calculate the month to date in Obiee11g

Hello
I use OBIEE11g, im a problem-

Problem: results of the beginning of this month beginning of previous month results vs.
example: If Im in 12 February the calculation is * February 1 to February 12 * also * January 1 to January 12 *.

-Zia

Can you give your month there is calculation please?

It is similar to the AGO (SH. Salesfacts. "" Quantity ", SH. TimesDim. ("" Month ", 1).

Tags: Business Intelligence

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I would like to ask for your help on how to calculate the months on growth rate previous/[(last month-previous month) month * 100%] in a query OBIEE. This report should always be calculated for the last 2 months available.

I have the following query:
Month0 | Months1. Month2
Produces a 500 | 100. 200
Product B 600 | 300. 150

I would like to add month on month column as follows:
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Produces a 500 | 100. 200 | + 100.00%
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Hello

The best way to solve this problem is by using the function Ago. With this, you can create a logical column for the previous month. Then you will have 2 columns available with which you can do your calculations.

If you want to do this with a calculated item (don't know if formatting will work for you), but you can do the more general calculation using $1 for the $2 for the 2 column and 1 column in your calculation. If ($x) columns will change with the columns in your report.

Concerning

Select records based on the monthly anniversary date
Hello

I have a table with a field of date_added and I want to select records from monthly birthday of this field.

for example. ID, Date_added
1, DECEMBER 10, 2012
2, 11 NOVEMBER 2012
3, 10 MARCH 2012
4, FEBRUARY 28, 2012
5, 30 DECEMBER 2012

So for January 10, 2013, I would like to return the documents 1 and 3 only

I started watching the extract function, but this little falls down to the records at the end of the month. For example, on 28 February, I also include records where 29, 30 or 31 is the day of date_added. So, in the above table, I won't return documents 4 and 5, but retrieves only returns 4.

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I use 11g

Thank you
user11259857 wrote:
Hi Al,.

No I don't think that there is a flaw. For example, the 30 Jan is not the last day of the month, but is June 30. Thus, on 30 January I want only to interrogate all the dates falling on the 30th - 28 February, after already being interrogated on the 28 Jan. However, on 30 June, I would query the 30th of each month, as well as the 31 January, March, may, July, Aug, Oct, dec. February is a special case, being both shorter and a leap year

Thank you

Richard

Hi Richard,

so, we can assume that your needs are:

If the day is the last day of that month, interview the day of the other months being > = day of the current month.

That is to say:

Jan-30 = query only 30 other months
Jan-31 = query only 31 of the other months

Feb-28 (non-leap years) = question 28, 29, 30 and 31 other months
29 February (leap year) = query 29,30,31 to other months

Mar-30 = query only 30 other months
Apr-30 = query 30 and 31 other months

This hypothesis is correct?

If so, then something like this should work:
```
WITH mydates(id, date_added) AS
(
   SELECT 1, TO_DATE('10-DEC-2012', 'DD-MON-YYYY') FROM DUAL UNION ALL
   SELECT 2, TO_DATE('11-NOV-2012', 'DD-MON-YYYY') FROM DUAL UNION ALL
   SELECT 3, TO_DATE('10-MAR-2012', 'DD-MON-YYYY') FROM DUAL UNION ALL
   SELECT 4, TO_DATE('28-FEB-2012', 'DD-MON-YYYY') FROM DUAL UNION ALL
   SELECT 5, TO_DATE('30-DEC-2012', 'DD-MON-YYYY') FROM DUAL
)
SELECT id, date_added
  FROM mydates
 WHERE (:input_date = LAST_DAY(:input_date) AND TO_CHAR(:input_date,'DD') <= TO_CHAR(date_added, 'DD'))
    OR TO_CHAR(:input_date,'DD') = TO_CHAR(date_added, 'DD');

with input_date = 28/02/2013

id   date_added
4    28/02/2012
5    30/12/2012
```
Kind regards.
Al

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Fix

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TIMESTAMPDIFF (sql_tsi_month, col1, col2)

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sum of the months of data first weeks
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January 3 09 2
:
:
:
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```
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Hello

Assuiming the due date is the same day of the month (if possible) as start_date:
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I found a Julian Georgian so far.

