How do I remove leading zeroes from output of 'date' or avoid octal interpretation of such decimal numbers?

Question

I have this:

date +"%H hours and %M minutes"

I use festival to say it up.. but it says like: "zero nine hours".. I want it to say "nine hours"!

but date always give me 09... so I wonder if bash can easly make that become just 9?

in the complex script I tried like

printf %d 09

but it fails.. not octal :(

any idea?

OP, your accepted answer is great - for your specific case, but it doesn't match the question title. — einpoklum, Sep 13 '16 at 08:47
@ilkkachu the old title was key in me finding this question, looking for forcing decimal interpretation of octal-format numbers. — muru, Jun 04 '18 at 07:55
@muru, yeah, I thought if I could stick the note about octal there, but didn't come up with a nice phrasing. You could change it of course, at least you know what you searched for? The answers and question still mainly seem to be about removing the leading zero. — ilkkachu, Jun 04 '18 at 08:03

LiuYan 刘研 · Accepted Answer · 2018-04-08T03:23:47.447

58

In your case, you can simply disable zero padding by append - after % in the format string of date: %-H
By default, date pads numeric fields with zeroes. The following optional flags may follow '%':
- - (hyphen) do not pad the field
- _ (underscore) pad with spaces
- 0 (zero) pad with zeros
- ^ use upper case if possible
- # use opposite case if possible
See date manual
If you want to interpret number in different base, in bash
- 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 base is a decimal number between 2 and 64 representing the arithmetic base, and n is a number in that base
So, to interpret a number as decimal, use 10#n form, eg. 10#09
```
echo $((10#09*2))
18
```
See Arithmetic Evaluation section of bash manual.

edited Apr 08 '18 at 03:23

answered Apr 03 '13 at 03:49

LiuYan 刘研

3,910
5
31
34

2

thx, very detailed! btw, `echo $((011)) $((10#011))` is another good example that outputs `9 11` – Aquarius Power Apr 29 '13 at 21:10
1

This does not work for negative numbers: `lat=048 ; latd=${lat#0} ; echo $((latd-1))` works, but `lat=-048 ; latd=${lat#0} ; echo $((latd-1))` doesn't. – Luis A. Florit Feb 12 '16 at 00:17
`plus=$((10#01))` – DrBeco Mar 24 '17 at 23:21

Dickster · Answer 2 · 2015-12-19T19:08:48.437

If you're trying to do comparisons in decimal with date values, I've found this method to be very effective:

let mymin=$(date '+1%M') ; let mymin=$mymin%100

That always yields a decimal value. So I can do this:

if [[ $mymin -le 1 ]]; then   # only 0 and 1 are true.

There's no problem with 08 or 09 this way. Using %10 instead of %100 gives you ten-minute ranges from 0 through 9. I also find the following gives decimal values, without leading zeros:

echo "$((10#$(date +%M)))"

score 2 · Answer 3 · edited Dec 22 '16 at 18:48

Portably, you can easily remove a leading 0 from a variable. This leaves the variable unchanged if there is no leading 0.

eval $(date +"h=%H m=%M")
h=${h#0}
m=${m#0}
say "$h hours and $m minutes"

In bash, ksh or zsh, you can use ksh's additional glob patterns to remove any number of leading 0s. In bash, run shopt -s extglob first. In zsh, run setopt kshglob first.

eval $(date +"h=%H m=%M")
h=${h#+(0)}
m=${m#+(0)}
say "$h hours and $m minutes"

score 1 · Answer 4 · answered Apr 03 '13 at 03:08

In general in bash:

test ${#number} -eq 2 && number=${number#0}

resulting in

date +"%H %M" | 
{ read hours minutes
hours=${hours#0}
minutes=${minutes#0}
echo "${hours} hours and ${minutes} minutes"; }

Or, different:

date +"%H hours and %M minutes" | sed -n -e 's/0\([0-9]\)/\1/g' -e p

I am surprised that date does not have appropriate format options.

How do I remove leading zeroes from output of 'date' or avoid octal interpretation of such decimal numbers?

4 Answers4

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