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extract part of string using sed

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# 1  
Old 07-12-2012
extract part of string using sed
Hi,

I have the following string and I need to extract the date from it:

TextHere,2012/07/11,1

I tried using sed:
Code:
sed -n 's#^.*\([0-9]{4}/[0-9]{2}/[0-9]{2}\).*$#\1#p'

But it doesn't return anything. The following line doesn't even return '2012':
Code:
sed -n 's/^.*\([0-9]{4}\).*$/\1/p'

I used to use grep -o for this, but it's not supported on this system unfortunately.
# 2  
Old 07-12-2012
Hi


Code:
$ echo TextHere,2012/07/11,1 | awk -F, '{print $2}'
2012/07/11

# 3  
Old 07-12-2012
Quote:
Originally Posted by guruprasadpr
Hi


Code:
$ echo TextHere,2012/07/11,1 | awk -F, '{print $2}'
2012/07/11

Thanks, this is not sufficient unfortunately. The string doesn't always look like this so I need some way to just match the date.
# 4  
Old 07-12-2012
Code:
# echo 'TextHere,2012/07/11,1' | sed 's#^.*\([0-9]\{4\}/[0-9]\{2\}/[0-9]\{2\}\).*$#\1#'
2012/07/11

This User Gave Thanks to balajesuri For This Post:
Subbeh
# 5  
Old 07-12-2012
Hi

Code:
sed -n 's#^.*,\([0-9]\{4\}/[0-9]\{2\}/[0-9]\{2\}\).*$#\1#p'

Guru.
This User Gave Thanks to guruprasadpr For This Post:
Subbeh
# 6  
Old 07-12-2012
Alternatively, you can use grep too
Code:
echo 'TextHere,2012/07/11,1' | grep -oE '[0-9]{4}/[0-9]{2}/[0-9]{2}'

# 7  
Old 07-12-2012
Thanks, that's it!
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