how to fix date

[thanikondharish]

data s1 ;
name=22122018 ;
run;

proc sql ;
select input(put(name,best.),ddmmyy10.) as date format=ddmmyyd10. from s1 ;
quit;

if run above program it gives 22-12-2020 but i want as it is date 

[Ksharp]

data s1 ;
name=22122018 ;
run;

proc sql ;
select input(put(name,best. -l),ddmmyy10.) as date format=ddmmyyd10. from s1 ;
quit;

[thanikondharish]

what is best. -|

[data_null__]

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@thanikondharish wrote:
what is best. -|

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-L left justify.

[thanikondharish]

how it works could you explain that

[Ksharp]

Your code would keep lead blanks .

Like :   ___20122018  .So when you input() it ,your code only read the first 10 characters ___201220, that lead to wrong date.

if you use put(x,best. -L) would get rid of lead blanks and you would get:

20122018 and input() will correctly read it .

[Kurt_Bremser]

So you have a 8-digit number that should be a date. Convert it with

date = input(put(name,z8.),ddmmyy8.);

The z8. format makes sure that leading zeroes are not dropped.

What format you assign to the new variable depends on how you want it displayed.