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Hi Roger, Etbin

Thanks for your replies.

I found the solution on the month to display between two dates.

To calculate that I used the loop as below

declare

ST DATE: =: date_start;--TO_DATE('24-MAY-2014','DD-MON-YYYY');

ED DATE: =: date_end;--TO_DATE('02-JUL-2014','DD-MON-YYYY');

number of v_t_amount;

CURSOR C1 IS

SELECT more GRAND (ST, ADD_MONTHS (TRUNC (st, 'MONTH'), (LEVEL - 1))) dt.

LESS (ed,

LAST_DAY (ADD_MONTHS (TRUNC (st, 'MONTH'), (LEVEL - 1)))

) ld

OF THE DOUBLE

CONNECT BY LEVEL<>

MONTHS_BETWEEN (TRUNC (ed, 'MONTH'), TRUNC (st, 'MONTH'))

+ 1

ORDER BY 1;

V_AMOUNT NUMBER: = 0;

V_T_AMOUNT NUMBER: = 0;

NUMBER OF V_DAYS;

BEGIN

FOR I IN C1 LOOP

V_AMOUNT: = (nvl(:SALARY_DIFF,0) + nvl(:SPECIAL_ALLOWANCE,0) + nvl(:CF_OTHERS,0)

+ NVL(:Computer,0) + nvl(:PHONE_ALLOW,0) + nvl(:WORK_NATURAL,0)

(+ nvl(:TRANSPORTATION,0) * (i.1d - i.dt + 1) / TO_CHAR (LAST_DAY (i.1d), 'dd'));

V_T_AMOUNT: = V_AMOUNT;

end loop;

end if;

dbms_output.put_line (v_t_amount);

end;

To calculate the amount on the basis of the date of

Hello
We must consider the functionality of oracle when it sees in 365 days a year. (for leap years 366 days)
calendar year of 365 days. calendar year of 12 months. Months will track the number of calendar days in the month. For example the Feb has 28 days and Oct. has 31 days.
Example:
For the period from 27/06/12 to 20/08/12
Annual amount of $3214.
Calculation
$3214 / 12 months = $267,83 per month
For June, $267,83 / 30 days in a month = $8.93 daily rate
6/27-6/30 is 4 days of service. x 4 = $35,72 $8.93
7/1-7/31 is considered to be a full month. = $267,83
For August, 8/1 /-8/20 is considered to be 20 days. Per day = 267.83 / 31 = 8.64
20 days = $172.79
TOTAL = $476,34
Please help me on how to do the following calculations.
Concerning
Suresh

Another solution, works better than the previous:

with the data as

(

Select to_date (June 27, 2013 ',' DD/MM/YYYY ') from_dt, to_date (August 20, 2013 ',' DD/MM/YYYY ') to_dt, amt double 3214

)

Select sum (DM)

de)

Select from_dt, to_dt, start_dt, last_dt,

(last_dt - start_dt + 1) *--> multiply applicable days of the month at the daily amount, calculated below

Sum)

round)

(amt/12) /--> calculate a monthly amount

TO_CHAR)

LAST_DAY (curr_dt)

, 'DD '.

)--> Find last day of the month and divide the monthly amount with the last day and calculate the daily amount

2

)--> Around the daily amount to 2 decimal places

) on sm--> sum the amount daily (partition to_char (curr_dt, 'MM'))

de)

Select from_dt, to_dt, TN,

less (add_months (from_dt, level - 1), to_dt) curr_dt,.

decode (level 1, less (add_months (from_dt, level - 1), to_dt), trunc (less (add_months (from_dt, level - 1), to_dt), 'MM')) start_dt;

less (last_day (add_months (from_Dt, level - 1)), to_dt) last_dt

from the data

connect by level<= months_between(trunc(to_dt,="" 'mm'),="" trunc(from_dt,="" 'mm'))="" +="">

)

);

SUM (DM)

