Basic Skills Back

Match Literal Text

  • Problem

    Create a regular expression to exactly match the sentence: The punctuation characters in the ASCII table are: !"#$%&'()*+,-./:;<=>?@[]^_`{|}~.

  • Solution:

    /The punctuation characters in the ASCII table are: !"#$%&'()*+,-./:;<=>\?@[\]\^_`{|}~/

  • Discussion

    Any regular expression that does not include any of the dozen character $()*+.?[\^{| simply matches itself. These 12 punctuation characters (標點符號) are called metacharacters, which you should add a blackslash \ before them, if you want to match them literally.

Match Nonprintable Characters

  • Problem

    Match a string of the following ASCII control characters: bell (07), escape (1B), form feed (0C), line feed (0A), carriage return (0D), horizontal tab (09), and vertical tab (0B).

  • Solution


  • Discussion

    Seven of the most commonly used ASCII control characters have dedicated escape sequences, which all consists of a blackslash \ followed by a letter. JavaScript does not support \a and \e, and so for JavaScript we need a separate solution.

  • Variation

    In JavaScript, we can also match those seven control characters by: /\cG\x1B\cL\cJ\cM\cI\cK/, and using \cA through \cZ, you can match one of the 26 control characters too. The c must be lowercase, and the letter following the c is case insensitive, which is recommended to use an uppercase letter.

    Besides, we can also use the 7-bit character set to match control characters: /\x07\x1B\x0C\x0A\x0D\x09\x0B/. It's recommended to use the Unicode code rather than to use \x80 through \xFF, because various regex engine will interpret various meanings for them.

    ASCII table with 7-bit characters are described as followed:

0 1 2 3 4 5 6 7 8 9 A B C D E F
2 SP ! " # $ % & ' ( ) * + , - . /
3 0 1 2 3 4 5 6 7 8 9 : ; < = > ?
4 @ A B C D E F G H I J K L M N O
5 P Q R S T U V W X Y Z [ \ ] ^ _
6 ` a b c d e f g h i j k l m n o
7 p q r s t u v w x y z { | } ~ DEL

Match One of Many Characters

  • Problem

    Create one regular expression to match all common misspelling of calendar, and allow an a or e to be used in each of the vowel (元音) position.

    Create another regular expression to match a single hexadecimal character.

    Create a third regex to match a single character that is not a hexadecimal character.

  • Solution

    • Calendar with misspellings


    • Hexadecimal character


    • Non-hexadecimal character


  • Discussion

    The notation using square brackets [] is called a character class, and a character class matches a single character out of (來自) a list of possible characters.

    Inside such a character class, only four characters have a special function: \, ^, -, and ]. As same as metacharacters, they should follow after \ when you try to match them literally.

    A caret (^) placed immediately after the opening bracket [ means to match any character that is not in the list.

    A hyphen (-) creates a range when it's placed between two characters. To know exactly what they're, you may have to check the ASCII or Unicode character table. For example, /[A-z]/ is as same as /[A-Z[\]\^_`a-z]/. Actually, it's recommended to create ranges only between to digits or between two letters that are both uppercase or lowercase.

    Note that: Reversed ranges, such as [z-a] are not permitted

  • Variation

    Actually, there are also some shorthand character class in JavaScript like \d, \D, \w, \W, \s, and \S.

    /\d/ matches any numeric character

    /\D/ matches any non-numeric character

    /\w/ is always identical to /[a-zA-Z0-9_]/

    /\W/ is identical to /[^a-zA-Z0-9_]/

    /\s/ matches any whitespace character, which includes spaces, tabs, and line brakes

    /\S/ matches any character not matched by /\s/

    In addition, we can remove case sensitivity, by using /i like /[A-F0-9]/i or /[a-z0-9]/i for matching hexadecimal character.

Match Any Character

  • Problem

    Match a quoted character.

    Provide one solution that allows any single character, except a line break, between the quotes.

    Provide another that truly allows any character, including line breaks.

  • Solution

    • Any character except line breaks


    • Any character including line breaks


  • Discussion

    The dot notation . is one of the oldest and simplest regular expression features. Though it means to match any single character, it does not include a line break at all.

    If you do want to allow your regular expression to span multiple lines, you can solve it by combining \s and \S into /[\s\S]/.

    Since the dot has been so convenient, it is the most abused regular expressions feature. Therefore, use it only when you really want to allow any character. When not, use a character class of negated (否定的) character class.

Match Something at the Start and/or the End of a Line

  • Problem

    Match the world alpha, but only if it occurs at the very beginning of the subject text.

    Match the world omega, but only if it occurs at the very end of the subject text

    Match the world begin, but only if it occurs at the beginning of a line.

    Match the world end, but only if it occurs at the end of a line.

