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[Solved] sed : last character replace

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# 1  
Old 07-07-2011
[Solved] sed : last character replace
Hi all ,

I have to write a shell script that takes a number as input , like 123
and the output will be 6 ,i.e the output will be the sum of digits of the input.

I have an idea as follows,
Code:
echo "123"|fold -1|tr '\n' '+'|bc

But the problem is after " echo "123"|fold -1|tr '\n' '+' " the output is 1+2+3+ which is not working with bc.

Is there is any way to replace the last character of a line to "\n" ???

Thanking You ,
M.Choudhury Smilie
Last edited by Franklin52; 07-07-2011 at 04:26 AM.. Reason: Please use code tags for code and data samples, thank you
# 2  
Old 07-07-2011
Code:
 
$ echo "123" | nawk ' BEGIN{FS=""} {for(i=1;i<=length($0);i++) {a=a+substr($0,i,1)}} END{print a}'
6

# 3  
Old 07-07-2011
Or:
Code:
echo "123" | sed -e 's/./&+/g' -e 's/.$//' | bc

This User Gave Thanks to Franklin52 For This Post:
M.Choudhury
# 4  
Old 07-07-2011
Alternatively..
Code:
echo "123"|fold -1|tr '\n' '+'| sed 's/+$/\n/'| bc
echo "123"|awk 'BEGIN{FS=""} {print $1"+"$2"+"$3}'| bc

This User Gave Thanks to michaelrozar17 For This Post:
M.Choudhury
# 5  
Old 07-07-2011
Code:
echo "123" | awk '$1=$1' FS= OFS=+ | bc

This User Gave Thanks to Scrutinizer For This Post:
M.Choudhury
# 6  
Old 07-07-2011
Thanks everyone ... Problem Solved Smilie
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