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

476.36

This avoids looping of data for all of the Dates if opposed to a loop for each month, which increases its performance. See plans to explain to the two queries:

explain plan for

with the data as

(

Select to_date (June 27, 2013 ',' DD/MM/YYYY ') from_dt, to_date (August 20, 2013 ',' DD/MM/YYYY ') to_dt, amt double 3214

),

div_data as

(

Select from_dt, to_dt, annual_amount from the amt Tower (amt/12, 2) monthly_amount

from the data

)

Select sum (per_day_amt)

de)

Select to_char (level - 1, 'Lun' + from_dt) curr_dt,.

sum (round (monthly_amount/to_number (substr (to_char (last_day (level - 1 + from_dt), 'DDMMYYYY'), 1, 2)), 2)) per_day_amt

of div_data

connect by level<= (to_dt="" -="" from_dt)="" +="">

To_char Group (level - 1, 'Lun' + from_dt)

);

Hash value of plan: 1042012692

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

| ID | Operation | Name                        | Lines | Bytes | Cost (% CPU). Time |

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

|   0 | SELECT STATEMENT |                             |     1.    13.     5 (20) | 00:00:01 |

|   1. TRANSFORMATION OF THE TEMPORARY TABLE.                             |       |       |            |          |

|   2.   LOAD SELECT ACE |                             |       |       |            |          |

|   3.    QUICK DOUBLE |                             |     1.       |     2 (0) | 00:00:01 |

|   4.   GLOBAL TRI |                             |     1.    13.            |          |

|   5.    VIEW                          |                             |     1.    13.     3 (34) | 00:00:01 |

|   6.     HASH GROUP BY.                             |     1.    16.     3 (34) | 00:00:01 |

|* 7 |      CONNECT TO WITHOUT FILTERING.                             |       |       |            |          |

|   8.       COUNT                      |                             |       |       |            |          |

|   9.        VIEW                      |                             |     1.    16.     2 (0) | 00:00:01 |

| 10.         TABLE ACCESS FULL | SYS_TEMP_0FD9D86CF_5F8521E2 |     1.    44.     2 (0) | 00:00:01 |

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

Information of predicates (identified by the operation identity card):

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

7 filter (LEVEL<>

explain plan for

with the data as

(

Select to_date (June 27, 2013 ',' DD/MM/YYYY ') from_dt, to_date (August 20, 2013 ',' DD/MM/YYYY ') to_dt, amt double 3214

)

Select sum (DM)

de)

Select from_dt, to_dt, start_dt, last_dt,

(last_dt-start_dt + 1) * sum (round ((amt/12)/to_char (last_day (curr_dt), 'JJ') 2)), in the sm (partition to_char (curr_dt, 'MM'))

de)

Select from_dt, to_dt, TN,

less (add_months (from_dt, level - 1), to_dt) curr_dt,.

decode (level 1, less (add_months (from_dt, level - 1), to_dt), trunc (less (add_months (from_dt, level - 1), to_dt), 'MM')) start_dt;

less (last_day (add_months (from_Dt, level - 1)), to_dt) last_dt

from the data

connect by level<= months_between(trunc(to_dt,="" 'mm'),="" trunc(from_dt,="" 'mm'))="" +="">

)

);

Hash value of plan: 1669308594

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

| ID | Operation | Name | Lines | Bytes | Cost (% CPU). Time |

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

|   0 | SELECT STATEMENT |      |     1.    13.     3 (34) | 00:00:01 |

|   1. GLOBAL TRI |      |     1.    13.            |          |

|   2.   VIEW                           |      |     1.    13.     3 (34) | 00:00:01 |

|   3.    KIND OF WINDOW.      |     1.    30.     3 (34) | 00:00:01 |

|   4.     VIEW                         |      |     1.    30.     2 (0) | 00:00:01 |

|* 5 |      CONNECT TO WITHOUT FILTERING.      |       |       |            |          |

|   6.       QUICK DOUBLE |      |     1.       |     2 (0) | 00:00:01 |

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

Information of predicates (identified by the operation identity card):

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

5 filter (LEVEL<>

Post edited by: PurveshK
Updated with explanation.

calculate the month to date in Obiee11g

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