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How to change the date format

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# 1  
Old 07-21-2011
How to change the date format
Hi Guys,

Can someone help me on how to change the date format using sed or any unix command to give my desired output as shown below.

INPUT FILE:
Code:
69372,200,20100122T17:56:02,2
53329,500,20100121T11:50:07,2
48865,100,20100114T16:08:16,2
11719,200,20100108T13:32:20,2



DESIRED OUTPUT FILE:
Code:
69372,200,22-JAN-10 05.56.02 PM,2
53329,500,21-JAN-10 11.50.07 AM,2
48865,100,14-JAN-10 04.08.16 PM,2
11719,200,08-JAN-10 01.32.20 PM,2


Thanks in advance.


Br,
Pinpe
# 2  
Old 07-21-2011
Manually (and noisy) with awk (without GNU extensions):

Code:
awk -F, 'BEGIN {
  split( "JAN,FEB,MAR,APR,MAY,JUN,JUL,AUG,SEP,OCT,NOV,DEC", m )
    }
  { 
    am = substr( $3, 10, 2) > 12 ? "PM" : "AM" 
    $3 = substr($3, 7, 2) "-" m[substr($3, 5, 2) + 0] "-" \
	substr( $3, 3, 2) " " ( am == "PM" ? sprintf( "%0.2d", substr( $3, 10, 2) - 12 ) \
	: sprintf( "%0.2d", substr( $3, 10, 2) ) ) \
	"." substr( $3, 13, 2 ) "." substr( $3, 16, 2 ) " " am 	
    }3' OFS=, infile

This User Gave Thanks to radoulov For This Post:
pinpe
# 3  
Old 07-21-2011
Quote:
Originally Posted by radoulov
Manually (and noisy) with awk (without GNU extensions):

Code:
awk -F, 'BEGIN {
  split( "JAN,FEB,MAR,APR,MAY,JUN,JUL,AUG,SEP,OCT,NOV,DEC", m )
    }
  { 
    am = substr( $3, 10, 2) > 12 ? "PM" : "AM" 
    $3 = substr($3, 7, 2) "-" m[substr($3, 5, 2) + 0] "-" \
	substr( $3, 3, 2) " " ( am == "PM" ? sprintf( "%0.2d", substr( $3, 10, 2) - 12 ) \
	: sprintf( "%0.2d", substr( $3, 10, 2) ) ) \
	"." substr( $3, 13, 2 ) "." substr( $3, 16, 2 ) " " am 	
    }3' OFS=, infile


Perfect! Thanks so much radoulov! Smilie Smilie
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