Backreference Pattern
Regular Expressions: Backreference Pattern
What is backreferencing in regular JavaScript expressions?
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// Backreference by number: \N
let str = `He said: "She's the one!" "She's the one!".`;
let regexp = /(['"])(.*?)\1/g;
console.log(str.match(regexp)); // "She's the one!"
// Backreference by name: \k<name>
let str = `He said: "She's the one!".`;
let regexp = /(?<quote>['"])(.*?)\k<quote>/g;
console.log(str.match(regexp)); // "She's the one!"
How many backreferences can a single regular expression have?
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Here's a simple JavaScript example with three backreferences...
let str = "The fat cat sat on the hat.";
let regex = /(\b\w{3}\b).*?(\b\w{3}\b).*?(\b\w{3}\b)/;
let match = regex.exec(str);
if (match != null) {
console.log(`Entire match: ${match[0]}`); // The fat cat sat
console.log(`Match 1: ${match[1]}`); // fat
console.log(`Match 2: ${match[2]}`); // cat
console.log(`Match 3: ${match[3]}`); // sat
}
In this code, the regex (\b\w{3}\b).*?(\b\w{3}\b).*?(\b\w{3}\b) contains three capturing groups, each capturing a 3-letter word in the string. The match array will contain the entire match at index 0 and each group match at subsequent indices.
Remember, the backreference within the same regular expression should be in the form of \1, \2, \3, etc. But when used in the replacement string of the String.prototype.replace() function, you should use $1, $2, $3, etc., instead.
let str = "abc123def456";
let regex = /(\d+)/g;
let newStr = str.replace(regex, '$1$1'); // "abc123123def456456"
console.log(newStr);
In this code, the regex (\d+) captures one or more digits as a group, and the replace() function uses $1$1 to replace each match with the match followed by itself. So "abc123def456" becomes "abc123123def456456".
Note that the actual number of backreferences supported may vary depending on the JavaScript engine you are using.
What is the use of backreferences in regular expressions?
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Can we backreference within the same regular expression?
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let str = "Hello hello";
let regex = /(\b\w+\b)\s+\1/;
let match = regex.exec(str);
if (match != null) {
console.log(`Matched repeated word: ${match[0]}`); // Hello hello
}
In this example, (\b\w+\b) is a capturing group that matches a word. \1 is a backreference that matches the exact same content as in the first group. So the regex matches a word followed by one or more spaces followed by the same word, hence it detects repeated words.
How are backreferences numbered in a regex pattern?
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Can we use backreferences in the replacement part of a regex replace operation?
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Here's an example:
let str = "abc 123";
let regex = /(\w+)\s+(\d+)/;
let newStr = str.replace(regex, '$2 $1');
console.log(newStr); // "123 abc"
In this code, (\w+) and (\d+) are capturing groups that match a word and a number, respectively. In the replace() function, $2 and $1 are backreferences that refer to the second and first group, respectively. So the function swaps the word and the number in the string.
How does the performance of a regular expression change when using backreferences?
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How does regex treat unmatched backreferences?
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Can nested capture groups be backreferenced in a regex?
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let str = "abc abc cba";
let regex = /((\b\w{3}\b)\s+\2)\s+\1/;
let match = regex.exec(str);
if (match != null) {
console.log(`Matched string: ${match[0]}`); // abc abc abc abc
}
In this example, (\b\w{3}\b) is a capturing group (group 2) that matches a 3-letter word. The outer group ((\b\w{3}\b)\s+\2) (group 1) matches a 3-letter word followed by a space and the same word again. The backreference \1 refers to the entire group 1, so the regex matches the same two words four times in a row.
Are backreferences always supported in all regex flavors?
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Here is a simple comparison table of backreference support in various regular expression engines.
| Regex Engine/Flavor | Backreference Support |
|---|---|
| JavaScript | Yes |
| Python | Yes |
| Perl | Yes |
| Java | Yes |
| .NET | Yes |
| PHP (PCRE) | Yes |
| Ruby | Yes |
| POSIX ERE | No |
| RE2 (Google) | No |
| Go (standard library) | No |
Again, some engines may support backreferences but with certain limitations or variations, so always check the documentation specific to your use case.
Please note that this information is accurate as of March 2021. For the most accurate and up-to-date information, refer to the official documentation for each engine.
Can a backreference pattern match more than one string?
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let str = "cat bat cat bat";
let regex = /(\b\w{3}\b)\s+\1/g;
let match;
while ((match = regex.exec(str)) != null) {
console.log(`Matched pair: ${match[0]}`);
}
In this case, the (\b\w{3}\b) capturing group matches a 3-letter word, and \1 backreference matches the same word again. So the regex matches repeated words.
In the string "cat bat cat bat", it will first match "cat cat", and in the next iteration, it will match "bat bat". So while the backreference matched different strings ("cat" and "bat") in different match operations, within each operation it matched the same string.
How does the backreference pattern help in data validation?
