I have a date in following format in my DB.
10/16 - mm/yy
I need to convert it to:
October/16
Is this possible?
If it's not possible then please tell me why.
This is not a date, it's missing the day, it's a bad way to store year/month. There should be a 4 digit year to avoid confusion and the year should be listed first to enable correct sorting, e.g. '2016/10' or a numeric value 201610.
You can cast it to a DATE first and then use a FORMAT to disply only month/year:
set dateformat myd;
select format(cast(mystupidcolumn + '/1' as date), 'MMMM/yy')
Or SUBSTR the month part and use a CASE.
try this format,
SELECT DATENAME(month, DATEADD(month, #mydate-1, CAST('2008-01-01' AS datetime)))
You can display date by using this code
select datename(month, YourColumnName) + '/' + right(YEAR(YourColumnName),2)
FROM yourTableName
Simply change yourColumnName with name of your table column and yourTableName with name of table.
Yes you can, and it depend in what database you use to call date functions
If you column Datetime format
SQL server DATENAME(Month, GETDATE())
MySQL database MONTHNAME(now())
otherwise
convert it will in your choice at database or you code logic
split the value and lookup at month enum or fake the date to be accepted and complete date format like 01/10/16
so do something like SELECT DATENAME(Month, datecolumn) + '/' + YEAR (datecolumn)
also you can use instead of Year function DATEPART(yy,datecolumn)
the way you do it with format will look like
CONVERT(VARCHAR(11),GETDATE(),106)
but excepted to get first 3 char of month JUN
Related
I have a date saved in an nvarchar type and I want to split the day, month and year into separate nvarchar variables (that means three variables). The date looks as follows: exposure_date ='2018-12-04' and the format is yyyy-dd-mm
any help please?
My whole project is stuck on this.
The "correct" answer here is to fix your datatype. When storing data always choose an appropriate data type for the data you're storing. For a date (with no time part) then the correct datatype is date. if you're storing numerical data, then use a numerical datatype, such as int or decimal. (n)varchar is not a one size fits all datatype and using it to store data that has a data type designed for it is almost always a bad choice. I'm storing the data as an (n)varchar because I need it in a specific format is never an excuse; have your presentation layer handle to display format, not your RDBMS.
The first step, therefore would be to change your string representation yyyy-dd-MM of a date to the ISO format yyyyMMdd by doing:
UPDATE YourTable
SET exposure_date = LEFT(exposure_date,4) + RIGHT(exposure_date,2) + SUBSTRING(exposure_date,6,2);
Now you have a unambiguous representation, you can change the data type of your column without concerns of incorrect implicit casts or error:
ALTER YourTable ALTER COLUMN exposure_date date;
Then, finally, you can treat your data as what it is, a date, and use the DATEPART function:
SELECT DATEPART(YEAR,exposure_date) AS Exposure_Year,
DATEPART(MONTH,exposure_date) AS Exposure_Month,
DATEPART(DAY,exposure_date) AS Exposure_Day
FROM YourTable;
You can also try the following
Declare #myDate date
select #myDate= Cast(substring('2011-29-12', 1, 4)
+ '-' + substring('2011-29-12', 9, 2)
+ '-' + substring('2011-29-12', 6, 2)
as Date) --YYYY-MM-DD
Select #myDate as DateTime,
datename(day,#myDate) as Date,
month(#myDate) as Month,
datename(year,#myDate) as Year,
Datename(weekday,#myDate) as DayName
The output is as shown below
DateTime Date Month Year DayName
--------------------------------------------
2011-29-12 29 12 2011 Thursday
You can find the live demo here
You can try below -
select concat(cast(year(cast('2018-12-04' as date)) as varchar(4)),'-',
cast(month(cast('2018-12-04' as date)) as varchar(2)), '-',
cast(day(cast('2018-12-04' as date)) as varchar(2)))
from tablename
If you have fixed format, then you could use this simple query with substring method:
select substring(dt, 1, 4) + '-' +
substring(dt, 9, 2) + '-' +
substring(dt, 6, 2) [YYYY-MM-DD]
from (values ('2018-31-12')) tbl(dt)
Let's go directly to the main issue, which is you are using the wrong datatype to store dates, you should store them as DATE, the datatypes are there for a reason and you need to choose a proper one for your column.
