How to print four digit numbers in shell script-CodePudding

Hello i am trying to write a shell script that does this

Takes input 4
and prints
0000
0001
0002 until
9999

Thank you in advance.

CodePudding user response：

You can use seq to generate a series of numbers between a start and stop value with a printf format string:

width=4
start=1
stop=9999
seq -f "%0${width}g" "$start" "$stop"
0001
0002
0003
0004
...
9997
9998
9999

If your OS is one of the super rare ones that does not include seq (which is not a POSIX requirement) you can use awk which is a POSIX requirement:

awk -v w="$width" -v start="$start" -v stop="$stop" 'BEGIN{
     for(i=start;i<=stop;i  ) printf("%0*d\n", w, i)}'

Or use bc to replace seq and xargs to call printf in pipe:

echo "for (i = $start; i <= $stop; i =1) i" | bc -l | xargs printf "%0${width}d\n"

Your superpower here is printf with the correct format.

CodePudding user response：

Use a for loop to iterate over desired range.
Use printf to format the number of digits you want to print.

The solution for your question is

#!/bin/bash
read -p "Enter the number of digits: " num_digits
START=0
STOP=9999
for((i=${START};i<=${STOP};i  ))
do
    printf '%0'${num_digits}'d\n' $i
done

Save this file and execute it. When prompted for number of digits, enter the number of digits you want to the output to be formatted.