Includes() only in the first word of a String

Question

Hi I would like to run includes() method only on the first word of a String. I have found that there is an optional parameter fromIndex . I would rather need to specify toIndex which value would be the index of first whitespace but something like that does not seem to exist.

Do you have any idea how I could achieve this? Thanks!

Answer 1

You've said you have a toIndex in mind, so two options:

1. Use indexOf instead:

    var n = str.indexOf(substr); if (n != -1 && n < toIndex) { // It's before `toIndex` } 

2. Split off the first word (using split or substring or whatever) and then use includes on it:

    if (str.substring(0, toIndex).includes(substr)) { // It's before `toIndex` } 

(Adjust the use of toIndex above based on whether you want it inclusive or exclusive, of course.)

Answer 2

If it's a sentence, just split the string to get the first word

 myString = "This is my string"; firstWord = myString.split(" ")[0]; console.log("this doesn't include what I'm looking for".includes(firstWord)); console.log("This does".includes(firstWord)); 

Answer 3

You can try the following approach and create a new method

 let str = "abc abdf abcd"; String.prototype.includes2 = function(pattern,from =0,to = 0) { to = this.indexOf(' '); to = to >0?to: this.length(); return this.substring(from, to).includes(pattern); } console.log(str.includes2("abc",0,3)); console.log(str.includes2("abc",4,8)); console.log(str.includes2("abc")); console.log(str.includes2("abd")) 

Answer 4

You can split your string and pass to the include method with index 0

 var a = "this is first word of a String"; console.log(a.includes(a.split(' ')[0]));