I am quite new to regex and I right now Have a problem formulating a regex to match a string where the first and last letter are different. I looked up on the internet and found a regex that just does it's opposite. i.e. matches words that have same starting and ending letter. Can anyone please help me to understand if I can negeate this regex in some way or can create a new regex to match my requirements. The regex that needs to be modiifed or changed is:

^\s|^[a-z]$|^([a-z]).*\1$

This matches these Strings :

aba,
a,
b,
c,
d,
" ", 
cccbbbbbbac,
aaaaba

But I want it to match strings like:

aaabbcz,
zba,
ccb,
cbbbba

Can anyone please help me in this regard? Thank you.

Note: I will be using this with Python Regex, so the regex should be compataible to be used with Python.

CodePudding user response：

You don't need a regex for this, just use

s[0] != s[-1]

where s is your string. If you must use a regex, you can use this:

^(.).*(?!\1).$

This looks for

• ^ : beginning of string
• (.) : a character (captured in group 1)
• .* : some number of characters
• (?!\1). : a character which is not the character captured in group 1
• $ : end of string

Regex demo on regex101

CodePudding user response：

This part of your pattern ^([a-z]).*\1$ only accounts for chars a-z, but you also want to exclude " "

You can rewrite that pattern by putting the part after the capture group inside a negative lookahead.

^(.)(?!.*\1$). 

• ^ Start of string
• (.) Capture a single char (including spaces) in group 1
• (?!.*\1$) Negative lookahead, assert that the string does not end with the same character
• . Match 1 characters so that the string has a minimum of 2 characters

See a regex demo.

If the string should start and end with a non whitespace character to prevent / trailing trailing spaces, you can start the match with a non whitespace character \S and also end the match with a non whitespace character.

^(\S)(?!.*\1$).*\S$

See another regex demo.