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Capture first N Bytes from first line in a file

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# 1  
Old 08-31-2009
Bug Capture first N Bytes from first line in a file

Hi Guyz,

I need to capture first N Bytes from the first line of my file.

Eg. If i have following data in File1
414d51204541495052475731202020204a910846230e420c Hello 3621363663212 Help Required

Then, i want the value of first 48 Bytes to be stored in a variable.
That is, variable MyValue should contain :-

Help Anyone.
# 2  
Old 08-31-2009
something like this :

Myval=`awk -v va=$Pos '{ print substr($0,1,va) }'  file_name.txt`

echo "$Myval"

# 3  
Old 08-31-2009
couple of ways:

#  cut -c1-48 infile

#  awk 'NR==1{print substr($0,1,48)}' infile

#  sed 's/^\(.\{48\}\).*/\1/;q' infile

#  dd if=infile ibs=48 count=1 of=outfile
1+0 records in
0+1 records out

:-) HTH
# 4  
Old 08-31-2009

can you please let me know how the below one is working ?

#  dd if=infile ibs=48 count=1 of=outfile

Last edited by panyam; 08-31-2009 at 08:22 AM..
# 5  
Old 08-31-2009
if : input file
of : output file
ibs : tell how many bytes to read at a time, something like block size
count: read those blocks for maximum of this times,


# dd if=infile ibs=N count=M of=outfile
Will work if N * M = 48
# 6  
Old 08-31-2009
Thank You ,

Forgot to check the man page of "dd" Smilie

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