Data per day, per month, per year

Hello, there is a field in my table that includes dates.

I counted my recordings of data per day using the aggregator operator and

This simple select DATE, COUNT (*) FROM TABLE GROUP BY DATE

Can you tell me how I count my files per month or per year? Thank you.

Hello

You can use functions to retrieve information from the date and use it, for example;

Select to_char (hiredate, 'YYYY') there, sum (sal) from emp group by to_char (hiredate, 'YYYY');

So in OWB if you have an expression before the aggregation operator, you can define the expressions and use them in the aggregation.
See you soon
David

Tags: Business Intelligence

Similar Questions

Date and days for the entire year come in the column
Hi all

I have 2 column 1 is the Date and day 2

And I have a button, I need when I press enter then his show all the year date in the date and day column in the column of day for example

Date - day+.
* 1 - jan - 2009 - Thursday *.
* 2 - jan - 2009 - Friday *.
* 3 - jan - 2009 - Saturday *.

And like on

I hope that you guys understand

Concerning

Wasim Ismail
Hai,

This means, you must fill in the block record multi that has 2 columns (1 is the date, and the other is updated.).

For this try this in the trigger WHEN BUTTON PRESSED the button.
```
DECLARE
     Dt_Date DATE;
BEGIN
     GO_BLOCK ('');
     CLEAR_BLOCK(NO_VALIDATE);
     FIRST_RECORD;
     Dt_Date := TO_DATE('01-JAN-2009');
     LOOP
          :.TI_DATE := TO_CHAR(Dt_Date, 'DD-MON-YYYY');
          :.TI_DAY  := TO_CHAR(Dt_Date, 'DAY');
          EXIT WHEN TO_CHAR(Dt_Date, 'YYYY') <> TO_CHAR(Dt_Date + 1, 'YYYY');   -- If next year reaches, exit from the loop.
          Dt_Date := Dt_Date + 1;
          NEXT_RECORD;
     END LOOP;
END;          
```
Kind regards

Manu.

If this answer is useful or appropriate, please mark. Thank you.

How to select dates per year?
Hello

Let's say I have a table (itemID, datePurchase, itemPrice)
and I want to list the amount of itemPrice according to the year of datePurchase.
for example. I want that the total amount spent during the year 2007

DATEPURCHASE ITEMPRICE
------------ ----------------
4 OCTOBER 07 300
7 NOVEMBER 07 250
17 JANUARY 08 125
16 NOVEMBER 08 250
27 NOVEMBER 08 250
17 DECEMBER 09 200

How do I filter the date of the year?

Kind regards
Keith
Hey, Keith,

963214 wrote:
... I want to list each year. Is there a better way to do it, other than the awkward example below?

Yes, there is: GROUP BY:
```
SELECT    TO_CHAR ( TRUNC (datepurchase, 'YEAR')
              , 'YYYY'
            )          AS year
,       SUM (itemprice)       AS total
FROM       table_x
GROUP BY  TRUNC (datepurchase, 'YEAR')
ORDER BY  year
;
```
In any case I use Oracle SQL Developer (3.2.09)

This is the version of your front end . It may be useful to know that, but it is much more important to know what your version of the database . If you are not sure, he'll tell you:
```
SELECT  *
FROM    v$version;
```
Who appear about 5 lines of commands, for example
```
BANNER
--------------------------------------------------------------------------------
Oracle Database 11g Enterprise Edition Release 11.2.0.1.0 - Production
PL/SQL Release 11.2.0.1.0 - Production
CORE    11.2.0.1.0      Production
TNS for 32-bit Windows: Version 11.2.0.1.0 - Production
NLSRTL Version 11.2.0.1.0 - Production
```
The important part is the bunch of numbers towards the end of the first line, 11.2.0.1.0 in this example.

Get start and end date of week by month and year

How I need the week is here - a week always begins on a Sunday so from February, it will be

begin the week of end of week

12-02-1 02/12/04
02/12/05 02/12/11
02/12/12-02/12/18
02/12/19 02/12/25
02/12/26-29/12/29

So, because the first week does not start on a Sunday he's leaving

1 (Wednesday) to 4th (saturday) just like a calendar.

