How to get weekly data in OBIEE

I had a requirement to extract the data each week for 10 weeks, please let me know how to apply this logic in RPD.

Am using OBIEE 11 G

Thank you

Nadia.

Hello Nadia,.

Try to use the Pivot table view, it will regroup by the measure.

What happened to your last thread? Is it resolved?

Thank you

Sasi nanou...

Tags: Business Intelligence

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Hi, I am using the DateTimePicker.

There are several question about this command:

1. how to get the Date, month and year?

-With the help of myDateTime.value, I get this string ' Wed May 08 14:45 ICT 2013 '-> I have to manually analyze this? Because I need in format yyyy/mm/dd. Or can set the format of dateTimePicker.value?

2 format on dateTimePicker self (on the user interface) is d/m/YY, I can't change to another format?

Thank you

dateTimePicker.value return a QDateTime not a string.

https://developer.BlackBerry.com/Cascades/reference/bb__cascades__datetimepicker.html#property-value

See more:

https://developer.BlackBerry.com/Cascades/reference/QDateTime.html#date

https://developer.BlackBerry.com/Cascades/reference/QDate.html

You can get
```
day = dateTimePicker.value.date().day()
month = dateTimePicker.value.date().month()
year= dateTimePicker.value.date().year()
dateYYYYMMDD = dateTimePicker.value.date().toString(Qt::ISODate)
dateYYYYMMDD2 = dateTimePicker.value.date().toString("YYYY/MM/DD")
dateYYYYMMDD3 = dateTimePicker.value.toString("YYYY/MM/DD")
```

How to get the date for the first Monday of each month

Dear members,

How to get the date for the first Monday of each month.

I wrote the following code

SELECT decode (to_char (trunc (sysdate + 30, 'MM'), 'DAY'), 'MONDAY', trunc (sysdate + 30, 'MM'), NEXT_DAY (trunc (sysdate + 30, 'MM'), "LUN")) FROM DUAL

But he looks at complex bith.

Abhishek

Published by: 9999999 on 8 March 2013 04:30

Use the IW format - it will make independent NLS solution. And all you need is truncated 7^th day of each month using IW:

select  sysdate current_date,
        trunc(trunc(sysdate,'mm') + 6,'iw') first_monday_the_month
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/

CURRENT_D FIRST_MON
--------- ---------
08-MAR-13 04-MAR-13

SQL>

Here is the list of the first Monday of the month of this year:

with t as(
          select  add_months(date '2013-1-1',level-1) dt
            from  dual
            connect by level <= 12
         )
select  dt first_of_the_month,
        trunc(dt + 6,'iw') first_monday_the_month
  from  t
/

FIRST_OF_ FIRST_MON
--------- ---------
01-JAN-13 07-JAN-13
01-FEB-13 04-FEB-13
01-MAR-13 04-MAR-13
01-APR-13 01-APR-13
01-MAY-13 06-MAY-13
01-JUN-13 03-JUN-13
01-JUL-13 01-JUL-13
01-AUG-13 05-AUG-13
01-SEP-13 02-SEP-13
01-OCT-13 07-OCT-13
01-NOV-13 04-NOV-13

FIRST_OF_ FIRST_MON
--------- ---------
01-DEC-13 02-DEC-13

12 rows selected.

SQL>

SY.

How to get the date - 7 days (week)
Hi all

I am beginner in Java and JSP

I'm JSP search form (VO) where I need to set a default value for the start_date - 7 days

OK I get this display today date but how to get that displays a date-7 days?

< %
DateFormat df = new SimpleDateFormat ("JJ. MM.yyyy");
Date date = new Date();
String default = 'StartDate =' + df.format (new Date());
% >

ID
Thank you
Simple approach if you're not worried about leap years: -.
```
Date todayMinus7 = new Date(System.currentTimeMillis() - 7 * 24 * 60 * 60 * 1000);
```
Alternatively take a look at the GregorianCalendar class and the add() method.

How to get the date of Sunday of the week
Hello
I have a column date 'start_date', I need to create a report for all the weeks from Sunday, for example:

I need to get the dates of the week in the form:

mm/dd/aa-23/01/11, 01/16/11, 09/01/11 and so on...

