Retrieve the first 9 digits from the numeric value

sollins · #1 (**permalink**) 06-30-2008

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

chihung · #2 (**permalink**) 06-30-2008

Code:

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

flysen · #3 (**permalink**) 06-30-2008

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

zaxxon · #4 (**permalink**) 06-30-2008

All in one sed:

Code:

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

Your example works:

Code:

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:

Code:

echo "000000001200820060734"| cut -c8-17

ghostdog74 · #5 (**permalink**) 06-30-2008

Code:

# a=000000001200820060734
# echo ${a:0:9}
000000001

radoulov · #6 (**permalink**) 06-30-2008

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

Code:

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

fpmurphy · #7 (**permalink**) 06-30-2008

And another way

Code:

#!/bin/ksh93

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

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