Thanks in advance

If far from Solomons help I have this

This is for January
WITH t AS)
SELECT TRUNC (TRUNC (to_date('01-01-12','MM-DD-YY'), 'YYYY')-1, 'W') + (LEVEL-1) * 7 week_start_date
OF THE DOUBLE
CONNECT BY LEVEL < = 53
)
SELECT week_start_date,
LEAST (LAST_DAY (to_date('01-01-12','MM-DD-YY')), week_start_date + 6) week_end_date
T
WHERE week_start_date > = TRUNC (to_date('01-01-12','MM-DD-YY'), 'MM')
AND week_start_date < ADD_MONTHS (TRUNC (to_date('01-01-12','MM-DD-YY'), 'MM'), 1)

begin the week of end of week
01/12/02-01/12/08
12/01/09-01/12/15
12/01/16-01/12/22
01/12/23-01/12/29
01/12/30-01/12/31

If satrts Sunday of the week:

WITH t AS (
           SELECT TRUNC(TRUNC(&dt,'MM') + 1,'IW') - 1 + (LEVEL - 1) * 7 week_start_date
            FROM  DUAL
            CONNECT BY TRUNC(TRUNC(&dt,'MM') + 1,'IW') - 1 + (LEVEL - 1) * 7 <= LAST_DAY(&dt)
          )
SELECT  GREATEST(week_start_date,TRUNC(&dt,'MM')) week_start_date,
        LEAST(LAST_DAY(&dt),week_start_date + 6) week_end_date
  FROM  t
  WHERE week_start_date <= LAST_DAY(&dt)
/

For example:

SQL> SET VERIFY OFF
SQL> DEFINE dt="DATE '2012-01-17'" -- January
SQL> WITH t AS (
  2             SELECT TRUNC(TRUNC(&dt,'MM') + 1,'IW') - 1 + (LEVEL - 1) * 7 week_start_date
  3              FROM  DUAL
  4              CONNECT BY TRUNC(TRUNC(&dt,'MM') + 1,'IW') - 1 + (LEVEL - 1) * 7 <= LAST_DAY(&dt)
  5            )
  6  SELECT  GREATEST(week_start_date,TRUNC(&dt,'MM')) week_start_date,
  7          LEAST(LAST_DAY(&dt),week_start_date + 6) week_end_date
  8    FROM  t
  9    WHERE week_start_date <= LAST_DAY(&dt)
 10  /

WEEK_STAR WEEK_END_
--------- ---------
01-JAN-12 07-JAN-12
08-JAN-12 14-JAN-12
15-JAN-12 21-JAN-12
22-JAN-12 28-JAN-12
29-JAN-12 31-JAN-12

SQL> SET VERIFY OFF
SQL> DEFINE dt="DATE '2012-02-25'" -- February
SQL> WITH t AS (
  2             SELECT TRUNC(TRUNC(&dt,'MM') + 1,'IW') - 1 + (LEVEL - 1) * 7 week_start_date
  3              FROM  DUAL
  4              CONNECT BY TRUNC(TRUNC(&dt,'MM') + 1,'IW') - 1 + (LEVEL - 1) * 7 <= LAST_DAY(&dt)
  5            )
  6  SELECT  GREATEST(week_start_date,TRUNC(&dt,'MM')) week_start_date,
  7          LEAST(LAST_DAY(&dt),week_start_date + 6) week_end_date
  8    FROM  t
  9    WHERE week_start_date <= LAST_DAY(&dt)
 10  /

WEEK_STAR WEEK_END_
--------- ---------
01-FEB-12 04-FEB-12
05-FEB-12 11-FEB-12
12-FEB-12 18-FEB-12
19-FEB-12 25-FEB-12
26-FEB-12 29-FEB-12

SQL> SET VERIFY OFF
SQL> DEFINE dt="DATE '2012-03-07'" -- March
SQL> WITH t AS (
  2             SELECT TRUNC(TRUNC(&dt,'MM') + 1,'IW') - 1 + (LEVEL - 1) * 7 week_start_date
  3              FROM  DUAL
  4              CONNECT BY TRUNC(TRUNC(&dt,'MM') + 1,'IW') - 1 + (LEVEL - 1) * 7 <= LAST_DAY(&dt)
  5            )
  6  SELECT  GREATEST(week_start_date,TRUNC(&dt,'MM')) week_start_date,
  7          LEAST(LAST_DAY(&dt),week_start_date + 6) week_end_date
  8    FROM  t
  9    WHERE week_start_date <= LAST_DAY(&dt)
 10  /

WEEK_STAR WEEK_END_
--------- ---------
01-MAR-12 03-MAR-12
04-MAR-12 10-MAR-12
11-MAR-12 17-MAR-12
18-MAR-12 24-MAR-12
25-MAR-12 31-MAR-12

SQL>

SY.