  • Solution

    • Start of the subject


    • End of the subject


    • Start of the line


    • End of the line


  • Discussion

    The regular expression tokens ^, $ are called anchors, which is used to match at certain position in JavaScript, and ^ is used for matching the start of the subject text, while $ is used for matching the end. However, if there is line break before, it will lose its function.

    So what if we do want to match the start or the end of each line, we can use \n.

  • Variation

    Actually, if you want to match the start or the end through lines, we can turn on the mode of searching in multiple lines with using /m like /^begin/m or /end$/m.

Match Whole Words

  • Problem

    Create a regex that matches cat in My cat is brown, but not in category or bobcat.

    Create another regex that matches cat in staccato, but not in My cat is brown, category or bobcat.

  • Solution

    • Word boundaries


    • Non-boundaries


  • Discussion

    \b is another anchor, which is called a word boundary, and used to match at the start or the end of a word. What we should remember is that line break characters are non-word characters, and it means that \b will match after a line break if the it's immediately followed by a word character like matching cat in the sentence My cat\n. In addition, \b is unaffected by the multiple line mode \m.

    \B matches at every position in the subject text where \b does not match. If you want to match staccato, category, bobcat excluding My cat is brown, we can combine \B like /\Bcat|cat\B/.

    In JavaScript, we view only ASCII characters as word characters. That's why /\w/ is identical to /[a-zA-Z0-9_]/.

Unicode Code Points, Categories, Blocks, and Scripts

  • Problem

    Use a regular expression to find the trademark sign (™) by specifying its Unicode code point rather than copying and pasting an actual trademark sign. If you like, what you paste is just another literal character.

    Create a regular expression that matches any character in the "Currency Symbol" Unicode category.

    Create a regular expression that matches any character in the "Greek Extended" Unicode block.

    Create a regular expression that matches any character that, according to the Unicode standard, is part of the Greek script. Unicode block.

    Create a regular expression that matches a grapheme (書寫位), or what is commonly thought of as a character: a base character with all its combining marks.

  • Solution

    • Unicode code point


    • Unicode category (JavaScript not supported)

    • Unicode block (JavaScript not supported)
    • Unicode script (JavaScript not supported)
    • Unicode grapheme (JavaScript not supported)
  • Discussion

    The Unicode code point U+2122 represents the "trademark sign" character.

  • Variation

    If we do want to match all characters in the Unicode category, we can use /[\u1F00-\u1FFF]/

Match one of Several Alternatives

  • Problem

    Create a regular expression that when applied repeatedly to the text Mary, Jane, and Sue went to Mary's house will match Mary, Jane, Sue, and then Mary again.

  • Solution


  • Discussion

    The vertical bar | splits the regular expression into multiple alternatives.

    JavaScript has also used a regex-directed engine as same as other programming languages, which means that all possible permutations (交換) of the regular expression are attempted at each character position in the subject text, before the regex attempts at the next character position.

    The other kind of engine is a text-directed engine. The key difference between those two kinds of engines is that the text-directed engine visits each character in the subject text only once, while the regex-directed may visits many times. This kind of engine is much faster, but support expressions only in the mathematical sense.

    The order of the alternatives in the regular expression does not matter only when two of them can match the same position in the string. For example, /Jane|Janet/ will match Jane in the sentence "My name is Janet.", while /Janet|Jane/ will match Janet. Since Janet has been a word, we should use /\bJane\b|\bJanet\b/ to solve this problem.

Group and Capture Parts of the Match

  • Problem

    Improve the regular expression for matching Mary, Jane, or Sue by forcing the match to be a whole world.

    Create a regular expression that matches any date in yyyy-mm-dd format, and separately capture the year, the month, and the day.

  • Solution



  • Discussion

    Since the alternative operator | has had the lowest precedence, /\bMary|Jane|Sue\b/ are separated into three regexes: /\bMary/, /Jane/, and /Sue\b/. Therefore, you have to group your alternatives with (), called a capturing group.

    As in /\b(\d\d\d\d)-(\d\d)-(\d\d)\b/, the regex /\b\d\d\d-\d\d-\d\d\b/ does exactly the same, because there are no alternatives in this regex. What if we want to extract values of the year, the month, and the day? In this case, we should use such capturing groups.

  • Variation

    Actually, using group notation () can also avoid capturing values when it's not needed, as long as we add a notation ?: following the open parentheses like \b(?:Mary|Jane|Sue)\b, which won't capture any value for you in the subject text. The main benefit of this approach is that we can add them to an existing regex without upsetting the references to numbered capturing groups. Another benefit is performance.

Match Previously Matched Text Again

  • Problem

    Create a regular expression that matches "magical" dates in yyyy-mm-dd format. A date is magical if the year minus the century, the month, and the day of the month are the same numbers, like "2008-08-08".