So, you need to ALTER your table and change the column datatype to DATE instead of NVARCHAR datatype.
ALTER <Table Name Here>
ALTER COLUMN <Column Name Here> DATE;
Then all things will easy, you just run the following query to get the desired output
SELECT YEAR(<Column Name Here>) TheYear,
MONTH(<Column Name Here>) TheMonth,
DAY(<Column Name Here>) TheDay
FROM <Table Name Here>
Which is the right and the best solution.
You can also (if you are not going to alter your table) do as
CREATE TABLE Dates(
StrDate NVARCHAR(10)
);
INSERT INTO Dates VALUES
(N'2018-12-04'),
(N'Invalid');
SELECT LEFT(StrDate, 4) StrYear,
SUBSTRING(StrDate, 6, 2) StrMonth,
RIGHT(StrDate, 2) StrDay
FROM Dates;
OR
SELECT YEAR(StrDate) StrYear,
MONTH(StrDate) StrMonth,
DAY(StrDate) StrDay
FROM (
SELECT TRY_CAST(StrDate AS DATE) StrDate
FROM Dates
)T
In SQL. How to convert a column A from (YYYY-MM-DD) to (YYYYMM)? I want to show the dates in YYYYMM format instead of YYYY-MM-DD.
Data type is TIMESTAMP. Using Teradata Studio 15.10.10.
For Teradata either use
to_char(tscol, 'YYYYMM') -- varchar result
or
extract(year from tscol) * 100 + extract(month from tscol) -- integer result
In Teradata you can format dates pretty much at will. To get YYYYMM, you would use
select <your date> (format 'yyyymm') (char(6))
Your date column needs to be actual date for this, not a string.
There are 3 functions you'll need.
MONTH() function. Returns the MONTH for the date within a range of 1 to 12 ( January to December). It Returns 0 when MONTH part for the date is 0.
YEAR() function. Returns a 4 digit YEAR.
CONCAT() function is used to concatenate two or more strings together.
So here's an example of combining the 3 functions.
SELECT CONCAT(YEAR('1969-02-18'),MONTH('1969-02-18'))
or you can do it in one with
select DATE_FORMAT('1969-02-18','%Y%m')
So to answer your question if it is referring to column A, you can use
SELECT DATE_FORMAT(A,'%Y%m')
SQL Fiddle:
http://www.sqlfiddle.com/#!9/a6c585/48362
You can use DATEPART to get the year and month parts of the date, cast to a varchar, pad and the concaternate.
SELECT DATEPART(YEAR,GETDATE())
SELECT DATEPART(MONTH,GETDATE())
SELECT CAST(DATEPART(YEAR,GETDATE()) AS VARCHAR(4)) + RIGHT('00' + CAST(DATEPART(MONTH,GETDATE()) AS VARCHAR(2)),2)
I am working on a project in which dates and times ar stored as a varchar e.g. "30-11-2017,7:30" first date in dd-mm-yyy format and then time separated with a comma. I am trying to filter on it but it is not working correctly kindly guide me how to filter data on date.
select *
from timetrack
where startDateAndTime >= '30-11-2017,7:30'
In attached image records have been shown. When I apply above query it shows no records
You can easily convert your date to SQL datatype datetime uisng parse function, for example select parse('30-11-2017,7:30' as datetime using 'it-IT').
So, in your case, you can apply this function in where clause, so you can easily apply comparison between dates:
select *
from timetrack
where parse(startDateAndTime as datetime using 'it-IT') >= '2017-11-30 07:30:00.000'
Your format is apparently italian :) But you have to specify your own date in the format convertable to datetime, as I have done in above example.
NOTE: parse is available starting with SQL Management Studio 2012.
Unless you are using ISO date format (yyyy-MM-dd HH:mm:ss or close) applying ordering (which inequalities like greater than or equal use) will not work: the date order is disconnected from the string ordering.
You'll need to parse the date and times into a real date time type and then compare to that (details of this depend on which RDBMS you are using).
If, you want to just filter out the date then you could use convert() function for SQL Server
select *
from timetrack
where startDateAndTime >= convert(date, left(#date, 10), 103)
Else convert it to datetime as follow
select *
from timetrack
where startDateAndTime >= convert(datetime, left(#date, 10)+' ' +
reverse(left(reverse(#date), charindex(',', reverse(#date))-1)), 103)
You need the date in a datetime column, Otherwise you can't filter with your current varchar format of your date.
