Bash Script to add numbers which has 0 prefix

kingston · 07-09-2015, 08:19 AM

Hi Guru's

I have to write a bash script which has requirement as follows. Can someone help me?

Input:

015
009

Output:

009
010
011
012
013
014
015

To be precise, when i give inputs as follows, the script should give an output as

kingston-vm-009
kingston-vm-010
kingston-vm-011
kingston-vm-012
kingston-vm-013
kingston-vm-014
kingston-vm-015

I tried couple of combinations to add/subtract using "bc" and "expr". But didn't get succeeded.

Thanks in Advance.

Regards

Kingston S

pan64 · 07-09-2015, 08:23 AM

would be nice to see what has you really tried

kingston · 07-09-2015, 08:54 AM

Script is as follows

#!/bin/bash
FN=king-ston-vm-005
LN=king-ston-vm-008
FN_N=`echo $FN |awk -F "vm-" '{print $2}'`
echo $FN_N
LN_N=`echo $LN |awk -F "vm-" '{print $2}'`
echo $LN_N
TN_N=`bc<<<$LN_N-$FN_N`
TN_N_1=`expr 1 + $TN_N`
echo -e "Total No Nodes : $TN_N_1"

FN1=`echo $FN |awk -F "-" '{print $1"-"$2"-"$3"-"}'`
LN1=`echo $LN |awk -F "-" '{print $1"-"$2"-"$3"-"}'`
#echo $FN1
#echo $LN1

c=$FN_N
d=1
while [ $c -le $LN_N ]
do
host="$FN1$FN_N"
echo -e "Node$d : $host"
c=$(($c+1))
d=$(($d+1))
done

Output is :

[kvedanay@one scripts]$ ./test1
005
008
Total No Nodes : 4
Node1 : king-ston-vm-005
Node2 : king-ston-vm-005
Node3 : king-ston-vm-005
Node4 : king-ston-vm-005

Expected output is :

005
008
Total No Nodes : 4
Node1 : king-ston-vm-005
Node2 : king-ston-vm-006
Node3 : king-ston-vm-007
Node4 : king-ston-vm-008

Regards

Kingston S

Ser Olmy · 07-09-2015, 09:08 AM

Please put your scripts/code inside [code][/code] tags, as that greatly improves readability by preserving whitespace and indentation.

Here's some information that you may find useful:

To prevent numbers with leading zeros from being interpreted as octal by bash, prefix them with 10#, which means "the following number is in base 10". For instance, echo $((010 + 1)) will produce 9 as octal "10" is 8 in decimal, while echo $((10#010 + 1)) returns 11.
To print integers prefixed with a specific number of zeros, use printf (for instance, printf "%04d%s" $a will add up to three zeros, ensuring that $a is at least 4 digits long)

grail · 07-09-2015, 10:22 PM

In addition I would suggest looking at parameter expansion and for loops. There should be no reason to call on external commands such as awk.

berndbausch · 07-10-2015, 01:34 AM

The seq command might be an alternative. It doesn't interpret leading zeroes as octal. Unfortunately it doesn't print them, so you have to add the leading zeroes back in.

HMW · 07-10-2015, 01:56 AM

Here is how you can loop using the arguments passed to your script. Note that you can NOT use $1 and $2 directly, this:

Code:

for i in {$1..$2}

...will NOT work.

But this works:

Code:

#!/bin/bash

BEGIN=$1
END=$2

for ((i=$BEGIN; i<=$END; i++)); do
    printf "%03d%s\n" $i
done

Here's the output:

Code:

./for_loop.sh 6 12
006
007
008
009
010
011
012

Also, this:

Quote:

Originally Posted by Ser Olmy

To prevent numbers with leading zeros from being interpreted as octal by bash, prefix them with 10#, which means "the following number is in base 10". For instance, echo $((010 + 1)) will produce 9 as octal "10" is 8 in decimal, while echo $((10#010 + 1)) returns 11.

...is very useful information. Thanks!

HMW · 07-10-2015, 02:01 AM

Quote:

Originally Posted by berndbausch

The seq command might be an alternative. It doesn't interpret leading zeroes as octal. Unfortunately it doesn't print them, so you have to add the leading zeroes back in.

That depends...

