extract date portion from file


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# 1  
extract date portion from file

Hi,

I have a file where there is a date field (single line variable length file)

how to extract just the date portion from it

the position of date field may vary anywhere in the line
but will always have the format mm-dd-yyyy

for eg .

xxxxxxxxxxxxxxx09-10-2006xxxxxxxxxxxxxxxxxxxx
# 2  
Try this..

Code:
$ echo "xxxxxxxxxxxxxxx09-10-2006xxxxxxxxxxxxxxxxxxxx" | sed 's/^.*\([0-9]\{2\}-[0-9]\{2\}-[0-9]\{4\}\).*$/\1/'
09-10-2006

# 3  
Hi Mona

The date which i gave was just for example
it may be any date in general so i cannot hardcode; all i know is the format will be the same
# 4  
Python alternative:

Code:
import time
mm,yy = time.strftime("%m,%Y",time.localtime()).split(",")
s[ s.find(mm) : s.find(yy) + len(yy) ]


Last edited by ghostdog74; 10-06-2006 at 11:42 AM..
# 5  
as i mentioned the date given was just an example
It will be any date in general; so cannot hardcode
# 6  
Quote:
Originally Posted by misenkiser
Hi Mona

The date which i gave was just for example
it may be any date in general so i cannot hardcode; all i know is the format will be the same
No where i have hard coded the date in my solution. If the format MM-DD-YYYY occurs anywhere in the line then it will output.

Code:
$ cat samp
xxxxxxxxxxxxxxx09-10-2006xxxxxxxxxxxxxxxxxxxx
sfsd12-12-2007dfdfd
08-30-2003dfdfdfdf
fgfgfgf01-01-2001
$ sed 's/^.*\([0-9]\{2\}-[0-9]\{2\}-[0-9]\{4\}\).*$/\1/' samp
09-10-2006
12-12-2007
08-30-2003
01-01-2001
$

HTH,
Mona
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