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Jorius
Asked: 2020-04-22 07:19:28 +0800 CST 2020-04-22 07:19:28 +0800 CST

Quantifier to match 1 or more characters in Regular Expressions

  • 772

I have the following regular expression, which works for a 2 letter subdomain:

var subdomains = [
    "es.stackoverflow.com", // valido
    "ru.stackoverflow.com", // valido
    "12.stackoverflow.com", // invalido
    "30.stackkoverflow.com" // invalido
];

const regex = "\^[a-z]{2}.stackoverflow.com$";

for (var i = 0; i < subdomains.length; i++) {
  console.log((subdomains[i].match(regex)) ? 'valido' : 'invalido');
}

But I want it to match any subdomain of any length , just letters a to z , nothing else.

This is what I tried:

var subdomains = [
    "esssss.stackoverflow.com",    // valido
    "ru.stackoverflow.com",        // valido
    "12.stackoverflow.com",        // invalido
    "30.stackkoverflow.com",       // invalido
    "es.stackoverflow.com",        // valido  
    "subdominio.stackoverflow.com" // valido
];   

const regex = "\^[a-z].stackoverflow.com$";

for (var i = 0; i < subdomains.length; i++) {
    console.log((subdomains[i].match(regex)) ? 'valido' : 'invalido');
}

But if I remove the occurrence of N characters {2}, it doesn't work for me. How can I correct it?

javascript

1 Answers

  1. Best Answer

    Something to correct before starting:

    • Regular expressions in JavaScript go with each /other, without using quotes. That builds directly to a RegExp object .
    • You shouldn't escape the ^with a \to match the start of the text (it goes directly into the regex).
    • Periods match any character (one) that is not a newline, unless you escape them. To match a dot literal, then you do have to escape them as \.. More info on all special characters: Safe way to escape user input to be processed by regular expressions in JavaScript
    • To just see if it matches or not, you may want to use the RegExp.test() method .

    So let's start with the expression:

    const miRegex = /^[a-z]{2}\.stackoverflow\.com$/;


    How to quantify in regex

    Quantifiers in regular expressions apply to the previous subpattern. You used it in what you're trying: in [a-z]{2}, it {2}quantizes to [a-z](repeats it 2 times).

    If no quantifier is used, such as in /^[a-z]\.stackoverflow\.com$/, then it matches texts that have only 1 character as a subdomain.

    What you are trying to do is match 1 or more characters in the subdomain. For that, we could use the quantifier {1,}(1 to infinities) or, what is the same, the quantifier+ .

    /^[a-z]+\.stackoverflow\.com$/


    Code

    var subdomains = [
    "esssss.stackoverflow.com",    // valido
    "ru.stackoverflow.com",        // valido
    "12.stackoverflow.com",        // invalido
    "30.stackkoverflow.com",       // invalido
    "es.stackoverflow.com",        // valido  
    "subdominio.stackoverflow.com",// valido
    "ES.stackoverflow.com"         // invalido (mayúsculas)
];   

const regex = /^[a-z]+.stackoverflow.com$/;

for (var i = 0; i < subdomains.length; i++) {
    console.log(subdomains[i], regex.test(subdomains[i]) ? 'valido' : 'invalido');
}


    Other quantifiers

      Cuant.   Descripción                             
 -------- ------------------------------------------------------------------ 
  ?        1 o 0                                                             
  ??       0 o 1                                                             
  *        infinito a 0                                                      
  *?       0 a infinito                                                      
  +        infinito a 1                                                      
  +?       1 a infinito                                                      
  {n}      exactamente n repeticiones                                      
  {m,n}    hasta n repeticiones, mínimo m                         
  {m,n}?   mínimo m, hasta n repeticiones
  {m,}     entre m e infinitas repeticiones, cuantos más puedan coincidir    
  {m,}?    entre m e infinitas repeticiones, cuantas menos puedan coincidir
  {,n}     entre 0 y n repeticiones (sólo Python y Ruby)

    Examples:

      Cuant.    Descripción                             
 --------- --------------------------------------------------------------------
  \w+       1 o más alfanuméricos (incluye "_"), cuantas más puedan coincidir
  \d{6}     exactamente 6 dígitos consecutivos
  x{3,6}    3 a 6 letras "x" consecutivas, cuantas más puedan coincidir
  @{2,5}?   2 a 5 arrobas consecutivas, cuantas menos puedan coincidir


    Other options for this case

    Alternatively, if you wanted it to match case as well, we use the (CASE INSENSITIVE) modifier :i

    /^[a-z]+\.stackoverflow\.com$/i

    Or that it matches, whether or not it has a subdomain, we could group the first part with (?:... ), and make it optional with the quantifier {0,1}, which is the same as the quantifier? .

    /^(?:[a-z]+\.)?stackoverflow\.com$/i

    more resources

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