Shell Programming and Scripting

Hi, I am extracting a date string from the source file like this :

Code:

06/05/2014 16:04:00

I want to change it to

Code:

  05-JUN-14 04.05.00.000000000 PM

I basically store the date in a variable. I got solutions to change date in dd-mmm-yyyy format using tr but I guess it works only with the "date" command and not with a variable that stores the old date.

The command is

Code:

 date '+%d-%h-%Y | tr 'a-z' 'A-Z'

But it is not working when I replace "date" with my variable 'tdate'.
Is there a way in UNIX to get the date and timestamp from a stored variable?

Moderator's Comments:

Please use code tags next time for your code and data. Thanks

What have you tried so far?

Have you read the formatting options in the date manual page?

Code:

man date

What are you trying to achieve? If it's just set the alpha characters (month only) comes out as all capitals? If it's in a variable, you could:-

Code:

tdate="05-Jun-14 04.05.00.0000000000 PM"
typeset -u tdate
echo "$tdate"

Does that do it?

Robin

Are you adding a minute to get from 16:04:00 to 04.05.00.000000000 PM?
And, I don't think you can . If you're lucky, you can use the date command as rbatte1 proposes, if not, you need to work on every single field...

Code:

echo $tdate | tr 'a-z' 'A-Z'

BTW, You have unbalanced quotes in your original command. Highlighted in RED.

What I want to do is convert a date-time format to As I earlier said I cannot use the "date" command as this date is not the system date but a value that I am extracting from the input file like this :

Code:

tdat=`head -1 ${INPUTFILE} | cut -f9-10 -d'|' | sed 's/|/ /g' | cut -c1-19`

So basically I want to covert the format of this tdat variable.

What have I tried so far? I have used this variable and queried it in an oracle database which works fine, but I want to do this purely in UNIX.

---------- Post updated at 12:34 AM ---------- Previous update was at 12:33 AM ----------

What I want to do is convert a date-time format to As I earlier said I cannot use the "date" command as this date is not the system date but a value that I am extracting from the input file like this :

Code:

tdat=`head -1 ${INPUTFILE} | cut -f9-10 -d'|' | sed 's/|/ /g' | cut -c1-19`

So basically I want to covert the format of this tdat variable.

What have I tried so far? I have used this variable and queried it in an oracle database which works fine, but I want to do this purely in UNIX.

That's irrelevant if your date supports the -d --date option

Code:

date -d "06/05/2014 16:04:00" "+%d-%b-%y %I.%M.%S.%N %p"

Code:

FORMAT="+%d-%b-%y %I.%M.%S.%N %p"
date -d "$tdat" "$FORMAT"

My system does not support -d