bash get a regular expression number up to 2 digits

Question

I simply need to know how to make a regex that accepts a number that has up to 2 digits

All I have just now is

^[0-9]{2}$

Which would match a number with exactly 2 digits, but I don't know how to specify "match a number which has up to 2 digits".

Also, if there is a way to make sure that this number isn't 0 then that would be a plus, otherwise I can check that with Bash.

Thanks ! :)

Note that the input variable comes from read -p "make a choice" Number

EDITING MY POST - SHOWING CODE IN CONTEXT :

while true; do
    read -p "Please key in the number of the engineer of your choice, or leave empty to select them all: " Engineer
    if [ -z "$Engineer" ]; then
        echo "No particular user specified, all engineers will be selected."
        UserIdSqlString="Transactions.Creator!=0 "
        break
    else
        if [[ $Engineer =~ ^[0-9]{1,2}$ && $Engineer -ne 0 ]]; then
            echo "If you want a specific engineer type their number otherwise leave blank"  
        else
            echo "yes"
            break
        fi
    fi
done

I've edited my post. I don't mind involving grep or sed at all :) — Bluz, Nov 08 '13 at 17:12
@Bluz, you have the `if-else` backward. `if [[ $Engineer =~ ^[0-9]{1,2}$ && $Engineer -ne 0 ]]; ` returns true when you have a non-zero two digit number set into `Engineer` so `echo "yes"` should follow this — iruvar, Nov 08 '13 at 17:18
lol I am a doughnut...shows that I need a coffee break!! Thanks mate! :) — Bluz, Nov 08 '13 at 17:20
Your test still has a problem when `Engineer`'s value is `08` or `09`. Please see my comment in 1_CR's answer below. — gniourf_gniourf, Nov 08 '13 at 18:15

iruvar · Accepted Answer · 2013-11-08T18:18:42.697

10

the bash [[ conditional expression supports extended regular expressions.

[[ $number =~ ^[0-9]{,2}$ && $number -ne 0 ]]

or as the inimitable @gniourf_gniourf points out in his comments, the following is needed to handle numbers with leading zeroes correctly

[[ $number =~ ^[0-9]{,2}$ ]] && ((number=10#$number))

edited Nov 08 '13 at 18:18

answered Nov 08 '13 at 16:51

iruvar

22,736
7
53
82

mmmhhh....doesn't seem to work :( I can add a zero. Note that the variable comes from read -p . I don't know if that is relevant but I'll edit my question just in case. – Bluz Nov 08 '13 at 17:00
@Bluz, have you tried `if [[ $number =~ ^[0-9]{,2}$ && $number -ne 0 ]]; then echo 'yes'; fi` after `read -p "make a choice" number`? – iruvar Nov 08 '13 at 17:04
yes I have, let me edit again to show you my code. Also, is that not supposed to be {1,2} instead of {,2} as you put in your expression ? Did you do that on purpose or is this a typo? I tried with {1,2} as well but still not working :( – Bluz Nov 08 '13 at 17:08
@Bluz, `{,2}` is as intended, that's an accepted form of "interval expression" – iruvar Nov 08 '13 at 17:13
2

Be aware that this solution is broken in the case when the value of `number` is `08` or `09`, since in the checking `$number -ne 0`, bash will try to interpret `08` or `09` as a number in radix 8 (because prefixed by a `0`) and this will throw an error. – gniourf_gniourf Nov 08 '13 at 18:11
4

Maybe `[[ $number =~ ^[[:digit:]]{1,2}$ ]] && ((number=10#$number))` would be better: it will assign to `number` its value in radix 10, even if there are leading zeros, and at the same time check it's not zero. `:)` – gniourf_gniourf Nov 08 '13 at 18:13
1

@gniourf_gniourf, looking forward to your book on `bash` ;-) Incorporated in the answer – iruvar Nov 08 '13 at 18:19
Thank you 1_CR! Though, I'm not planning to write any book yet, there's still so much to learn! – gniourf_gniourf Nov 08 '13 at 18:24
hey thanks guys for your additional comments and for going the extramile! I did actually notice the same thing and solved it by : $Engineer =~ ^[0-9]{1,2}$ && $Engineer -ne 0 && $Engineer =~ ^[^0] And because the maximum of digits allowed is 2 then 001 wouldn't work either :) – Bluz Nov 08 '13 at 18:30

score 7 · Answer 2 · answered Nov 08 '13 at 16:51

7

^([1-9]|[1-9]{1}[0-9]{1})$

matches every number from 1,2,3...99

answered Nov 08 '13 at 16:51

EverythingRightPlace

1,197
12
33

score 6 · Answer 3 · edited May 03 '22 at 12:46

6

The answer that I found is:

^[0-9]{1,2}$

edited May 03 '22 at 12:46

Wai Ha Lee

8,598
83
57
92

answered Nov 08 '13 at 16:48

Bluz

5,980
11
32
40

score 0 · Answer 4 · answered Dec 30 '18 at 00:38

0

#!/bin/bash
if [ "$1" -gt 0 ] 2>/dev/null ;then 
    echo "$1 is number." 
else 
    echo 'no.' 
fi

answered Dec 30 '18 at 00:38

Jagger Yu

577
4
5

bash get a regular expression number up to 2 digits

4 Answers4