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Finding first 6 characters

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# 1  
Old 06-26-2013
Finding first 6 characters
Hi ,
I'm using KSH88

I tried the following example to get the last 6 characters from a string
Code:
 echo 'abcdefghids' | sed 's/.*\(.\{6\}\)$/\1/'

What chages i need to do to get the first 6 characters from the string
my desired output should be abcdef

Thank you
# 2  
Old 06-26-2013
why not cut
A simple cut would be
Code:
echo 'abcdefghids' | cut -c -6

This User Gave Thanks to krishmaths For This Post:
smile689
# 3  
Old 06-26-2013
Hello,

Could you please try to use the folliwng to get the last 6 characters as per your request.

Code:
 
awk '{print substr($0,(length($0)-6))}' FILE Name

Please let me know in case you have any queries.



Thanks,
R. Singh
This User Gave Thanks to RavinderSingh13 For This Post:
smile689
# 4  
Old 06-26-2013
parameter expansion

Code:
 
var=abcdefghids
echo ${var:$((-6))}

This User Gave Thanks to vidyadhar85 For This Post:
rbatte1
# 5  
Old 06-26-2013
grep
Hey,

Try sth like this,

If your grep supports -o option, It will give you first 6 characters of the string.

Code:
echo 'abcdefghij'|grep -o '^.\{6\}'

Cheers!!
-R
# 6  
Old 06-26-2013
To smile689,

vidyadhar85 has the best solution as it does not spawn another process. If you have a lot of data to process, spawning a process each time can be very time consuming. If you shell doesn't quite like his syntax (or you want something similar, but in just ksh) you can use this slightly messier one:-
Code:
var=abcdefghis
newvar="${var%${var##??????}}"
echo $newvar



Robin
Liverpool/Blackburn
UK
# 7  
Old 06-26-2013
Quote:
Originally Posted by vidyadhar85
parameter expansion

Code:
 
var=abcdefghids
echo ${var:$((-6))}

Unfortunately ${var:offset} is not available in ksh88
Quote:
Originally Posted by rbatte1
To smile689,

vidyadhar85 has the best solution as it does not spawn another process. If you have a lot of data to process, spawning a process each time can be very time consuming. If you shell doesn't quite like his syntax (or you want something similar, but in just ksh) you can use this slightly messier one:-
Code:
var=abcdefghis
newvar="${var%${var##??????}}"
echo $newvar

[..]
Good solution, if you add double quotes the subsitution gets protected from possible special characters in the replacement string :
Code:
var=abcdefghis
printf "%s\n" "${var%"${var##??????}"}"


--
(still thanks for that, @alister)
Last edited by Scrutinizer; 06-26-2013 at 03:39 PM..
This User Gave Thanks to Scrutinizer For This Post:
rbatte1
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