

  Hi all
  
  I created the following table named myenterprise
  

  CITY       STOREID    MONTH_NAME TOTAL_SALES            
  ---------- ---------- ---------- ---------------------- 
  paris      id1        January    1000                   
  paris      id1        March      7000                   
  paris      id1        April      2000                   
  paris      id2        November   2000                   
  paris      id3        January    5000                   
  london     id4        Janaury    3000                   
  london     id4        August     6000                   
  london     id5        September  500                    
  london     id5        November   1000

  If I want to find which is the total sales by city? I'll run the following query

  SELECT city, SUM(total_sales) AS TOTAL_SALES_PER_CITY
  FROM myenterprise
  GROUP BY city
  ORDER BY city, TOTAL_SALES_PER_CITY;

  that works very well and produces the expected result, i.e.

  CITY       TOTAL_SALES_PER_CITY   
  ---------- ---------------------- 
  london     10500                  
  paris      17000            

  Now in one of my books SQL (Mastering Oracle SQL) I found another method by using the SUM, but this time as an analytic function. Here's what the method of the book suggests as an alternative to the problem:

  SELECT city, 
         SUM(SUM(total_sales)) OVER (PARTITION BY city) AS TOTAL_SALES_PER_CITY
  FROM myenterprise
  GROUP BY city
  ORDER BY city, TOTAL_SALES_PER_CITY;

  I know that the analytic functions are executed after the GROUP BY clause has been transformed completely and Unlike regular aggregate functions, they return their result for each line belonging to the partitions specified in the partition clause (if there is a defined partition clause).
  
  Now my problem is that I do not understand what we have to use two functions SUM? If we only use one only, i.e.

  SELECT city, 
         SUM(total_sales) OVER (PARTITION BY city) AS TOTAL_SALES_PER_CITY
  FROM myenterprise
  GROUP BY city
  ORDER BY city, TOTAL_SALES_PER_CITY;

  This generates the following error:

  Error starting at line 2 in command:
  SELECT city, 
         SUM(total_sales) OVER (PARTITION BY city) AS TOTAL_SALES_PER_CITY
  FROM myenterprise
  GROUP BY city
  ORDER BY city, TOTAL_SALES_PER_CITY
  Error at Command Line:2 Column:11
  Error report:
  SQL Error: ORA-00979: not a GROUP BY expression
  00979. 00000 -  "not a GROUP BY expression"
  *Cause:    
  *Action:

  The error is generated for the line 2 column 11 which is, for the expression SUM (total_sales), well it's true that total_sales does not appear in the GROUP BY clause, but this should not be a problem, it has been used in an analytical function, so it is evaluated after the GROUP BY clause.
  
  So here's my question:
  
  Why use SUM (SUM (total_sales)) instead of SUM (total_sales)?
  
  
  Thanks in advance!
  :)
  
  
  
  
  
  In case you are interested, that's my definition of the table:

  DROP TABLE myenterprise;
  CREATE TABLE myenterprise(
  city VARCHAR2(10), 
  storeid VARCHAR2(10),
  month_name VARCHAR2(10),
  total_sales NUMBER);
  
  INSERT INTO myenterprise(city, storeid, month_name, total_sales)
    VALUES ('paris', 'id1', 'January', 1000);
  INSERT INTO myenterprise(city, storeid, month_name, total_sales)
    VALUES ('paris', 'id1', 'March', 7000);
  INSERT INTO myenterprise(city, storeid, month_name, total_sales)
    VALUES ('paris', 'id1', 'April', 2000);
  INSERT INTO myenterprise(city, storeid, month_name, total_sales)
    VALUES ('paris', 'id2', 'November', 2000);
  INSERT INTO myenterprise(city, storeid, month_name, total_sales)
    VALUES ('paris', 'id3', 'January', 5000);
  INSERT INTO myenterprise(city, storeid, month_name, total_sales)
    VALUES ('london', 'id4', 'Janaury', 3000);
  INSERT INTO myenterprise(city, storeid, month_name, total_sales)
    VALUES ('london', 'id4', 'August', 6000);
  INSERT INTO myenterprise(city, storeid, month_name, total_sales)
    VALUES ('london', 'id5', 'September', 500);
  INSERT INTO myenterprise(city, storeid, month_name, total_sales)
    VALUES ('london', 'id5', 'November', 1000);

  Edited by: dariyoosh on April 9, 2009 04:51

  It is clear that thet Analytics is reduntant here...
  You can even use AVG or any analytic function...

  SQL> SELECT city,
    2         avg(SUM(total_sales)) OVER (PARTITION BY city) AS TOTAL_SALES_PER_CITY
    3  FROM myenterprise
    4  GROUP BY city
    5  ORDER BY city, TOTAL_SALES_PER_CITY;
  
  CITY       TOTAL_SALES_PER_CITY
  ---------- --------------------
  london                    10500
  paris                     17000
  

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  I was very discouraged when I saw the big red warning to the top of the page:

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  In the Groups of Contact with Vista, only contacts who are also included as normal Contacts will be imported into Windows Live Mail. Contacts that have been created for the Contact Group only will not be imported into Windows Live Mail. You will need to write these down and add them manually in Windows Live Mail.
  ---------------

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• Doubts formidable wrt "select", "join" and "group by" clause to calm a while...