Help, please

Thank you...
Hello

To find the last Sunday before or equal to start_date:
```
SELECT     TRUNC ( start_date + 1
           , 'IW'
           ) - 1          AS week_start_date
...
```
You can also use an expression like that in a GROUP BY clause.

It's not you NLS parameters. Always ISO weeks start on Monday. Your week begins 1 day earlier, that is why the above expression is the + 1 and -1.
(According to the parameters of the NLS)
```
SELECT     TRUNC ( start_date
           , 'D'
           )          AS week_start_date
...
```
could do the same thing, but I still have rocommend with "IW"; a few extra touches is not big thing to pay for a guarantee that it will still work.)

I hope that answers your question.
If not, post a small example of data (CREATE TABLE and INSERT statements) and the results desired from these data.

How to get the date of end of week max and its rows of a table

Hello

Table [temp2]
 
   id          name        dt
  123        a             2-mar-2010
  124        b            1-feb-2010
  125        c             3-apr-2010
  123        a             13-mar-2010
  125        c             13-mar-2010
  123        a             12-feb-2010

This table how to get this line by id - name (id and the name of the composite key) whose date is the date of last weekend (last Saturday).
for example, suppose to have IDs 123 values date February 12, 2010, 13-mar-2010 this date falls on the last Saturday 13-mar-2010, how to get the value of the line on Saturday last for all the unique id and the name?

So I wrote this but probably there a few traps...

select id,name,dt from (select id,name,dt,max(dt) over(partition by id,name) as 
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,'27-mar-2010','20-mar-2010','13-mar-2010')

Here I have spent in the in clause 4 last weekend dates manually (although either has been calculated by programming) in front.
Thank you

Here's another option:

SELECT     *
FROM
(
     SELECT     ID
     ,     NAME
     ,     MAX(CASE WHEN TO_CHAR(DT,'DY','NLS_DATE_LANGUAGE=ENGLISH') = 'SAT' THEN DT END) AS DT
     FROM     TEMP2
     GROUP BY ID, NAME
)
WHERE DT IS NOT NULL

It is always useful to provide the following information:

1. oracle version (SELECT * FROM V$ VERSION)
2. examples of data in the form to CREATE / INSERT commands.
3. expected results
4 explanation of the expected results (alias "business logic")
5. use.

 tags for #2 and #3. See FAQ (Link on top right side) for details.

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var length = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
var days = ["Sun", "Mon", "Tue", "Wed", "Game", "Fri", "Sam"];
Count = 0;
day = 0;
months = 0;
week = 5;
Str = "";
While (count < 365)
{
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day = 0;
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}
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Tanks!

Try this,

app.activeDocument.viewPreferences.rulerOrigin = RulerOrigin.pageOrigin;

app.activeDocument.zeroPoint = [0,0];

{with (App.activeDocument.viewPreferences)}

horizontalMeasurementUnits = MeasurementUnits.points;

verticalMeasurementUnits = MeasurementUnits.points;

rulerOrigin = RulerOrigin.pageOrigin;

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day = 0;

months = 0;

week = 5;

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{

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}

for (i = 0; i<>

{

var myTextframe = document.pages [i].textFrames.add ({geometricBounds:[0,0,50,150]});})

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Vandy

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Hello

Please help me how to get the day of the week using another date field.

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Thanks in advance!

Kind regards
Florian...

Published by: Florian on October 29, 2010 03:07
Maybe it's
```
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<>http://www.NI.com/LVData">
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Shankar

Hi, Shankar,

Sorry, we don't know what you want.

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;

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----

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1-01 - 7.-10 1er January 11

1 February 1, 11 1 July 11
"
"
"
o/p:

1 1 January 10 01 - aug - 10

1 02-aug-10 (first_date + 1) 31-aug-10(second_date-1) - dates of discontinuity

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1 January 2, 11 January 31, 11

1 February 1, 11 1 July 11

'
'
'
Please let me know is possible

SQL> select  *
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        ID FIRST_DAT SECOND_DA
---------- --------- ---------
         1 01-JAN-10 01-AUG-10
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         1 01-FEB-11 01-JUL-11

SQL> with t1 as (
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 15             )
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 19             first_date
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        ID FIRST_DAT SECOND_DA
---------- --------- ---------
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         1 02-AUG-10 31-AUG-10
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         1 02-JAN-11 31-JAN-11
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How to get weekly data in OBIEE

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