Regex to fetch string until the occurrence of a special character in javascript

Using split and slice to find the second to last split group.
lastIndexOf solution, to just search from the back. Find first =, then continue to the next =, slice between them.

str = "I have candy='4' and ice cream = 'vanilla'"
console.log(
str.split('=').slice(-2)[0]
)

console.log(
str.slice(str.lastIndexOf('=',x=str.lastIndexOf('=')-1)+1,x<-1?undefined:x)
)

str = "and ice cream = 'vanilla'"
console.log(
str.split('=').slice(-2)[0]
)

console.log(
str.slice(str.lastIndexOf('=',x=str.lastIndexOf('=')-1)+1,x<-1?undefined:x)
)

str = "I have cand'4' and ice cream 'vanilla'"
console.log(
str.split('=').slice(-2)[0]
)

console.log(
str.slice(str.lastIndexOf('=',x=str.lastIndexOf('=')-1)+1,x<-1?undefined:x)
)

You can probably try:

.*=(.*)=.*|(.*)=.*

Explanation of the above regex:

.* - Matches everything except a newline before a = symbol greedily.

= - Matches = literally.

=(.*)= - Represents a capturing group capturing everything between =. You can use multiple symbols if you want by using them as character classes something like [=%#].

|(.*)=.* - Represents an alternate capturing group representing left hand side of the equal sign in case there is no = after the latest =.

$1$2 - You can use this as the replacement option for replacing the complete string with the required output.

You can find the demo of the above regex in here.

const regex = /.*=(.*)=.*|(.*)=.*/g;
const str = `"I have candy='4' and ice cream = 'vanilla'"
I have candy='4' and ice cream = 'vanilla' and house='big'
Regex is fun
Regex is beauty =
Regex is awesome = test test test
`;
const subst = `$1$2`;

// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);
// You can use trim if you want to get rid of previous or after spaces.
console.log(result);

Ultimately I think you could just use split and reverse with an extra trim:

str = "I have candy='4' and ice cream = 'vanilla' and house='big'"
leftOfEqual = str.split('=').reverse()[1].trim();
console.log(leftOfEqual)

But must you use regex one could use a pattern like:

[^=]+(?==[^=]*$)

See the Online Demo

• [^=]+ - Negated equal sign, one or more times
• (?==[^=]*$) - Positive lookahead for equal sign followed by negated equal sign zero or more times up to end string ancor.

This should also tick:

"In case there is no presence = to the left of latest =, I should get the complete string till the beginning then."

In your current pattern \S+(?= *=) you are matching 1+ non whitespace chars and assert what is on the right is an equals sign.

You might also use a capturing group and match the last = at the end of the string.

If there is no group 1 value available, then return the match, matching any char except a newline or =. Else return the group 1 value.

^(?:.*=([^\r\n=]+)=[^=\r\n]+$|[^\r\n=]+)

Explanation

• ^ Start of string
• .*= Match any char except a newline 0+ times until the last occurrence of =
• [^\s=]+ Match any char except - or a whitespace char
• \s*=\s* Match = between optional whitespace chars
• ( Capture group 1
  • [^\r\n=]+ Match 1+ times any char except = or a newline
• ) Close group 1
• =[^=\r\n]+$ Match the last = at the end of the string.
• | Or
• [^\r\n=]+ Match 1+ times any char except a newline or =

Regex demo

const pattern = /^(?:.*=([^\r\n=]+)=[^=\r\n]+$|[^\r\n=]+)/;
[
  "I have candy='4' and ice cream = 'vanilla'",
  "I have candy='4' and ice cream = 'vanilla' and house='big'",
  "test test ='test test test'"
].forEach(s => {
  console.log(s.match(pattern)[1] || s.match(pattern)[0]);
});

You can use lookaround

const findData = (str) => {
  let data = str.match(/(?<==)([^=]+)(?==[^=]*$)/)
  return data ? data[1].trim() : ''
}

let str1 = "I have candy='4' and ice cream = 'vanilla' and house='big'"
let str2 = "I have candy='4' and ice cream = 'vanilla'"

console.log(findData(str1))
console.log(findData(str2))

// You can use split as well

console.log(str1.split('=').slice(-2)[0].trim())
console.log(str2.split('=').slice(-2)[0].trim())