Published by: Solomon Yakobson, January 12, 2012 13:19

Is it $167 per month and $1999 per year? It seems overpriced. I suspect that Adobe refers to a dollar and sixty-seven cents per month. Until it is clearly written, I've maybe wrong. Where can I get clear unequivocal negative cost.

Is it $167 per month and $1999 per year? It seems overpriced. I suspect that Adobe refers to a dollar and sixty-seven cents per month. Until it is clearly written, I've maybe wrong. Where can I get clear clear of the cost?

Adobe Creative Cloud prices are here

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No shows the price you quoted so I wonder what page you read exactly?

Is it true that you can use Creative cloud €19.66 per month each year when you are a teacher or a student?

Is it true that you can use Creative cloud €19.66 per month each year when you are a teacher or a student?
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Which is correct?

Hello

Promotional price is valid for the first year, then you renew at the price standard student/teacher edition.

Kind regards

Bev

Acrobat Professional DC membership on an annual basis with ONE payment per year?

I would like to subscribe to Acrobat Professional DC on an annual basis with ONE payment per year (same as the creative cloud subscription, I already have) and not on the monthly fee. Is it possible, and where can I find the link to download?
Thanks for help, I've been browsing for this already too long.

Hi ReguBee,

The available option is to Acrobat Pro Dc subscription match monthly Plans and prices | Adobe Acrobat DC. There is no one time annual payment option.

Kind regards
Nicos

I had Adobe Cloud (all applications) to 32.98 USD more than two years now, I have wanted to renew but new contract price is $79.98 - I only use Dreamweaver maybe three times a year, maybe pro adobe acrobat 4 times per year and photoshop maybe 2 times. Is

I had Adobe Cloud (all applications) to 32.98 USD more than two years now, I have wanted to renew but new contract price is $79.98 - I only use Dreamweaver maybe three times a year, maybe pro adobe acrobat 4 times per year and photoshop maybe 2 times. Is there a plan months I can go that is close to the $32 per month?

Hi Justin,

The full price creative Cloud for individuals is $49.99 / month.

If you add the broth, it's when you pay $79.98 / month.

Please visit:-Creative Cloud price and membership plans | Adobe Creative Cloud

For support:-contact us | Adobe

split with day, month and year in a date

Hi all
I have this point of view
with (id, dt) t (select 1, to_date('02/06/2010','MM/DD/YYYY') of all the double union)
Select 2, to_date('11/29/2001','MM/DD/YYYY') of all the double union
Select 3, to_date('02/06/2011','MM/DD/YYYY') of all the double union
Select option 4, double to_date('10/10/2011','MM/DD/YYYY')
)
--
-end of test data
--
SELECT id, dt
floor (months_between(sysdate,dt)/12) years
, floor (mod (months_between (sysdate, dt), 12)) months
-The day is ambiguous because of the number of days in a different month
-you could design your own algorithm to give the desired results, but it is a good approximation
, floor (sysdate-(add_months (dt, floor (months_between (sysdate, dt))) as dys
t
/
ID DT YEARS MONTHS DYS
---------- ----------- ---------- ---------- ----------
1-6 FEBRUARY 2010 5 8 22
2 29 NOVEMBER 2001 13 10 29
3-6 FEBRUARY 2011 4 8 22
4-10 OCTOBER 2011 4 0 18
As a result, 26 years old, 26 months, 91 days.

And subtract years, MONTHS and DYS as format
erase years, from 1 to 12 months and the days of the 1 to 30 as
91 days = 30 x 3 + 1 = > 1 day over 3 months
26 months + 3 months = 29 months = > 24 months + 5 months = 2 years + 5 months
and 26 years + 2 years = 28
desire to result: 28 years, 5 months and 1 day.
Is there a function that give me this result from the query above?
Kind regards
Gordan

Hello Gordan,.

I understand that you are the SUM of all the lines.

Here are two ideas:

-1 - using a date
(a) SUM (TRUNC (sysdate) - dt)
I would give you the total number of days.
(b) adding the number of days to, for example, DATE ' 1900-01-01' would give another date...
(c) simply subtract the years 1900 and it takes several years,
subtract 1 from the month (1-12 from January to December) and you have the number of months
subtract 1 from the day of the month (1-31) and you have the number of days.
It is of course 'approximate' as the months do not have the same number of days...

-2 - using only the number of days
Maybe you want something else, for example: the total number of days divided by 365.25 for many years, the recall divided by 30 to get the number of months, the reminder being the number of days.

(a) SUM (TRUNC (sysdate) - dt)

I would give you the total number of days.