  • Solution


  • Discussion

    Firstly, we can use a capturing group to capture the value of the year minus the century. After that, we can match the same text anywhere in the regex using a backreference, like \1 for the first group, and \10 for the group 10. With that backreference, the first capturing group will store a value to match later text literally.

    If a capturing group is repeated, either by a quantifier {} or by backtracking. The stored value will overwritten each time the capturing group matches something, and the backreference will also matches only the text that was last captured by the group.

    Note that, a backreference only works after a capturing group. For example, the regular expressions /\b\d\d\1-(\d\d)-\1\b/ and /\b\d\d\1-\1-(\d\d)\b/ can never match anything generally. Nevertheless, in JavaScript, backreferences before a capturing group will always be successful to capture, because a group that has captured a zero-length match will cause /1 to succeed like /(^)\1/. It means that, /\b\d\d\1-(\d\d)-\1\b/ will match 20-12-12, or /\b\d\d\1-\1-(\d\d)\b/ will match 20--12.

Repeat Part of the Regex a Certain Number of Times

  • Problem

    Create regular expressions that match the following kinds of numbers:

    • A googol (A decimal number with 100 digits.)
    • A 32-bit hexadecimal number.
    • A 32-bit hexadecimal number with an optional h suffix.
    • A floating-point number with an optional integer part, a mandatory fractional (小數) part, and an optional exponent (指數冪).
  • Solution

    • Googol


    • Hexadecimal number


    • Hexadecimal number with optional suffix


    • Floating-point number


  • Discussion

    The quantifier {n} means to repeat the preceding regex token n number of times, and /ab{1}c/, therefore, is same as /abc/. In addition, /ab{0}c/ is same as /ac/.

    The quantifier {n,m} means that a repeat range from n to m, in which m should be greater than n. What if n and m are equal, we will have fixed repetition. For example, /\b\d{100,100}\b/ is same as /\b\d{100}\b/.

    The quantifier {n,} means that n or more characters should be matched, and /\d{1,}/, therefore, is same as /\d+/, and /\d{0,}/ is same as /\d*/.

    When it comes to optional character, sometimes we can choose the way of using ? following the optional character. Actually, /h?/ does the same as /h{0,1}/.

    In addition to repetition of a character, we can also repeat a group with the quantifier, like /(?:abc){3}/, which does the same as /abcabcabc/. When considering about repeating a capturing group, what we should remember, mentioned above, is that the capturing group should stored the last captured value. For example, /(\d\d){3}/ is same as /\d\d\d\d(\d\d)/. If you do want to capture all values, you should place the capturing group notation outside your repetition like ((?:\d\d){3}).

Choose Minimal or Maximal Repetition

  • Problem

    Match a pair of <p> tags, and text between them.

  • Solution


  • Discussion

    All the quantifiers used in the last section are greedy, which means that they try to repeat as many times as possible. This can make it hard to pair tags. Consider the following simple example:

      The very <em>first</em> task is to find the beginning of a paragraph.
      Then you have to find the end of the paragraph.

    We can't use a regex which simply stop when it encounters a < character.

    Take a look at one incorrect solution for this problem: /<p>.*<\/p>/.

    The only difference from the solution is that this lacks the extra question mark, causing the regex uses the same greedy asterisk (*), which means that both all paragraphs will be matched by this regex, until the end of the subject text. That also means that the regex will match text, which starts from the first <p> , and ends to the last </p>.

    Why? The greedy * grabs as much text as possible. With each repetition of a quantifier beyond the quantifier's minimum, the regular expression stores a backtracking position, for the engine to go back in when the regex following the quantifier fails.

    To solve this problem, there are lazy quantifiers. You can make any quantifier lazy by placing a question mark after it: /*?/, /+?/, /??/, and /{7,42}?/. A lazy quantifier repeats as few times as needed, stores one backtracking position, and allows the regex to continue. When matches, it will does nothing but procrastinate (拖延) and not to search any more.

    In order to know how exactly the operation of greedy and lazy repetition does behind, we can comparing how /\d+\b/ and /\d+?\b/ act on a couple of different subject texts.

    If we use /\d+\b/ on 1234, \d+ will match all the digits, \b then matches. If we use /\d+?\b/, \d+? will first matches only 1, and \b fails between 1 and 2, then \d+? expands to 12, and \b still fails, until matching all digits, and \b succeeds.

    What if the subject text is 1234X. If using /\d+\b/, \d+ still match 1234, but then \b fails. \d+ backtracks to 123, \b still fails. This continues until all fails.

    If using /\d+?\b/, \d+? first matches only 1, and expands to 12 after \b failed to match between 1 and 2. Continue to 1234, \b still fails, and then \d+? tries to match 1234X, but fails at the end.

    With these descriptions, we can know that a regex with a greedy quantifier will first match as much text as possible, while a regex with a lazy quantifier will first match as less as possible.

