Calendars Primavera / calculation of monthly working hours

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• [8i] help with function with parameters (for the calculation of the work)

  Let me start by saying, I've never written a function before, and I do not have access to create a feature in my database (that is, I can't test this feature). I am trying to achieve a function I can ask my IT Department to add for me. I hope that someone can take a look at what I wrote and tell me if this should work or not, and if it's the right way to go to solve my problem.
  
  I'm creating a function to make a very simple calculation of work (add/subtract a number of days to a date in the calendar).
  
  The database, I work with has a table with the schedule of work. Here is a sample table and sample data, representative of what is in my work table calendar:
  

  CREATE TABLE caln
  (     clndr_dt     DATE,
       shop_days     NUMBER(5)
       CONSTRAINT caln_pk PRIMARY KEY (clndr_dt)
  );
  
  INSERT INTO     caln
  VALUES (To_Date('01/01/1980','mm/dd/yyyy'),0);
  INSERT INTO     caln
  VALUES (To_Date('01/02/1980','mm/dd/yyyy'),1);
  INSERT INTO     caln
  VALUES (To_Date('01/03/1980','mm/dd/yyyy'),2);
  INSERT INTO     caln
  VALUES (To_Date('01/04/1980','mm/dd/yyyy'),3);
  INSERT INTO     caln
  VALUES (To_Date('01/05/1980','mm/dd/yyyy'),3);
  INSERT INTO     caln
  VALUES (To_Date('01/06/1980','mm/dd/yyyy'),3);
  INSERT INTO     caln
  VALUES (To_Date('01/07/1980','mm/dd/yyyy'),4);
  INSERT INTO     caln
  VALUES (To_Date('01/08/1980','mm/dd/yyyy'),5);
  INSERT INTO     caln
  VALUES (To_Date('01/09/1980','mm/dd/yyyy'),6);
  INSERT INTO     caln
  VALUES (To_Date('01/10/1980','mm/dd/yyyy'),7);
  INSERT INTO     caln
  VALUES (To_Date('01/11/1980','mm/dd/yyyy'),8);
  INSERT INTO     caln
  VALUES (To_Date('01/12/1980','mm/dd/yyyy'),8);
  INSERT INTO     caln
  VALUES (To_Date('01/13/1980','mm/dd/yyyy'),8);
  INSERT INTO     caln
  VALUES (To_Date('01/14/1980','mm/dd/yyyy'),9);

  The table includes since 01/01/1980 but 31/12/2015.
  
  I have written (and validated) this parameter query that performs the calculation of my working day (mday):

  SELECT     cal.clndr_dt
  FROM     CALN cal
  ,     (
       SELECT     cal.shop_days+:mdays     AS new_shop_days
       FROM     CALN cal
       WHERE     cal.clndr_dt     =:start_date
       ) a
  WHERE     cal.shop_days     = a.new_shop_days
  AND     ROWNUM          =1
  ORDER BY     cal.clndr_dt;

  Based on this request, I created the following function (and I have no idea if it works or if the syntax is right, etc..):

  CREATE OR REPLACE FUNCTION add_mdays 
       (start_date     IN DATE,
       mdays          IN NUMBER(5))
  RETURN     DATE 
  IS
       new_date DATE;
  BEGIN
  
       SELECT     cal.clndr_dt
       FROM     CALN cal
       ,     (
            SELECT     cal.shop_days+mdays     AS new_shop_days
            FROM     CALN cal
            WHERE     cal.clndr_dt     =start_date
            ) a
       WHERE     cal.shop_days     = a.new_shop_days
       AND     ROWNUM          =1
       ORDER BY     cal.clndr_dt;
  
       RETURN     new_date;
  
  END add_mdays;  //edit 9:31 AM - noticed I left off this bit

  I'm also not sure how to do to have the function handle results that would return a date outside the range of dates that appear in the table (prior to 01/01/1980 or after until 31/12/2015 - or, another way to look at what was, before the caln.clndr_dt or the caln.clndr_dt MAX value MIN value).
  
  My goal is to be able to use the function in a situation similar to the following:
  
  First of all, here is a sample table and data:

  CREATE TABLE orders
  (     ord_no          NUMBER(5),
       plan_start_dt     DATE,
       CONSTRAINT orders_pk PRIMARY KEY (ord_no)
  );
  
  INSERT INTO orders
  VALUES (1,To_Date('01/08/1980','mm/dd/yyyy'));
  INSERT INTO orders
  VALUES (2,To_Date('01/09/1980','mm/dd/yyyy'));
  INSERT INTO orders
  VALUES (3,To_Date('01/10/1980','mm/dd/yyyy'));

  And here's how I would use my function:

  SELECT     orders.ord_no
  ,     orders.plan_start_dt
  ,     add_mdays(orders.plan_start_dt, -3) AS prep_date
  FROM     orders

  Thus, the function would allow me to come back, for each command in my table of orders, the date is 3 days working (mdays) before the start of the plan of each order.
  
  I go about it the right way? I have to create a function to do this, or is there a way for me to integrate my request (which makes my mday calculation) in the example query above (eliminating the need to create a function)?
  
