Retrieve the first 9 digits from the numeric value

Shell Programming and Scripting
1

Hi Buddy....

I want to retrieve the first 9 digits from the below numeric value.moreover i want to truncate prefix zero's from the retrived number

000000001200820060734

Retrived number should be passed to one more Shell Script.Pls guide me how to proceed.I'm new to Unix Shell Script...

Thanks
Soll

2 
num=000000001200820060734
echo $num | cut -c1-9 | sed -e 's/^[0]*//'
3

echo $num|sed -e 's/^[0]*//'|cut -b 1-9

4

All in one sed:

sed 's/^0*\(.\{9\}\).*/\1/g'

Your example works:

000000001200820060734
        ^^^^^^^^^

## If number of didigts in front changes, ie. if the last leading zero is not a prefix:
000000000200820060734
         ^^^^^^^^^

So this can go wrong and you might want to use the following, based on a fix number/position of digits:

echo "000000001200820060734"| cut -c8-17
5 
# a=000000001200820060734
# echo ${a:0:9}
000000001
6

What shell you are using?
With the latest bash you could write something like this:

% n=000000001200820060734
% printf -vn "%.9s" $((10#$n))
% echo $n
120082006
7

And another way

#!/bin/ksh93

n=000000001200820060734
n=${n/*(0)({9}(\d))*(\d)/\2}
print $n
8

And Z-Shell:

zsh-4.3.4% typeset -i n=000000001200820060734
zsh-4.3.4% print $n[1,9]
120082006
9

another way

# n=100400001200820060734
# read -n 9 a <<<$n
# echo $a
100400001
10

another option:
typeset -L9 a
a=<the number>

11

This is nice, I suppose it's ksh (ksh93?) feature (also supported by zsh)

zsh-4.3.4% typeset -L9 n=000000001200820060734
zsh-4.3.4% print $n
000000001

Given the OP requirement it should be (assuming integer value, of course):

zsh-4.3.4% typeset -iL9 n=000000001200820060734
zsh-4.3.4% print $n
120082006