CODEWITHSUNDEEP
bash
Charlie
Charlie

Reputation: 2271

Remove leading zeros from number

How to convert all numbers in Bash/shell?

VAR=00005
VAR=00010
VAR=00601
VAR=00550

to

echo $VAR #5
echo $VAR #10
echo $VAR #601
echo $VAR #550

Upvotes: 3

Views: 7324

Answers (7)

wsdzbm
wsdzbm

Reputation: 3660

echo $((10#${VAR})) can do what you want

Works at least for bash and mksh.

From the bash manual:

Constants with a leading 0 are interpreted as octal numbers. A leading ‘0x’ or ‘0X’ denotes hexadecimal. Otherwise, numbers take the form [base#]n, where the optional base is a decimal number between 2 and 64 representing the arithmetic base, and n is a number in that base. If base# is omitted, then base 10 is used.

The docs for mksh are more concise regarding the leading zeros:

Prefixing with "10#" forces interpretation as decimal, even with leading zeros.

Upvotes: 2

Jake Ireland
Jake Ireland

Reputation: 652

The reason people are having issues with 08 and 09 is because numbers with trailing zeros are treated by the shell as octals. You can do something like this:

VAR="08"
let "VAR=10#${VAR}"
echo $VAR

to remove the trailing leading zeros.

Upvotes: 3

Kent
Kent

Reputation: 195029

or like this:

kent$  echo "00005
00010
00601
00550"|awk '$0*=1'
5
10
601
550

for your updated question (with VAR)

first of all, you should have different variable names, not all same as VAR.

see the example below:

kent$  VAR=00601

kent$  VAR=$((VAR+0))

kent$  echo $VAR
601

EDIT

for the comment.(08, 09 didn't work):

08, 09 worked here, might be something with my shell to do. I have zsh. I tested followings under bash, they worked. hope helps:

under zsh:

kent$  v=08

kent$  v=$((v+0))

kent$  echo $v
8

under bash, below worked

kent@7PLaptop:/tmp$ bash -version
GNU bash, version 3.2.48(1)-release (i486-pc-linux-gnu)
Copyright (C) 2007 Free Software Foundation, Inc.
kent@7PLaptop:/tmp$ v=08
kent@7PLaptop:/tmp$ v=$(sed 's/^0*//'<<< $v)
kent@7PLaptop:/tmp$ echo $v
8

Upvotes: 6

Charlie
Charlie

Reputation: 2271

to: ДМИТРИЙ МАЛИКОВ

printf "%d\n" 0000
printf "%d\n" 0001
printf "%d\n" 0002
printf "%d\n" 0003
printf "%d\n" 0007
printf "%d\n" 0008
printf "%d\n" 0009
printf "%d\n" 0010
printf "%d\n" 0011
printf "%d\n" 0012

result

0
1
2
3
7
0 line 8: printf: 0008: invalid octal number
0 line 9: printf: 0009: invalid octal number
8 !error - correctly 10
9 !error - correctly 11
10 !error - correctly 12

Upvotes: -2

choroba
choroba

Reputation: 241748

With extglob, you do not need any external process:

shopt -s extglob                       # Enable extended globbing
for i in 00005 00010 00601 00550; do
    echo ${i##+(0)}
done

Upvotes: 2

ДМИТРИЙ МАЛИКОВ
ДМИТРИЙ МАЛИКОВ

Reputation: 21972

$> cat text    
00005
00010
00601
00550

$> sed -r 's/0*([0-9]*)/\1/' text    
5
10
601
550

Using printf:

$> while read n; do printf "%0d\n" $((10#$n)); done < text
5
10
601
550

Note, when a numerical format expects a number, the internal printf-command will use the common Bash arithmetic rules regarding the base. In order to force decimal representation and as a side effect also remove any leading zeros for a Bash variable we should use $((10#$n))

Upvotes: 6

Ricardo
Ricardo

Reputation: 77

You can use printf to add or remove leading zeros:

$ printf "%05d\n" 5    
00005

$ printf "%d\n" 00005
5

$ printf "%010d\n" 00005
0000000005

Upvotes: -2

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