strange regex behaviour in bash

Question

I'm trying to find every number where more than two individual digits double. Such as:

0000000012
1837488809

My code is:

if [[ $number =~ (.)\1{3,10} ]];
then
    echo "$number found"
fi

It is working but not as expected: It echo’s out numbers containing on 11, and just them.

What am I doing wrong?

what do you mean containing on 11?, also two double numbers means it should match 1122 right? — Pruthvi Raj
– Pruthvi Raj, Commented Sep 4, 2015 at 18:32
1122 should not appear. i made my question a bit more precise, hopefully. — ǝlpoodooɟƃuooʞ
– ǝlpoodooɟƃuooʞ, Commented Sep 4, 2015 at 18:39
Add set -x to your script and see what the [[ command is actually seeing. — Etan Reisner
– Etan Reisner, Commented Sep 4, 2015 at 18:47
The upper bound on the brace match isn't necessary. If the lower bound matches it doesn't matter how many more there are. — Etan Reisner
– Etan Reisner, Commented Sep 4, 2015 at 18:48

Cyrus · Accepted Answer · 2015-09-04 18:47:06Z

1

I propose GNU grep:

grep -oE '(.)(\1){2,9}' <<< "1837488809"

Output:

answered Sep 4, 2015 at 18:47

Cyrus

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ǝlpoodooɟƃuooʞ · Accepted Answer · 2015-09-04 19:16:54Z

0

I went with:

check="grep -oE '(.)(\1){2,9}'"
if [ $check -gt 1 ]; then ...

answered Sep 4, 2015 at 19:16

ǝlpoodooɟƃuooʞ

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chepner · Accepted Answer · 2015-09-04 19:28:22Z

0

bash regular expressions don't support back references, so you are simply looking for a string containing at least one character before a string of 3-10 1s. A correct regular expression would be

# Tedious, yes, but not really any longer than a piece of code to generate it
regex="0{3,10}|1{3,10}|2{3,10}|3{3,10}|4{3,10}|5{3,10}|6{3,10}|7{3,10}|8{3,10}|9{3,10}"
if [[ $number =~ $regex ]];
then
    echo "$number found"
fi

answered Sep 4, 2015 at 19:28

chepner

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