Examples
Search a string for "welcome":
let text = "Hello world, welcome to the universe.";
let result = text.indexOf["welcome"];
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Search a string for "Welcome":
let text = "Hello world, welcome to the universe.";
let result = text.indexOf["Welcome"];
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Find the first occurrence of "e":
let text = "Hello world, welcome to the universe.";
text.indexOf["e"];
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Find the first occurrence of "e", starting at position 5:
let text = "Hello world, welcome to the universe.";
text.indexOf["e", 5];
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Find the first occurrence of "a":
let text = "Hello world, welcome to the universe.";
text.indexOf["a"];
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Definition and Usage
The indexOf[]
method returns the position of the first
occurrence of a value in a string.
The indexOf[]
method returns -1 if the value is not found.
The indexOf[]
method is case sensitive.
Syntax
string.indexOf[searchvalue, start]
Parameters
Parameter | Description |
searchvalue | Required. The string to search for. |
start | Optional. The position to start from [default is 0]. |
Return Value
Type | Description |
A number | The first position where the search-value occurs. -1 if it never occurs. |
The Differense Between
String indexOf[] and String search[]
The indexOf[]
method cannot search against a regular expression.
The search[]
cannot take a start position argument.
Browser Support
indexOf[]
is an ECMAScript1 [ES1] feature.
ES1 [JavaScript 1997] is fully supported in all browsers:
Chrome | IE | Edge | Firefox | Safari | Opera |
Yes | Yes | Yes | Yes | Yes | Yes |
The indexOf[]
method, given one argument: a substring to search for, searches the entire calling string, and returns the index of the first occurrence of the specified substring. Given a second argument: a number, the method returns the first occurrence of the specified substring at an index greater than or equal to the specified number.Try it
Syntax
indexOf[searchString]
indexOf[searchString, position]
Parameters
searchString
Substring to search for, coerced to a string.
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
OptionalThe 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]
returns4
— because it causes the method to behave as if the second argument were0
, and the first occurrence ofo
at a position greater or equal to0
is at position4
.'hello world hello'.indexOf['world', 12]
returns-1
— because, while it's true the substringworld
occurs at index6
, that position is not greater than or equal to12
.'hello world hello'.indexOf['o', 99]
returns-1
— because99
is greater than the length ofhello world hello
, which causes the method to not search the string at all.
Return value
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.
Description
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
Checking occurrences
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'
Examples
Using indexOf[]
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
indexOf[] and case-sensitivity
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
Using indexOf[] to count occurrences of a letter in a string
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
Specifications
ECMAScript Language Specification # sec-string.prototype.indexof |
Browser compatibility
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