how to sort the date ?

Question

here is my sample code

create table [sample](T_id int identity(1,1),date varchar(50));

04-05-2013
07-03-2013
10-01-2013
20-08-2013
18-02-2013
09-07-2013
07-06-2013
05-12-2013
05-12-2013
05-12-2013

i want to sort the date based on DD-MM-YYYY fomat:

i m using below query

select distinct convert(varchar,date,103),convert(datetime,date,103)as date from [sample] order by date desc;

i m getting below output,

[no column name]       date
05-12-2013        2013-12-05 00:00:00.000
20-08-2013         2013-08-20 00:00:00.000
09-07-2013       2013-07-09 00:00:00.000
07-06-2013         2013-06-07 00:00:00.000
04-05-2013         2013-05-04 00:00:00.000
07-03-2013         2013-03-07 00:00:00.000
18-02-2013         2013-02-18 00:00:00.000
10-01-2013         2013-01-10 00:00:00.000

how do i get the column name for fist one...? can anyone help my question..

Hari Bharadwaj Vangala · Answer

select convert(varchargetdate()103) as date1convert(datetimegetdate()103)as date from [sample] order by date desc; Use the same as for first column like u used in second. Hope this was expected answer. Hope this helps..

selva kumar · Answer

it displays YYYY-MM-DD format...but this not my solution.... i want to display DD-MM-YYYY... tell me another way......

Hari Bharadwaj Vangala · Answer

http://stackoverflow.com/questions/15194968/sql-sort-month-ascending-and-months-descending Y ou should declare ur date1 also as datetime then u can use the feature as present in above link... If u declare ur date 1 as datetime then u get the above expected result... select date1 from (select distinct convert(datetimedate103) as date1convert(datetimedate103)as date2 from [sample])t order by MONTH(date1) desc Please convert the o/p in the design page while u display.Hope this helps....refining my knowledge:P...tq.

selva kumar · Answer

ur query gives following output: 20-08-2013 18-02-2013 10-01-2013 09-07-2013 07-06-2013 07-03-2013 05-12-2013 04-05-2013 but i am expecting like this; sorting based on month vice.... 05-12-2013 20-08-2013 09-07-2013 07-06-2013 04-05-2013 07-03-2013 18-02-2013 10-01-2013 tell me another way.....

Pankaj Pandey · Answer

select date1 from (select distinct convert(varchardate103) as date1convert(datetimedate103)as date2 from [sample]) order by date1 desc;

Hari Bharadwaj Vangala · Answer

select date1 from (select distinct convert(varchardate103) as date1convert(datetimedate103)as date2 from [sample]) t order by date1 desc; This will work perfectly...

selva kumar · Answer

select distinct convert(varchardate103)as dateconvert(datetimedate103)as date1 from [sample] order by date desc; this is my query. if i m using this... i m getting two columns.....like this... date1 date 05-12-2013 2013-12-05 00:00:00.000 20-08-2013 2013-08-20 00:00:00.000 09-07-2013 2013-07-09 00:00:00.000 07-06-2013 2013-06-07 00:00:00.000 04-05-2013 2013-05-04 00:00:00.000 07-03-2013 2013-03-07 00:00:00.000 18-02-2013 2013-02-18 00:00:00.000 10-01-2013 2013-01-10 00:00:00.000 i want to display first column........

Pankaj Pandey · Answer

hiiii select convert(varchargetdate()103) as Date_varcharconvert(datetimegetdate()103)as Date_datefrom [sample] order by date desc; mark as answered if helpful

selva kumar · Answer

i get the column name for first one ...then i want to display the first column results....