Add recurring entries for days of the week

Hello,
I'm trying to setup a daily schedule for my dog walking business. I'd like to have each dog's name automaticaly listed under the day of the week which they are to walk on.
For example: "Scarlett" walks on Tuesdays, Wednesdays, and Fridays.
If B1 says "Tuesday" can I have one of the cells in that column automatically read "Scarlett" each week?
I'd like to have it so that all I have to do is put Tuesday up at the top and it automatically creates a list of my Tuesday dogs for me, Lulu, Scarlett, Mark, Dani, Zoe, Sharky, etc.

Ah, those famous words, "All I have to do..."
Here's a partial solution. Unfortunately, it doesn't cover the "all I have to do" part. More about that in the description. Click the image for a larger version.
Table 1 (Data) is the table on which the name of each dog is entered (column A) and the days that dog is to be walked recorded using check box cells.
Table 2 (Index) creates a set of lookup tables from the information recoded on Data. Each table contains a column for the day listing a serial number to represent each dog scheduled to be walked that day and the column containing the names of the dogs, plus any intervening columns. The list of dogs has bee moved to the right as I was considering using VLOOKUP, which requires the search values to be listed to the left of the result values.
Tables 3 through 9 are the daily lists. These were created as a single seven-column table, then split into separate single column tables for each day.
Formulas:
Data: No formulas on this table.
Index:
A2: =IF(Data::B,MAX(A$1:A1)+1,"")
Fill down to row 10, then select A2:A11 and fill right to column G.
H2: =Data::A
H2 (enhanced): =IF(LEN(Data::A)>0,Data::A,"")
Fill down to H11
(Enhanced version prevents showing a zero on rows where no dog's name has been entered on Data.)
Days:
Make a seven column table with one header row.
A1: Sunday
Fill right to G1.
A2: =IF(ROW()-1>MAX(Index :: A),"",LOOKUP(ROW()-1,Index :: A,Index :: $H))
Fill down to A11. Select A2:A11, Fill right to column G.
Separate the table into seven single column tables, one for each day.
To do this:
Click on any cell to make the table active and show the column and row reference tabs.
Click one of the column reference tabs to select that column.
Click the same tab a second time and drag down until the column separates from the rest of the table. Continue to drag until the column is completely separated from the table, then release the mouse button.
Repeat for the rest of the columns to be separated.
Each of these tables may be placed on a separate sheet to make printing a single list easier if desired.
Regards,
Barry

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PARTITION p7_sun VALUES LESS THAN ()
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VJ,
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CREATE TABLE YY_EVENT
PART_KEY       DATE                           NOT NULL,
SUBPART_VALUE NUMBER                             NULL,
EVENT_NAME     VARCHAR2(30 BYTE)                  NULL,
EVENT_VALUE    NUMBER                             NULL
TABLESPACE TEST_DATA
PARTITION BY RANGE (PART_KEY)
PARTITION Y_EVENT_20090310 VALUES LESS THAN (TO_DATE(' 2009-03-11 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN'))
    LOGGING
    NOCOMPRESS
    TABLESPACE TEST_DATA,
PARTITION Y_EVENT_20090311 VALUES LESS THAN (TO_DATE(' 2009-03-12 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN'))
    LOGGING
    NOCOMPRESS
    TABLESPACE TEST_DATA
)Regards

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Rick,
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You had pre-filled Column A with 3-letter abbreviations of the Days of the Week. I am not sure this is important, since you could have formatted the Date (Column B) to show the same information, but I didn't like seeing the abbreviations for days that didn't have data entered yet so I replaced the text with an expression...
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I am new to Oracle and am looking for a procedure that will calculate the first day of the week given today's date.
Given 1/23/2000 I need to find Sunday 1/21/2000.

Hi there,
I don't think there's a built-in Oracle procedure for what you want. However...
In SQL (at least, Oracle's SQL), you have the function TO_CHAR and TO_DATE that allow you to convert a date to a string and vice-versa. They work pretty much in a similar way, that is :
TO_<something>( <data_to_convert>, <format> )
eg.: To have the day of the week for the current date, you can do
SELECT TO_CHAR( SYSDATE, 'D' ) FROM DUAL;
Wednesday, January 24th, 2001 would then return 4.
To have the day of the year, you would do
TO_CHAR( SYSDATE, 'DDD' )
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In one ugly line, you could do this by doing the following :
select to_date( (to_char(sysdate,'YYYY') | | to_char( to_number( to_char(sysdate,'DDD')) - to_number( to_char(sysdate,'D')) + 1)),'YYYYDDD') from dual;
Eurk.
Or you could write a function that would return the date of the first day of the week of a specific date, like :
create or replace
FUNCTION FDOW (P_DATE DATE) RETURN DATE
IS
R_FDOW DATE;
BEGIN
R_FDOW := to_date( (to_char(P_DATE,'YYYY') | | to_char( to_number( to_char(P_DATE,'DDD')) - to_number( to_char(P_DATE,'D')) + 1)),'YYYYDDD');
return R_FDOW;
END;
(FDOW : First Day Of Week - not really original, I know)
Then, you can call it wherever you want:
SELECT FDOW(SYSDATE) FROM DUAL;
Have fun!
Frederic Denis

From where to get "First day of the week" data for all the locales, is it present in CLDR spec 24?