Test for a Match without Adding It to the Overall Mathc

  • Problem

    Find any word that occurs between a pair of HTML bold tags, without including the tags in the regex match.

  • Solution


    Actually, JavaScript supports the lookahead (?=</b>), but not the lookbehind (?<=<b>).

  • Discussion

    Lookaround that looks forward is called lookahead, with (?=...) as its syntax. When the regular expression engine exits a lookaround group, it discards the text matched by the lookaround, because the text is discarded, any backtracking positions rembered by alternation or quantifiers inside the lookaround are also discarded. While the lookahead does not consume (消耗) any text, the regex engine will remember which part of the text was matched by any capturing groups inside the lookahead.

    If the lookahead is at the end of the regex, you will indeed end up with capturing groups that matches text not matched by the regular expression itself.

    If the lookahead is in the middle of the regex, you can end up with capturing groups that match overlapping (重複的) parts of the subject text.

    But what if we use a backreference outside the lookaround to a capturing group, which is created inside the lookaround. For example: apply /(?=(\d+))\w+\1/ on 123x12. The greedy \d+ matches 123, and store the value in the first capturing group. The engine then exits the lookahead, resetting the match-in-progress to the start of the string, discarding the backtracking positions remembered by the greedy plus but keeping the 123 stored in the first capturing group. Then, the greedy \w+ matches all 123x12, and \1 fails at the end of the string. Event with several backtracking attempts, it still fails to match. The final 12 would match \1 if the regex engine could return to the lookahead and give up 123 in favor of 12, but the regex engine doesn't do that.

    Since JavaScript has not suppported the lookbehind, there's another solution to solve it by using capturing groups: /(<b>)(\w+)(?=<\/b>)/. When applied to <b>cat<\/b>, the overall regex match will be <b>cat. The first capturing group will hold <b>, and the second, cat, exactly what we want.

    If the requirement is that you want to do a search-and-replace, simple use a backreference to the first capturing group to reinsert the opening tag into the replacement text, like string.replace(/(<b>)(\w+)(?=<\/b>)/, '$1replacement');.

    If you really want to simulate lookbehind:

      var mainRegexp = /\w+(?=<\/b>)/;
      var lookbehind = /<b>$/;
      if (match = mainRegexp.exec('<b>cat</b>')) {
          /** found a word before a closing tag </b> */
          var potentialMatch = mathc[0];
          var leftCotext = match.input.substring(0, match.index);
          if (lookbehind.exec(leftCotext)) {
               * lookbehind matched
               * potentialMatch occurs betwen a pair of <b> tags
          } else {
              /** loobehind failed: potentialMatch is no good */
      } else {
          /** unable to find a word before a closing tag </b>. */

Match One or Two Alternatives Based On a Condition

  • Problem

    Create a regular expression that matches a comma-delimited list of the words one, two, and three. Each word can occur any number of times in the list, but each one must occur at least once.

  • Solution


Insert Literal Text into the Replacement Text

  • Problem

    Search and replace any regular expression match literally with the eight characters: $%\*$1\1.

  • Solution

    string.replace(//, '$%\&$$1\1')

  • Discussion

    In JavaScript, dollar signs ($) need to be escaped only when they are followed by a digit, ampersand (&), backtick (\), straight quote, underscore (_), plus sign (+), or another dollar sign ($). To escape a dollar sign, precede it with another dollar sign.

Insert the Regex Match into the Replacement Text

  • Problem

    Replace a url with <a> wrapping it.

  • Replacement

    string.replace(/http:\S+/, '<a href="$&">$&</a>')

  • Discussion

    JavaScript has adopted the $& syntax to insert the regex match into the replacement text.

Insert Part of the Regex Match into the Replacement Text

  • Problem

    Match any contiguous sequence of 10 digits, such as 1234567890, and convert the sequence into a nicely formatted phone number, like (123) 456-7890.

  • Solution

    string.replace(/\b(\d{3})(\d{3})(\d{4})\b/, '($1) $2-$3')

  • Discussion

    JavaScript support $1 only in the replacement syntax. If a capturing group with the specified two-digit number exists in your regular expression, both digits are used for the capturing group, like $10 for the tenth capturing group, if it exists. Otherwise, only $1 make effect with a literal 0 following.

    Of course, if the capturing group does not exist, your replacement expression will act like an literal.

Insert Match Context into the Replacement Text

  • Problem

    Create replacement text that replaces the regex match with the text before the regex match, followed by the whole subject text, followed by the text after the regex match. For example, if Match is found in BeforeMatchAfter, replace the match with BeforeBeforeMatchMatchAfterAfter.

  • Solution

    string.replace(/Match/, $`$`$&$'$')

  • Discussion

    JavaScript use $` and $' for left and right context, while $& for the whole regex match.

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