String.prototype.indexOf()

searchString

Substring to search for. If the method is called with no arguments, searchString is coerced to "undefined". Therefore,'undefined'.indexOf() returns 0 — because the substring undefined is found at position 0 in the string undefined. But 'undefine'.indexOf(), returns -1 — because the substring undefined is not found in the string undefine.

position Optional

The method returns the index of the first occurrence of the specified substring at a position greater than or equal to position, which defaults to 0. If position is greater than the length of the calling string, the method doesn't search the calling string at all. If position is less than zero, the method behaves as it would if position were 0.

• 'hello world hello'.indexOf('o', -5) returns 4 — because it causes the method to behave as if the second argument were 0, and the first occurrence of hello at a position greater or equal to 0 is at position 4.
• 'hello world hello'.indexOf('world', 12) returns -1 — because, while it's true the substring world occurs at index 6, that position is not greater than or equal to 12.
• 'hello world hello'.indexOf('o', 99) returns -1— because 99 is greater than the length of hello world hello, which causes the method to not search the string at all.

The index of the first occurrence of searchString found, or -1 if not found.

Return value when using an empty search string

Searching for an empty search string produces strange results. With no second argument, or with a second argument whose value is less than the calling string's length, the return value is the same as the value of the second argument:

'hello world'.indexOf('') // returns 0
'hello world'.indexOf('', 0) // returns 0
'hello world'.indexOf('', 3) // returns 3
'hello world'.indexOf('', 8) // returns 8

However, with a second argument whose value is greater than or equal to the string's length, the return value is the string's length:

'hello world'.indexOf('', 11) // returns 11
'hello world'.indexOf('', 13) // returns 11
'hello world'.indexOf('', 22) // returns 11

In the former instance, the method behaves as if it found an empty string just after the position specified in the second argument. In the latter instance, the method behaves as if it found an empty string at the end of the calling string.

Strings are zero-indexed: The index of a string's first character is 0, and the index of a string's last character is the length of the string minus 1.

'Blue Whale'.indexOf('Blue')      // returns  0
'Blue Whale'.indexOf('Blute')     // returns -1
'Blue Whale'.indexOf('Whale', 0)  // returns  5
'Blue Whale'.indexOf('Whale', 5)  // returns  5
'Blue Whale'.indexOf('Whale', 7)  // returns -1
'Blue Whale'.indexOf('')          // returns  0
'Blue Whale'.indexOf('', 9)       // returns  9
'Blue Whale'.indexOf('', 10)      // returns 10
'Blue Whale'.indexOf('', 11)      // returns 10

The indexOf() method is case sensitive. For example, the following expression returns -1:

'Blue Whale'.indexOf('blue')  // returns -1

When checking if a specific substring occurs within a string, the correct way to check is test whether the return value is -1:

'Blue Whale'.indexOf('Blue') !== -1  // true; found 'Blue' in 'Blue Whale'
'Blue Whale'.indexOf('Bloe') !== -1  // false; no 'Bloe' in 'Blue Whale'

The following example uses indexOf() to locate substrings in the string "Brave new world".

const str = 'Brave new world'

console.log('Index of first w from start is ' + str.indexOf('w'))   // logs 8
console.log('Index of "new" from start is ' + str.indexOf('new'))   // logs 6

The following example defines two string variables.

The variables contain the same string, except that the second string contains uppercase letters. The first console.log() method displays 19. But because the indexOf() method is case sensitive, the string "cheddar" is not found in myCapString, so the second console.log() method displays -1.

const myString    = 'brie, pepper jack, cheddar'
const myCapString = 'Brie, Pepper Jack, Cheddar'

console.log('myString.indexOf("cheddar") is ' + myString.indexOf('cheddar'))
// logs 19
console.log('myCapString.indexOf("cheddar") is ' + myCapString.indexOf('cheddar'))
// logs -1

The following example sets count to the number of occurrences of the letter e in the string str:

const str = 'To be, or not to be, that is the question.'
let count = 0
let position = str.indexOf('e')

while (position !== -1) {
  count++
  position = str.indexOf('e', position + 1)
}

console.log(count)  // displays 4

• String.prototype.charAt()
• String.prototype.lastIndexOf()
• String.prototype.includes()
• String.prototype.split()
• Array.prototype.indexOf()