I am trying to get "First day of the week" data from CLDR spec24 but cannot find where to look for it in the spec. I need this data to calculate numeric value of "LOCAL day of the week".
This data to implement "c" and "cc" day formats that equals numeric local day of the week.
e.g if "First day of the week" data for a locale is 2 (Monday) , it means numeric value for local day of the week will be 1 if it is Monday that day, 2 if it is Tuesday that day and likewise.

Hi
If you want to week to be started with Sunday then use the following formula:
TimestampAdd(SQL_TSI_DAY, 1-DAYOFWEEK(Date'@{var_Date}'), Date'@{var_Date}') if it's retail week(starts from Monday) then the follow below:
TimestampAdd(SQL_TSI_DAY, 1-DAYOFWEEK(Date'@{var_Date}'), Date'@{var_Date}')
I'm assuming var_Date is the presentation variable for prompt...
Edited by: Kishore Guggilla on Jan 3, 2011 4:48 PM

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consider this situation,
Two or more productid will be accquired by same customerid, by same shipvia, on the same day of the week of shipped date. i want the simple query for this.
my tables are from northwind:
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i tried some but it is not exact, it gives wrong result.
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select ProductID,o.CustomerID,COUNT(productid) as no_of_time_purchased
from orders o join [Order Details] od on o.OrderID=od.OrderID group by ProductID,o.CustomerID
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join
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Hi,
I think I have a solution that will at least give you a clue how to go about it. I have simplified the tables you mentioned and created them as temporary tables on my side, with some fake data to test with. I have incldued the generation of these temporary
tables for your review.
In my example I have included:
1. A customer which has purchased the same product on the same day, using the same ship 3 times,
2. Another example the same as the first but the third purchase was on a different day
3. Another example the same as the first but the third purchase was a different product
4. Another example the same as the first but the third purchase was using a different "ShipVia".
You should be able to see that by grouping on all of the columns that you wich to return, you should not need to perform any subselects.
Please let me know if I have missed any requirements.
Hope this helps:
CREATE TABLE #ORDERS
OrderID INT,
CustomerID INT,
OrderDate DATETIME,
ShipVia VARCHAR(5)
CREATE TABLE #ORDERS_DETAILS
OrderID INT,
ProductID INT,
INSERT INTO #ORDERS
VALUES
(1, 1, GETDATE(), 'ABC'),
(2, 1, GETDATE(), 'ABC'),
(3, 1, GETDATE(), 'ABC'),
(4, 2, GETDATE() - 4, 'DEF'),
(5, 2, GETDATE() - 4, 'DEF'),
(6, 2, GETDATE() - 5, 'DEF'),
(7, 3, GETDATE() - 10, 'GHI'),
(8, 3, GETDATE() - 10, 'GHI'),
(9, 3, GETDATE() - 10, 'GHI'),
(10, 4, GETDATE() - 10, 'JKL'),
(11, 4, GETDATE() - 10, 'JKL'),
(12, 4, GETDATE() - 10, 'MNO')
INSERT INTO #ORDERS_DETAILS
VALUES
(1, 1),
(2, 1),
(3, 1),
(4, 2),
(5, 2),
(6, 2),
(7, 3),
(8, 3),
(9, 4),
(10, 5),
(11, 5),
(12, 5)
SELECT * FROM #ORDERS
SELECT * FROM #ORDERS_DETAILS
SELECT
O.CustomerID,
OD.ProductID,
O.ShipVia,
COUNT(O.ShipVia),
DATENAME(DW, O.OrderDate) AS [Shipped Day]
FROM #ORDERS O
JOIN #ORDERS_DETAILS OD ON O.orderID = OD.OrderID
GROUP BY OD.ProductID, O.CustomerID, O.ShipVia, DATENAME(DW, O.OrderDate) HAVING COUNT(OD.ProductID) > 1
DROP TABLE #ORDERS
DROP TABLE #ORDERS_DETAILS

Discovering the day of the week for given date

hi guys, im trying to discover the day of the week for a given date when i do this it works fine:
select to_char(sysdate,'FMDAY') from dual
however, when i try to pass in my own value as a parameter i get the error: invalid number
select to_char(:mydate,'FMDAY') from dual .
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Edited by: 786733 on 21-Feb-2011 03:41

786733 wrote:
hi guys, im trying to discover the day of the week for a given date when i do this it works fine:
select to_char(sysdate,'FMDAY') from dual
however, when i try to pass in my own value as a parameter i get the error: invalid number
select to_char(:mydate,'FMDAY') from dual .
the value im passing through to mydate is 21/02/2011 so i have no idea why it is not working. Any help would be greatly appreciated. thanksyou're probably passing in a string instead of a date, so you need to convert it to a date with the appropriate format.
select to_char(to_date(:mydate,'DD/MM/YYYY'),'FMDAY') from dual

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Is there a way to check if the entered date is a certain day of the week?
I just would like to alert the user if the date that was entered, which i have format "m/d/yy" on the format properties, that the date they entered is not a Sunday date. I tried to use the util.scand function but then was unable to figure out how to script it to look for the value to be Sunday.
I tried to put the following in the validation script:
var EFDT = event.value;
var day = util.scand("dddd", EFDT);
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day is null
4:Field:ValidateException in line 4 of function top_level, script Field:Validate
TypeError: day is null
4:Field:Validate
i am sure this means that my if criteria is not correct syntax, but i have run into a dead end when trying to search for more help.
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Edit: corrected code

Add recurring entries for days of the week

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