Code:

for num in $(seq -w 01 10); do echo $num; done
01
02
03
04
05
06
07
08
09
10

berndbausch · 07-10-2015, 03:19 AM

Quote:

Originally Posted by HMW

That depends...

Code:

for num in $(seq -w 01 10); do echo $num; done
01
02
03...

I somehow missed the -w option. Thanks for correcting me! Seems that this is the easiest solution.

HMW · 07-10-2015, 04:26 AM

Quote:

Originally Posted by berndbausch

I somehow missed the -w option. Thanks for correcting me! Seems that this is the easiest solution.

Not in this case, since we are dealing with two leading zeroes. seq is not useful for that:

Code:

seq -w 001 10
01
02
03
04
05
06
07
08
09
10

printf is probably the best solution.

Best regards,
HMW

Shadow_7 · 07-10-2015, 06:45 AM

printf is probably better, but you can treat the number (without leading zeroes) as a string. And concatenate it with a bunch of leading zeroes, and tail the length of string you wish to use.

Code:

for COUNT in 1 2 3 4 5 6 7
do
  STRING="0000000"$COUNT
  USEFUL=$(cat $STRING | tail -c 4)
  echo $USEFUL
done

But that's quirky as various versions of tail might be -c 4 or -c 3 for a 3 digit output. There's also length syntax and substring syntaxes built into bash that can do much of the same stuff in less intuitive ways. Which will fail if said script is not run under bash as not all distros default to bash and if you run the script with "sh script.sh" instead of "./script.sh" it will use that other default shell. Which might be dash for distros like debian which lacks those length and substring extras (or uses a different syntax).

kingston · 07-10-2015, 08:11 AM

Hi HMW

Your solution almost worked. But it is not working, when the second argument was given with the "0" prefix.
Can you have a look?

Code:

[kingston@one scripts]$ ./test2 002 12
002
003
004
005
006
007
008
009
010
011
012
[kingston@one scripts]$ ./test2 002 012
002
003
004
005
006
007
008
009
010

Regards

Kingston S

grail · 07-10-2015, 09:04 AM

Personally I would just strip the leading zeroes as well, seeing the original string contains other characters.

Code:

#!/usr/bin/env bash

FN=king-ston-vm-005
LN=king-ston-vm-008

start=${FN##*0}
end=${LN##*0}

for (( ind = start; ind <= end; ind++ ))
do
	printf "03d\n" "$ind"
done

HMW · 07-10-2015, 09:07 AM

Quote:

Originally Posted by kingston

Hi HMW

Your solution almost worked. But it is not working, when the second argument was given with the "0" prefix.
Can you have a look?

Ah! I didn't realize that you sent the leading zeroes in as arguments. I should have read more carefully.

Ok, it's the octal thingy again causing trouble. If it were me, I would strip the arguments ($1 and $2) of leading zeroes, and then use the loop.

Like this:

Code:

echo "012" | sed 's/^00\?//'
12

echo "002" | sed 's/^00\?//'
2

This regex also leaves the "proper" three-digit numbers untouched:

Code:

echo "135" | sed 's/^00\?//'
135

So, my solution (which may be a bit backward, but at least it seems to work now) is:

Code:

#!/bin/bash

BEGIN=$(echo $1 | sed 's/^00\?//')
END=$(echo $2 | sed 's/^00\?//')

for ((i=$BEGIN; i<=$END; i++)); do
    printf "03ds\n" $i
done

Examples:

Code:

./leading_zeroes.sh 002 012
002
003
004
005
006
007
008
009
010
011
012

Code:

./leading_zeroes.sh 089 100
089
090
091
092
093
094
095
096
097
098
099
100

Code:

./leading_zeroes.sh 1 3
001
002
003

As you can see from the above, you can now pass the arguments both WITH leading zeroes and without, and it will still work.

Best regards,
HMW

Edit: I see that Grail beat me to it. And as you can see doesn't use sed, take your pick!

kingston · 07-18-2015, 11:45 AM

Thanks everyone. Special thanks to HMW and Grail. Your solution is the one I was looking for.
Started building rest part of my script.

Before resolving this thread, I would like to highlight the following (sorry, I couldn't explain)

Following printf command didn't work

Code:

printf "03ds\n" $I

But this one is working fine

Code:

printf "%03d%s\n" $I