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Extract Date part from the filename

 
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# 1  
Old 04-01-2014
Extract Date part from the filename
Hi All,

I have incoming source files abcmmyy.txt I need to extract the mmyy part from the filename and pass that to a variable . I really appreciate your quick response on this.

Thanks
raj
# 2  
Old 04-01-2014
You could try something like:
Code:
sed 's/.*\(....\)/\..*/\1/'

# 3  
Old 04-01-2014
Code:
grep -oP '[0-9]{4}'

# 4  
Old 04-01-2014
Hi Franklin,

Thanks for the quick reply but where do I need to put the filename , Could you please be more elaborative ?

Thanks
raj
# 5  
Old 04-01-2014
Where or how do you have the file name (variable, ls output, text file)?
# 6  
Old 04-01-2014
Hi RudiC,

Its ls output . I need to list all the files in that directory and pick the latest file and get the datepart from that file. The filename can go like this abc_xyzmmyy.txt

Thanks
raj
# 7  
Old 04-01-2014
Quote:
Originally Posted by rajeevm
Hi Franklin,

Thanks for the quick reply but where do I need to put the filename , Could you please be more elaborative ?

Thanks
raj
Code:
variable=$(commands... | sed 's/.*\(....\)/\..*/\1/')

or
Code:
variable=`commands... | sed 's/.*\(....\)/\..*/\1/'`

 
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