  Dear all,

  Kindly advise on below 2 doubts which have been daunting for quiet some time. The bottom has forced our requests to take alternative routes to receive solutions.

  Doubt 1) is it possible for us to use 'select (*)' with a group and a join clause, if we have at least 2/3 tables with minimum 15-20 columns... given the constraint of having to add all the names column in the group by clause.

  Question 2) is it possible to use "select" (*) with the exception of a single column of the table?
  That is to say, I hate the output of a statement select (*) to have the entire field except a field in a table

  Ask your advisor for the same thing. Excuse me because I have no examples to illustrate this point in time. A successful idea will do.

  Thank you and best regards,

  Séverine Suresh

  Hey, Sebastian,

  Séverine Suresh - 3035408 wrote:

  Dear all,

  Kindly advise on below 2 doubts which have been daunting for quiet some time. The bottom has forced our requests to take alternative routes to receive solutions.

  Doubt 1) is it possible for us to use 'select (*)' with a group and a join clause, if we have at least 2/3 tables with minimum 15-20 columns... given the constraint of having to add all the names column in the group by clause.

  Sure.  If you had problems, you did it wrong.  Your postal code, examples of data (CREATE TABLE and INSERT statements) and the results desired from these data.

  Check out the Forum FAQ: Re: 2. How can I ask a question on the forums?

  Normally, you don't want to ' SELECT *...» "in a join.  Are most joins are equijoins, ' SELECT *...» "produced 2 copies of / columns used for Assembly.  In addition, many tables have columns (for example, modified_date) that are not necessary in most applications.

  All columns must be included explicitly in a GROUP BY clause. You cannot use * it.

  You can not have parentheses around the * in ' SELECT *...» ».  You might think 'SELECT COUNT (*) '

  Question 2) is it possible to use "select" (*) with the exception of a single column of the table?
  That is to say, I hate the output of a statement select (*) to have the entire field except a field in a table

  ...
  

  No.; If you don't want a particular column in the result set, you cannot use "SELECT *...» ».

  Your front end can have so as not to display a column in the result set.  For example, in SQL * more:

  COLUMN modified_date NOPRINT

  means that you won't see any column called modified_date.

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  Each rectangle is on another layer.

  Duplicate a rectangle at a time as I can, but not this way.

  I will duplicate the rectangles in the second document.

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  Hello

  I tried other ways but always appears a problem.

  This is the result

  I wrote that does not have problems positioning of objects between them and their layers.

  Example usage of three labels:

  var findLabels = new Array("Name","Photo","Company");
  var joinLb = findLabels.join();
  var result = myAttempt_18 (joinLb);
  alert("Duplicate objects: " + result);
  
  function myAttempt_18(joinLabels){
      var resultLength = 0;
  
      var sourceDoc = app.documents[0];
      var sourceObj;
  
      var destDoc = app.documents[1];
      var destpage = destDoc.pages[0];
      var destSameLayer = new Array();
      var destArrayObjs = new Array();
      var duplicateObj;
  
      with (sourceDoc.pages[0]){
          for (var h = pageItems.length-1; h >= 0; h --){
              sourceObj = pageItems[h];
              if (sourceObj.label != ""){
                  if (joinLabels.indexOf(sourceObj.label) > -1 ){
                    duplicateObj =  sourceObj.duplicate(destpage);
                    destArrayObjs.push(duplicateObj);
                    destSameLayer.push(sourceObj.itemLayer.name);
                      }
                  }
              }
          }
  
      if (destArrayObjs.length > 0 ){
          resultLength = destArrayObjs.length;
          destDoc.select(destArrayObjs, SelectionOptions.replaceWith);
          destGroup = destDoc.groups.add(destDoc.selection);
          destGroup.move ([0,0]);
          destDoc.select (destDoc.selection, SelectionOptions.removeFrom);
          destGroup.ungroup();
  
          for (var L = 0; L < destArrayObjs.length; L ++){
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              }
          }
        return resultLength;
      }
  

  This script will be part of another larger script.

  The code isn't beautiful, but you can use.

  Link to the document that I tested. (www.wetransfer.com)

  http://we.TL/6gpfa3zC2W

  Elvis

• I can't see second network interface integrated through Esxi 4.0 on IBM X 3200 M2

  Hi everyone, I can't see second network interface integrated through Esx 4.0 on IBM X 3200 M2, I installed and configured a first NIC and the third card PCI-express NETWORK Broadcom NetExtreme without any problem. Above their Esxi platform, I have installed two operating systems comments, but I need the second network adapter for vmware management. Now, I use the first network interface integrated to do this, but is not quite secure. I can't reinstall ESX because this system is in production.
  I have read the information, but I can't understand how do.

  Thanks, Claudio

  Welcome to the community,

  I checked the compatibility of your equipment with ESXi4.0 and result in not supported. (Check in HCL )

  IBM	IBM System x 3200 M3	Intel i3/i5 Clarkdale series	ESX 4.1 U1 4.1 4.1 U1 4.1 ...
  IBM	IBM System x 3200 M3	Intel Xeon 34xx series Lynnfield	ESX 4.1 U1 4.1 4.1 U1 4.1 ...

  Can you post here a model of all the network cards?

  Thank you

  Please, do not forget the points of call of the "useful" or "correct" answers

  Mauro Bonder - moderator