Is there a variance in outcomes between constructing a pattern with a string and constructing a pattern with a regular expression "literal" in JavaScript?

Question

Is there a variance in outcomes between constructing a pattern with a string and constructing a pattern with a regular expression "literal" in JavaScript?

Are there any distinctions between using RegExp literals versus strings?

String.prototype.lastIndexOf = function(pattern) {
    pattern = pattern + "(?![\s\S]*" + pattern + ")";
    var match = this.match(pattern);
    return (match == null) ? -1 : match.index;
}

function indexOfLastNewline(str) {
    var match = str.match(/\r?\n(?![\s\S]*(\r?\n))/);
    return (match == null) ? -1 : match.index;
}

var str = "Hello 1\nHello 2\nHello 3\nHello4";
alert(str.lastIndexOf("(\r?\n)")); // always returns the 1st newline (7)
alert(indexOfLastNewline(str)); // returns correctly (23)

Update

even if I utilize a RegExp object, the outcome remains the same

http://jsfiddle.net/yMMrk/2/

javascript regex

Answer 1

Answer №1

To properly handle the backslash character in a string, you should escape it by using double backslashes, like so:

String.prototype.lastIndexOf = function(pattern) {
    pattern = pattern + "(?![\\s\\S]*" + pattern + ")";
    var match = this.match(pattern);
    return (match == null) ? -1 : match.index;
}

function findLastNewlineIndex(str) {
    var match = str.match(/\r?\n(?![\s\S]*(\r?\n))/);
    return (match == null) ? -1 : match.index;
}

var str = "Hello 1\nHello 2\nHello 3\nHello4";
alert(str.lastIndexOf("(\\r?\\n)")); // correct result: 23
alert(findLastNewlineIndex(str)); // correct result: 23

You can test the code here.

Answer 2

To properly handle the backslash character in a string, you should escape it by using double backslashes, like so:

String.prototype.lastIndexOf = function(pattern) {
    pattern = pattern + "(?![\\s\\S]*" + pattern + ")";
    var match = this.match(pattern);
    return (match == null) ? -1 : match.index;
}

function findLastNewlineIndex(str) {
    var match = str.match(/\r?\n(?![\s\S]*(\r?\n))/);
    return (match == null) ? -1 : match.index;
}

var str = "Hello 1\nHello 2\nHello 3\nHello4";
alert(str.lastIndexOf("(\\r?\\n)")); // correct result: 23
alert(findLastNewlineIndex(str)); // correct result: 23

You can test the code here.

Answer 3

Answer №2

In most cases, when you are using a string to create a regular expression in JavaScript, you will need to escape the backslashes. However, if you are dealing with a regular expression literal, you will need to escape forward slashes if they appear in your pattern.

For instance, the following regex:

\s/

can be represented as a JavaScript string like this:

"\\s/"

and as a JavaScript regular expression literal like this:

/\s\//

Furthermore, there is a distinction in how mode modifiers are handled. For example, to create a case-insensitive regex, you would use a JavaScript string to construct a regex object like so:

var myRegex = new RegExp("[a-z]", "i");

On the other hand, with a regular expression literal, you can accomplish the same thing directly like this:

var myRegex = /[a-z]/i;

For more detailed information, refer to this guide.

Answer 4

In most cases, when you are using a string to create a regular expression in JavaScript, you will need to escape the backslashes. However, if you are dealing with a regular expression literal, you will need to escape forward slashes if they appear in your pattern.

For instance, the following regex:

\s/

can be represented as a JavaScript string like this:

"\\s/"

and as a JavaScript regular expression literal like this:

/\s\//

Furthermore, there is a distinction in how mode modifiers are handled. For example, to create a case-insensitive regex, you would use a JavaScript string to construct a regex object like so:

var myRegex = new RegExp("[a-z]", "i");

On the other hand, with a regular expression literal, you can accomplish the same thing directly like this:

var myRegex = /[a-z]/i;

For more detailed information, refer to this guide.

Is there a variance in outcomes between constructing a pattern with a string and constructing a pattern with a regular expression "literal" in JavaScript?

Answer №1

Answer №2

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