Regex match strings in bash

Question

I was curious if you can make regex to match 2nd character of a string with 2 from the back? For 1st and last its pretty easy and straightforward but i was curious if it can be done for any str length? I was playing with it in bash for the last hour and none of my solutions seams to work.

^(.).*\1$ thats the regex I have for 1st and last char it probably needs too be a little edited for this but i have no idea how.

Can you help me with the other one?

examples:
abcsba - match as b(index 1) == b(index -2)
regex - match as e(index 1) == a(index -2)
abba - match 
unix - not matcha as indexOf(n) != indexOf(i)
linux - not match as indexOf(i) != indexOf(u)

randomir · Accepted Answer · 2017-11-06 22:55:32Z

3

Just discard equal number of character from the beginning and from the end:

$ pat='^.(.).*\1.$'
$ [[ abcsba =~ $pat ]] && echo yes || echo no
yes
$ [[ unix =~ $pat ]] && echo yes || echo no
no

or, generally use {n} (eg. to match 3rd character with 3rd from the end):

pat='^.{2}(.).*\1.{2}$'
$ [[ abcdef =~ $pat ]] && echo yes || echo no
no
$ [[ abccef =~ $pat ]] && echo yes || echo no
yes

In the second example we use single-character-ERE duplication operator {m,n} defined in POSIX for both basic and extended regular expressions (ERE variant is only relevant here though, since =~ operator in bash uses ERE). The {n} form is a special case, equal to {n,n}, meaning repeat the preceding character (or group) exactly n times.

edited Nov 6, 2017 at 22:55

answered Nov 6, 2017 at 22:47

randomir

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8 Comments

Matt Over a year ago

When i'm using it to grep words that have the same 2nd and -2 string the same in grep -E '^.(.).*\1.$' it doesn't match all of the words like for example "did" so its like your regex doesnt like mine. When i run mine i have "did" in there and when im using mine and your together it doesnt match mine words.

randomir Over a year ago

That's because with ^.(.).*\1.$ we said that we want second character repeated as second to last. So, abba will match as well as adida.

Matt Over a year ago

Yes it works fine but it doesnt match words from previous grep. How can I join them so it will match both first last and 2nd and 2nd back?

randomir Over a year ago

You can combine them with alternation operator |, for example: ^(.).*\1$|^.(.).*\2.$. You need to use \2 in second part to match the second captured group.

randomir Over a year ago

That's simple, too. Try: ^(.)(.).*\2\1$. But please don't expand your original question in comments -- better ask a new one (and close this one by upvoting/accepting an answer). This is quite unreadable and not of much benefit for the next user that comes along.

|

user8854776 · Accepted Answer · 2017-11-07 05:48:38Z

0

As i understood you want to extract the characters . You can do it by awk command

echo "praveen" | awk '{print substr($1,1,2)}' pr 01HW497089:tmp Controller$

Here i am extracting column1 value from character 1 to character 3

answered Nov 7, 2017 at 5:48

user8854776

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