Without changing the existing columns, this can be achieved by making a computed column and making it persisted to optimize performance.
ALTER TABLE test add CstartDateTime
as convert(datetime, substring(startDateAndTime, 7,4)+ substring(startDateAndTime, 4,2)
+ left(startDateAndTime, 2) +' '+ right(startDateAndTime, 5), 112) persisted
Note: this require all rows in the column contains a valid date with the current format
Firstly, you need to check what is the data that is entered in the 'startDateAndTime' column,then you can convert that varchar into date format
If the data in 'startDateAndTime' column has data like '30-11-2017,07:30', you would then have to convert it into date:
SELECT to_date('30-11-2017,07:30','dd-mm-yyyy,hh:mm') from dual; --check this
--Your query:
SELECT to_date(startDateAndTime ,'dd-mm-yyyy,hh:mm') from timetrack;
Actually I have different date in SQL table when I pull those via SQL query, day of datetime field should have fixed day.
Example: (DD-MM-YYYY) day should be "7" > (7-MM-YYYY)
10-08-2007 > 07-08-2007
27-12-2013 > 07-12-2013
01-03-2017 > 07-03-2017
Can someone help me on this. Thanks in Advance.
Find the difference between 7 and the day of the original date and add that to the original date:
SELECT DATEADD(DAY, 7 - DAY(OriginalDate), OriginalDate)
Use DATEPART to take out the month and year parts. Cast those into varchar and concatenate with 07.
Query
select '07-' +
cast(DATEPART(mm, [date_column]) as varchar(2)) + '-' +
cast(DATEPART(yyyy, [date_column]) as varchar(4))
from your_table_name;
Assuming You might have to change the day number example
DECLARE #dayNum char(2)
SELECT #dayNum = '07'
select #dayNum + Right(convert(char(10),getdate(),105),8)
If that is not the case You could do this
select '07'+ Right(convert(char(10),'10-08-2007',105),8)
I'd go this way:
SELECT CONVERT(DATE,CONVERT(VARCHAR(6),GETDATE(),112)+'25',112);
CONVERT with format 112 will return the date as unseparated ISO (today we would get 20170407). Convert this to VARCHAR(6) will implicitly cut away the day's part (201704).
Now we add a day and use again CONVERT with 112, but now with DATE as target type.
One thing to keep in mind: The day you add must be two-digit. You can achieve this with
DECLARE #int INT=7;
SELECT REPLACE(STR(#int,2),' ','0');
Use DATEFROMPARTS: Updated ONLY works from 2012 - OP has tagged SQL-Server 2008
select DATEFROMPARTS ( year('10-08-2007'), month('10-08-2007'), 7 )
Assuming that your field is of datetime datatype and your fixed day is of integer type.
select datetimecolumn+(yourparamfixedday-datepart(dd,datetimecolumn))
I am writing a query where I need to calculate the number of days since a date that is stored in the database in the format "YYYYMMDD". Since this is not a Date datatype, I can't use native Date functions. What is the best way (performance-wise, readability-wise, etc.) to perform such a calculation in a SQL query.
Best? Convert that old table to use real date columns.
Next best? Write a database function to convert YYMMDD to a real date. Alan Campin's iDate can help. You'd end up with something akin to select cvty2d(date1)-cvty2d(date2) from ...
Better than nothing? Write ugly SQL to convert the number to character, split the character up, add hyphens and convert THAT to a real date. That beast would look something like
select
date(
substr(char(date1),1,4) concat
'-' concat
substr (char(date1),5,2) concat
'-' concat
substr(char(date1),7,2)
) -
date(
substr(char(date2),1,4) concat
'-' concat
substr (char(date2),5,2) concat
'-' concat
substr(char(date2),7,2)
)
from ...
Edit
The reason these gymnastics are necessary is that the DB2 DATE() function wants to see a string in the form of 'YYYY-MM-DD', with the hyphens being necessary.
Which version of SQL are you running? SQL2008 has no issues with that format as a date datatype
Declare #something nvarchar(100)
set #Something = '20120112'
select dateadd(dd, 1, #Something)
select datediff(dd, #Something, getdate())
2012-01-13 00:00:00.000
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