  Thank you very much in advance!
  
  Published by: user11033437 on February 2, 2010 08:55
  Fixed some typos in the last insert statements
  
  Published by: user11033437 on February 2, 2010 09:31 (fixed some syntax in the function)

  Hello

  Ah, referring to Oracle 8 and is not not able to test your own code makes me nostalgic for the good old days, when you have entered your cards and led to a window to the computer center and waited an hour for the work to be performed and then seen printing to find that you had made a typo.

  If you write functions, you should really test yourself. Like all codes, functions forge be written small not: write a line or two (or sometimes just a part of what would later become a single line), test, make sure it is running properly and repeat.
  Ideally, your employer must create a pattern of development in a development database that you can use.
  You can legally download your own instance of Oracle Express Edition free; just be careful not to use features that are not available in the database where the code will be deployed.

  You need a function to get the desired results:

  SELECT       o.ord_no
  ,       o.plan_start_dt
  ,       MIN (e.clndr_dt)     AS prep_date
  FROM       orders     o
  ,       caln          l
  ,       caln          e
  WHERE       l.clndr_dt     = o.plan_start_dt
  AND       e.shop_days     = l.shop_days - 3
  GROUP BY  o.ord_no
  ,            o.plan_start_dt
  ;
  

  It would be more effective (and somewhat simpler) If you've added a column (let's call it work_day) identified whether each line represents a work_day or not.
  For each value of shop_days, exactly 1 row will be considered as a working day.
  Then, the query may be something like:

  SELECT       o.ord_no
  ,       o.plan_start_dt
  ,       e.clndr_dt          AS prep_date
  FROM       orders     o
  ,       caln          l
  ,       caln          e
  WHERE       l.clndr_dt     = o.plan_start_dt
  AND       e.shop_days     = l.shop_days - 3
  AND       e.work_day     = 1
  ;
  

  You can use the analytic LAG function to populate the work_day column.

  A function would certainly be useful, although perhaps slower.

  The function you have posted has some errors:
  an argument can be stated under NUMBER (5); Just NUMBER.
  (b) when you SELECT in PL/SQL, as you do, you must SELECT a variable to store the results.
  (c) ROWNUM is arbitrary (making it useless in this problem) unless you draw a neat subquery. I don't think you can use ORDER BY in subqueries in Oracle 8. Use the ROW_NUMBER analytic function.
  (d) the service must end with an END statement.

  Given your current caln table, here's how I would write the function:

  CREATE OR REPLACE FUNCTION add_mdays
       ( start_date     IN           DATE          DEFAULT     SYSDATE,
         mdays          IN           NUMBER          DEFAULT     1
       )
  RETURN     DATE
  DETERMINISTIC
  IS
       --     add_mdays returns the DATE that is mdays working days
       --     after start_date.  (If mdays < 0, the DATE returned
       --     will be before start_date).
       --     Work days do not include Saturdays, Sundays or holidays
       --     as indicated in the caln table.
  
       new_date     DATE;          -- to be returned
  BEGIN
  
       SELECT     MIN (t.clndr_dt)
       INTO     new_date
       FROM     caln     f     -- f stands for "from"
       ,     caln     t     -- t stands for "to"
       WHERE     f.clndr_dt     = TRUNC (start_date)
       AND     t.shop_days     = f.shop_days + TRUNC (mdays)
       ;
  
       RETURN     new_date;
  END     add_mdays;
  /
  SHOW ERRORS
  

  Production code forge be robust (which includes "fool-proofing").
  Try to anticipate what people errors might appeal to your function and correct for them where possible.
  For example, if it only makes sense for start_date at midnight, mdays to be an integer, use TRUNC in the function where soembody passes a good value.
  Allow default arguments.
  Comment of your function. Put all comments within the service (i.e. after CREATION and before the END) so that they will remain in the data dictionary.
  If, given the same arguments, the function always returns the same value, mark it as DETERMINISTIC, for efficiency. This means that the system will remember the values transmitted rather than to call the function whenever it is said to.

  I wish I could score questions such as 'Correct' or 'useful '; you get 10 points for sure.
  You posted CREATE TABLE and INSERT statements (without even be begged).
  You gave a clear description of the problem, including the expected results.
  The code is well formatted and easy to read.
  All around, one of the more thoughtful and well written questions I've seen.
  Play well! Keep up the good work!

  Published by: Frank Kulash, February 2, 2010 13:10
  Added to my own version of the function.

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  When Jobs was alive, I do not remember Apple with Mickey$ oft problems like that.

  It just worked.

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  Hello
  
  I want to create a view of calendar in the apex is it possible? Maybe it uses Sql calendar.
  
  
  The weekly, monthly, and daily of Calenders account wizard.
  
  
  I'm under the Apex Version 2.0.0.00.49
  
  
  Thanks in advance,
  
  Huw.

  Hi Huw,

  You create a page template "Page Snippet" and apply it to the second page of schedule?

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  Not sure why FornCalc does not work, but the same logic works fine JavaScript.

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