(b) Division by 365.25 (le.25 is to take leap years into account, but you can choose for example "365")

(c) number of reminder of days (total - years * 365.25): divided by 30 is the number of months

(d) number reminder of days (total - years * 365.25 - months * 30): gives the number of days.

The following two options with your test data (by chance, this gives the same result in this case)

option 1:
with (id, dt) t (select 1, to_date('02/06/2010','MM/DD/YYYY') of all the double union)
Select 2, to_date('11/29/2001','MM/DD/YYYY') of all the double union
Select 3, to_date('02/06/2011','MM/DD/YYYY') of all the double union
Select option 4, double to_date('10/10/2011','MM/DD/YYYY')
)
target_date AS
(SELECT DATE ' 1900-01-01' + SUM (TRUNC (sysdate) - t.dt) FROM t trgt)
in_pieces AS
(SELECT TO_NUMBER (TO_CHAR (td.trgt, 'YYYY')) - 1900 y
, TO_NUMBER (TO_CHAR (td.trgt, 'MM')) - 1 m
, TO_NUMBER (TO_CHAR (td.trgt, 'DD')) - 1 d
Target_date TD
)
SELECT "result: ' |"
TRIM (CASE WHEN ip.y = 0 THEN NULL
WHEN ip.y = 1 THEN "1 year"
Of OTHER TO_CHAR (ip.y) | "years".
END |
CASE WHEN ip.m = 0 THEN NULL
WHEN ip.m = 1 THEN "1 month"
Of OTHER TO_CHAR (ip.m) | 'months '.
END |
CASE WHEN ip.d = 0 THEN NULL
WHEN ip.d = 1 THEN "1 day"
Of OTHER TO_CHAR (ip.d) | 'days '.
END
)
Of in_pieces ip
;
result: 28 years, 4 months and 28 days

option 2:

with (id, dt) t (select 1, to_date('02/06/2010','MM/DD/YYYY') of all the double union)
Select 2, to_date('11/29/2001','MM/DD/YYYY') of all the double union
Select 3, to_date('02/06/2011','MM/DD/YYYY') of all the double union
Select option 4, double to_date('10/10/2011','MM/DD/YYYY')
)
nb_days AS
(SELECT SUM (TRUNC (sysdate) - t.dt) n t)
, y AS
(SELECT Nb.n, FLOOR (nb.n / 365.25) y nb_days n. b.)
ym AS
(SELECT y.n, y.y, FLOOR ((y.n-y.y * 365,25) / 30) FROM a)
ymd AS
(SELECT ym.y, ym.m, ym.n - ym.y * 365.25 - ym.m * ym FROM 30 d)
SELECT "result: ' |"
TRIM (CASE WHEN ymd.y = 0 THEN NULL
WHEN ymd.y = 1 THEN "1 year"
Of OTHER TO_CHAR (ymd.y) | "years".
END |
CASE WHEN ymd.m = 0 THEN NULL
WHEN ymd.m = 1 THEN "1 month"
Of OTHER TO_CHAR (ymd.m) | 'months '.
END |
CASE WHEN ymd.d = 0 THEN NULL
WHEN ymd.d = 1 THEN "1 day"
Of OTHER TO_CHAR (ymd.d) | 'days '.
END
)
WAN
;
result: 28 years, 4 months and 28 days

Best regards

Bruno Vroman.

You can buy JUST Adobe Stock for three users in a company? The guy I chatted with that said I had to pay $199.99 per year over $ 69.99 per user. Is this correct?

You can buy JUST Adobe Stock for three users in a company? The guy I chatted with that said I had to pay $199.99 per year over $ 69.99 per user. Is this correct?
When I go on the website of plan, it says 199.99 per license as well. Do we buy 3 licenses then 3 plans to 69.99 each apps? It's very confusing.

Well I advise you to contact Adobe sales for the same because they can help you with the correct plan according to your needs.

Contact us | Adobe

I want to cancel my subscription (per year) of export Adobe PDF. How does it work? Hi R.Lauterbach

I want to cancel my subscription (per year) of export Adobe PDF.
How does it work?
greeting
R.Lauterbach

Cancel your subscription or membership. Creative cloud

https://helpx.Adobe.com/x-productkb/policy-pricing/cancel-membership-subscription.html

Set the calendar to display day, month and year

In addition to the time of day, my computer began giving the day of the week (Monday) only instead of day, month and year as it did before. Tried to go back to 29/12/2014 or something similar without success. Any suggestions?

