CODEWITHSUNDEEP
bashshelloutput-formatting
readdit
readdit

Reputation: 435

In bash, how could I add integers with leading zeroes and maintain a specified buffer

For example, I want to count from 001 to 100. Meaning the zero buffer would start off with 2, 1, then eventually 0 when it reaches 100 or more.

ex: 001 002 ... 010 011 ... 098 099 100

I could do this if the numbers had a predefined number of zeroes with printf "%02d" $i. But that's static and not dynamic and would not work in my example.

Upvotes: 30

Views: 45008

Answers (7)

R. Cocinero
R. Cocinero

Reputation: 388

To complement @DennisWilliamson's answer:

seq's -f parameter is a possible way to achieve the same:

$ seq -f "%03g" 1 100
001
002
003
...
098
099
100

Here, 3 sets the number of digits to pad with leading zeros as necessary, and can be set to any positive integer. This might require a sufficiently recent version of seq, though. See @AminAbbaspour's answer in another thread, who the credit goes to.

Above output could be used in both echo (if its before-mentioned brace expansion isn't an option) and printf, as well, if necessary. E.g.:

$ for i in $(seq -f "%03g" 1 100); do echo "file-${i}.log"; done
file-001.log
file-002.log
file-003.log
...
file-098.log
file-099.log
file-100.log

Of course, for such simple cases, the detour via for loop and printf/echo might be overkill – seq alone can produce the same output:

seq -f "file-%03g.log" 1 100

Upvotes: 0

khaverim
khaverim

Reputation: 3544

If you need to pad values up to a variable number with variable padding:

$values_count=514;
$padding_width=5;

for i in 0 `seq 1 $(($values_count - 1))`; do printf "%0*d\n" $padding_width $i; done;

This would print out 00000, 00001, ... 00513. (I didn't find any of the current answers meeting my need)

Upvotes: 0

Johannes Schaub - litb
Johannes Schaub - litb

Reputation: 506975

In Bash version 4 (use bash -version) you can use brace expansion. Putting a 0 before either limit forces the numbers to be padded by zeros

echo {01..100} # 001 002 003 ...
echo {03..100..3} # 003 006 009 ...

Upvotes: 15

juan
juan

Reputation: 41

# jot is available on FreeBSD, Mac OS X, ...    
jot -s " " -w '%03d' 5   
jot -s " " -w '%03d' 10  
jot -s " " -w '%03d' 50  
jot -s " " -w '%03d' 100

Upvotes: 4

SiegeX
SiegeX

Reputation: 140327

#!/bin/bash

max=100; 

for ((i=1;i<=$max;i++)); do 
    printf "%0*d\n" ${#max} $i
done

The code above will auto-pad your numbers with the correct number of 0's based upon how many digits the max/terminal value contains. All you need to do is change the max variable and it will handle the rest.

Examples:

max=10

01
02
03
04
05
06
07
08
09
10

max=100

001
002
003
004
005
006
...
097
098
099
100

max=1000

0001
0002
0003
0004
0005
0006
...
0997
0998
0999
1000

Upvotes: 6

Dennis Williamson
Dennis Williamson

Reputation: 360085

If by static versus dynamic you mean that you'd like to be able to use a variable for the width, you can do this:

$ padtowidth=3
$ for i in 0 {8..11} {98..101}; do printf "%0*d\n" $padtowidth $i; done
000
008
009
010
011
098
099
100
101

The asterisk is replaced by the value of the variable it corresponds to in the argument list ($padtowidth in this case).

Otherwise, the only reason your example doesn't work is that you use "2" (perhaps as if it were the maximum padding to apply) when it should be "3" (as in my example) since that value is the resulting total width (not the pad-only width).

Upvotes: 35

mob
mob

Reputation: 118605

If your system has it, try seq with the -w (--equal-width) option:

$ seq -s, -w 1 10
01,02,03,04,05,06,07,08,09,10

$ for i in `seq -w 95 105` ; do echo -n " $i" ; done
095 096 097 098 099 100 101 102 103 104 105

Upvotes: 35

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