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let dateStr = "01-01-2023";
let regex = /^(\d{2})-\1-\d{4}$/;
let isValid = regex.test(dateStr);
console.log(isValid); // true
Here, (\d{2}) is a capturing group that matches two digits. The backreference \1 ensures that the day part of the date is the same as the month part. The test() method returns true if the string matches the regular expression.
This is just one of many possible examples. In general, backreferences can help validate a wide range of patterned data, such as repeated words, mirrored strings, certain structured data formats, and more.
How does the regex engine handle multiple backreference patterns in the same expression?
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Can a backreference pattern be used to match repeating words or characters?
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Here is an example that uses a backreference to match a word that appears twice in a row.
let str = "cat cat are cute";
let regex = /(\b\w+\b)\s+\1/;
let match = regex.exec(str);
if (match != null) {
console.log(`Matched repeated word: ${match[0]}`); // cat cat
}
In this example, (\b\w+\b) is a capturing group that matches a word, and \1 is a backreference that matches the same word again. Thus, the regular expression matches any word followed by one or more spaces and the same word, which detects repeated words.
Similarly, you can match repeated characters:
let str = "Helloo World!";
let regex = /(\w)\1/;
let match = regex.exec(str);
if (match != null) {
console.log(`Matched repeated character: ${match[0]}`); // oo
}
In this example, (\w) is a capturing group that matches a word character, and \1 is a backreference that matches the same character again. Thus, the regular expression matches any character followed by the same character, which detects repeated characters.
How is a backreference pattern affected if the group it refers to does not participate in the match?
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Is it possible to use a backreference pattern within a character class?
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In the following example, the \1 within the character class is not treated as a backreference:
let str = "aa";
let regex = /(a)[\1]/; // This does not work as intended
let match = regex.exec(str);
if (match != null) {
console.log(`Match: ${match[0]}`); // Does not print 'aa'
} else {
console.log('No match'); // Prints 'No match'
}
In this case, the regular expression does not match 'aa' because [\1] is interpreted as attempting to match the characters '\' or '1', not as a backreference to the first capturing group.
Can a backreference pattern be used in a lookahead or lookbehind assertion?
View Answer:
let str = "123abc123";
let regex = /(\d+)(?=\D+\1)/;
let match = regex.exec(str);
if (match != null) {
console.log(`Match: ${match[0]}`); // 123
}
In this example, (\d+) is a capturing group that matches one or more digits, and (?=\D+\1) is a positive lookahead that asserts that these digits must be followed by one or more non-digit characters and then the same digits again.
Backreferences can be used in lookbehinds as well, but as of March 2021, lookbehinds are not fully supported across all browsers. Here's an example:
let str = "abc123abc123";
let regex = /(?<=\D+(\d+))\D+\1/;
let match = regex.exec(str);
if (match != null) {
console.log(`Match: ${match[0]}`); // abc123
}
In this example, (?<=\D+(\d+)) is a positive lookbehind that asserts that the match must be preceded by one or more non-digit characters and one or more digits, and \D+\1 matches one or more non-digit characters followed by the same digits. So the entire regex matches a string where a number is repeated with non-digit characters in between.
How does a regex engine treat a backreference pattern if the referred capture group matches an empty string?
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let str = "abc";
let regex = /(b?)(a)\1c/;
let match = regex.exec(str);
if (match != null) {
console.log(`Match: ${match[0]}`); // abc
}
In this regular expression, (b?) is a capturing group that matches zero or one 'b'. The 'b' is optional due to the question mark ?, so it could match an empty string. In the string "abc", (a)\1 matches 'a' followed by an empty string, because the backreference \1 refers to the first capturing group which matches an empty string (since there's no 'b' before 'a' in "abc"). Thus, the entire regular expression matches "abc".
What is a conditional backreference pattern in Regex?
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Even though JavaScript does not directly support conditional backreferences, you can often achieve similar functionality by using multiple regular expressions and controlling the logic with JavaScript code.
For up-to-date information, always refer to the official JavaScript or ECMAScript documentation or reliable sources that keep track of the evolving JavaScript language specifications.
Suppose we want to match strings of the format "abc" or "a(bc)", and we want to capture the "bc" part only if it's in parentheses. This is essentially a conditional backreference behavior.
We can achieve this by using two separate regular expressions and controlling the logic using JavaScript:
let str1 = "abc";
let str2 = "a(bc)";
let regex1 = /^a(bc)$/;
let regex2 = /^abc$/;
if (regex1.test(str1)) {
let match = regex1.exec(str1);
console.log(`Group: ${match[1]}`); // Does not print anything as there's no match
} else if (regex2.test(str1)) {
console.log(`Matched: ${str1}, but no group`); // Matched: abc, but no group
}
if (regex1.test(str2)) {
let match = regex1.exec(str2);
console.log(`Group: ${match[1]}`); // Group: bc
} else if (regex2.test(str2)) {
console.log(`Matched: ${str2}, but no group`); // Does not print anything as there's no match
}
In this example, we use two regular expressions: regex1 captures "bc" if it's within parentheses, and regex2 matches "abc" but doesn't capture anything. We first check regex1, and if there's no match, we check regex2. This way, we essentially conditionally capture the "bc" part.