Match specific part between curly brackets

Question

I have the following example string.

Hello {{ bob .name}} bob.name {{john.name}}

I have highlighted the part I'd like to match. It has to be between two curly brackets in each end and it has to be the first word before the dot.

I have tried this regex [^{{}}]+(?=\\}}) but it matches everything between the curly brackets like this

Hello {{ bob.name }} bob.name {{ john.name }}

It would also be very nice if I could provide a variable for which name to look for.

Answer 1

You can use this regex with a negated character class, a lookahead and a captured group:

/{{([^}.]+)(?=(?:(?!}}).)*}})/

RegEx Demo

• ([^}.]+) : [^}.]+ matches a character that is not a dot and not a } . (...) captures in group #1.
• (?=(?:(?!}}).)*}}) is a lookahead that asserts we have a }} ahead.

---

To replace bob that is inside {{...}} use:

repl = str.replace(/({{)bob(?=(?:(?!}}).)*}})/gi, '$1newVal'); 

To match bob inside {{...}} use:

{{bob(?=(?:(?!}}).)*}})

Answer 2

You can use {{([^.]*)[^}]*}} without any lookahead.

Regex101 Demo .

{{ }} ensures you are inside double brackets.

([^.]*) captures every character except a dot.

[^}]* matches the rest of what's inside your double brackets.

The first returning group is what you want.

Answer 3

Use

(?<={{)([^{}]+?)(?=(?:\.(.*))?}})

It matches only bob and have 2 matchied groups: first - bob and second - name .

You can use it like this (used ES6 syntax):

const str = 'Hello {{bob.name}} bob.name {{john.name}}';
str.replace(/(?<={{)([^{}]+?)(?=(?:\.(.*))?}})/g, (allMatch, person, property) => {
  console.log(`All match = ${allMatch}, person = ${person}, property = ${property}`);
}));