Remove leading zero from BASH array variables

Question

I am trying to remove leading zeroes from a BASH array... I have an array like:

echo "${DATES[@]}"

returns

01 02 02 03 04 07 08 09 10 11 13 14 15 16 17 18 20 21 22 23

I'd like to remove the leading zeroes from the dates and store back into array or another array, so i can iterate in another step... Any suggestions?

I tried this,

for i in "${!DATES[@]}"
do
    DATESLZ["$i"]=(echo "{DATES["$i"]}"| sed 's/0*//' ) 
done

but failed (sorry, i'm an old Java programmer who was tasked to do some BASH scripts)

@fedorqui, this is why: printf "%d" 08 ==> bash: printf: 08: invalid octal number — glenn jackman
– glenn jackman, Commented Jun 5, 2013 at 15:40

choroba · Accepted Answer · 2013-06-05 12:40:58Z

12

Use parameter expansion:

DATES=( ${DATES[@]#0} )

answered Jun 5, 2013 at 12:40

choroba

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2 Comments

l0b0 Over a year ago

You can use shopt -s extglob; DATES=("${DATES[@]##+(0)}") to remove any number of zeros.

Dave Over a year ago

Be careful when using @l0b0 suggestion. This will remove not only padded 0's but also the value "0" by itself as well. My script required supporting "0". Therefore I needed to build in more logic to recognize if "0" existed before using the shopt command, then re-adding "0" to my array at the end after all grooming was complete.

glenn jackman · Accepted Answer · 2013-06-05 12:46:57Z

4

With bash arithmetic, you can avoid the octal woes by specifying your numbers are base-10:

day=08
((day++))           # bash: ((: 08: value too great for base (error token is "08")
((day = 10#$day + 1))
echo $day              # 9
printf "%02d\n" $day   # 09

answered Jun 5, 2013 at 12:46

glenn jackman

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4 Comments

l0b0 Over a year ago

+1 for 10#$day, but how would you apply it to the current issue?

glenn jackman Over a year ago

Well, the question says "remove the leading zeroes...so i can iterate in another step". There's no need to remove the zeros in the first place.

l0b0 Over a year ago

I mean OP specifically asked about an array.

glenn jackman Over a year ago

I know, but sometimes we shouldn't give people exactly what they ask for (xy problem)

blue · Accepted Answer · 2013-06-05 12:41:49Z

1

You can use bash parameter expansion (see http://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html) like this:

echo ${DATESLZ[@]#0}

answered Jun 5, 2013 at 12:41

blue

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Comments

user2350426 · Accepted Answer · 2015-09-08 15:38:50Z

0

If: ${onedate%%[!0]*} will select all 0's in front of the string $onedate.

we could remove those zeros by doing this (it is portable):

echo "${onedate#"${onedate%%[!0]*}"}"

For your case (only bash):

#!/bin/bash
dates=( 01 02 02 08 10 18 20 21 0008 00101 )

for onedate in "${dates[@]}"; do
    echo -ne "${onedate}\t"
    echo "${onedate#"${onedate%%[!0]*}"}"
done

Will print:

$ script.sh
01      1
02      2
02      2
08      8
10      10
18      18
20      20
21      21
0008    8
00101   101

answered Sep 8, 2015 at 15:38

user2350426

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