Hello

Sorry for the late reply and I appreciate your patience.

User profile corruption might have caused the problem. As a solution, I suggest to create a new user profile and check if the date format displays the same thing.

To do this, see the link below.

http://Windows.Microsoft.com/en-us/Windows/fix-corrupted-user-profile#1TC=Windows-7.

It will be useful. Just answer us with the State, and we will be happy to help you.

Get the number of days in a month based on the month and year of fields

I have a column in my form which lists the days in a month. I want to configure a hidden field that calculates the total number of days in a month, based on the month and year of the field inputs. The number of days will determine what appears on the column. For example, if I put 4 months, and 2016 in the field of the year, I get 30 in the hidden field. Thus, on the column 'Day', I'll have numbers 1-30. Or if I put 2 months and 2016 in the field of the year, I get the 29 in the hidden field. If the numbers 1-29 appears in the column 'day '.
Found this on some forum javascript code:
//Month is 1 based function daysInMonth(month,year) { return new Date(year, month, 0).getDate(); } //July daysInMonth(7,2009); //31 //February daysInMonth(2,2009); //28 daysInMonth(2,2008); //29
I do not know how to convert this code in JavaScript to adobe and don't really know how to use it. All I know how to do is to configure the field values for the field month and year as variables. I am a novice programmer and would appreciate it really all the help I can get. Thank you in advance!

The code seems to be JavaScript and runs as needed by using the JavaScript console.

I would like to consider making more general code, so if you have a date string that includes at least the month and year we could just call the function and get the number of days for that month.

The following script will calculate the number of days in a month, by using at least the month and year values can display the result on the JavaScript console and all of the value field for the field that has this code as the custom calculation Script.

function daysInMonth (oDate) {}
return new Date (oDate.getFullYear (), oDate.getMonth () + 1, 0) .getDate ();
}

nMonth var = this.getField("Month").valueAsString; get the value of month;
nYear var = this.getField("Year").valueAsString; get the value of the year;

Event.Value = "";

If (nMonth! = "" & nYear!) = "") {}
var MyDate = util.scand ("' / mm/yyyy ', nMonth +" / "+ nYear); convert to date object;
var nDaysInMonth = daysInMonth (MyDate); get the number of days;

Console.Open (); Open the JavaScript console;

Console.clear(); clear the console;

Console.println ("Days in" + nMonth + ":" + nDaysInMonth); show days in month;

Event.Value = nDaysInMonth; Set the value of the field;

}

How to get the days and months when two dates are given
Hi all

I have a requirement where I need to return the number of months and days between two given dates.
I have no need to take account of the year as the difference in dates is always less than a year

I have 28 June 2012 and January 31, 2013, as dates.

I'm working on 10g

When I select the double months_between(:a,:b) and provide the dates he gives me 7.09677
I want to give only 7 months and 9 days, or what ever the exact days are.

Your contributions are appreciated
Try this:
```
WITH t AS
(SELECT TO_DATE('28-JUN-2012', 'DD-MON-YYYY') date1, TO_DATE('31-JAN-2013', 'DD-MON-YYYY') date2 FROM dual)
SELECT TRUNC(MONTHS_BETWEEN(date2, date1)) months,
       date2 - ADD_MONTHS(date1, TRUNC(MONTHS_BETWEEN(date2, date1))) days
FROM   t
```

problem to insert only month and year instead of the full date
```
select b.penjara_id, p.penj_lokasi, a.no_daftar, b.episod, b.nama1,to_char(a.trkh_mula_prl,'dd/mm/yyyy') as trkh_mula_prl, to_char(bulan_proses,'mm/yyyy') as bulan_proses,
        b.epd, b.lpd
from prl_daftar_proses a, senarai_pesalah b, penjara p

where a.no_daftar=b.no_daftar
and a.episod=b.episod
and b.penjara_id = p.penjara_id
and a.setuju_jplp is null
and a.bulan_proses between to_date(:FROM,'dd/mm/yyyy') and to_date(:TO,'dd/mm/yyyy')


order by b.penjara_id, a.bulan_proses,a.no_daftar
```
Hi, can someone help me how I can register only the month and year of the value which have full day in the database?

for example, the date is 18/09/2012, but I want to just insert 09/2012 as a parameter. If I want to insert only one setting, I can do... But I have problem when I want to insert two parameters...
Lina saleh wrote:
I already try before that and he invites me error like this...
"name of the invalid host/identification variable.

Change the name of the connection variable to: from_date and: to_date

Is a reserved keyword.

Data per day, per month, per year

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