Regex for Checking the Last Character

var regex = /^(.).*\1$/;
console.log(  regex.test('abcdsa'))console.log(  regex.test('abcdsaasaw'))

Regex to find last character of string

You were almost there - except for two minor mistakes.

string pattern = @"^[a-zA-Z]{2}[0-9]{3}.+[BPJ]$"

That should to the trick. Explanation:

1. ^[a-zA-Z]{2} - start of the string, followed by two letters
2. [0-9]{3} - three numeric characters at positions 2,3,4 (not two)
3. .+ - an arbitrary amount of characters, but at least one - you did not specify any limitations for this part. possibly you want to use [a-zA-Z0-9]+ instead
4. [BPJ]$ - uppercase letter B, P or J followed by end of the string. in your version, AB000A1| would have matched, too.

The length requirement is implicitly enforced because of the sizes of the parts: 2 + 3 + 1 or more + 1 = 7 or more.

Edit

I misread the "greater than 7" as "at least 7". In that case the pattern must be @"^[a-zA-Z]{2}[0-9]{3}.{2,}[BPJ]$", as Yong Shun pointed out.

Regex to match last character in string

Try this one:

^[a-z|A-Z|0-9]+[^I]\s?I{1}$

I think this is a more accurate solution.

Try demo

Regex expression to check if first and last char of set are different

You can use a capture group for the second char, and only match the last char if it is not the same as capture group 1.

^[ab]([ab])[ab]*(?!\1)[ab]$

• ^ Start of string
• [ab] Match a or b (Note that you can omit | as it means a pipe char in the character class
• ([ab]) Capture group 1, match either a or b
• [ab]* Optionally match a or b
• (?!\1) Negative lookahead, assert not the same value as captured in group 1 using the backreference \1
• [ab]$ match either a or b at the end of the string

Regex demo

Another option is immediately do the assertion after the capture group

^[ab]([ab])(?![ab]*\1$)[ab]*$

Regex demo

Or if supported, as negative lookbehind might also work. This page shows the compatibility for Javascript and lookbehinds

^[ab]([ab])[ab]*[ab]$(?<!\1)

Regex demo

Using Regex, how to check if second to last character is odd

I'd go with /(?<=[13579]{1})[05]|^[05]$/.

This utilises two conditionals. One that checks for the presence of an odd character in the second-to-last position when there's at least two characters in the string, and one that checks for a single character string.

Breaking this down:

• (?<=[13579]{1}) - does a positive lookbehind on exactly one odd character
• [05] - match a 0 or a 5 directly following the lookbehind
• | - denotes an OR
• ^ denotes the start of the string
• [05] - match a 0 or a 5
• $ - the end of the string

This can be seen in the following:

var re = /(?<=[13579]{1})[05]|^[05]$/;console.log(re.test('12345')); // 12345 should return `false`console.log(re.test('12335')); // 12335 should return `true`console.log(re.test('1')); // 1 should return `false`console.log(re.test('5')); // 5 should return `true`

// RegExp wayfunction getLastChar1(str){ var r = (/.$/).exec(str); if(r){  return r[0] } return str;}
// last string wayfunction getLastChar2(str){ if(str.length){  return str[str.length-1] } return str;}

console.log(getLastChar1("Hello Word!"))console.log(getLastChar2("Hello Word!"))

import re
string = 'aba'
string2 = 'no match'
pattern = re.compile(r'^(.).*\1$')

if re.match(pattern, string):
  print('ok')
else:
  print('nok')
if re.match(pattern, string2):
  print('ok')
else:
  print('nok')

ok
nok

^(.).*\1$

string = 'aba'
if string[0] == string[-1]:
  print 'same'

same