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user129715
Asked: 2020-09-09 19:53:23 +0800 CST 2020-09-09 19:53:23 +0800 CST

Count numbers and letters from a string with javascript

  • 772

From a string I need to know how many characters there are without counting spaces or numbers and send it to print, as well as another function that tells me if there are numbers and how many of that same string, even without spaces, this is what I have done so far.

I already get how many characters but I don't want it to count the numbers, I want them separately:

function  cadenaNumerosLetras(){
    var frase = document.getElementById(‘cadena’).value;
    var letterCount = frase.split(/\W/).join('').length;
    var contadorLetras = letterCount.toString();
    console.log(letterCount);
    document.getElementById(‘resultado’).value = contadorLetras;
    var numeros = "0123456789";
    if(!isNaN(frase)){
        for(i=0; i<frase.length;i++){
            if(numeros.indexOf(frase.charAt(i),0)!=-1){
                return 1;
                document.getElementById('numeros').innerHTML= frase;
            }
        }
        return 0;          
    }  
}
javascript

4 Answers

  1. Best Answer

    To get the length of a string without the spaces , just remove them and get the length of the resulting string:

    function lenWithoutSpaces( str ) {
  return str.replace( /\s/g, '' ).length;
}

console.log( lenWithoutSpaces( 'Hola y adiós, mundo cruel !' ) );

    To count numbers, you don't care if the string has spaces or not: a space is not a number . You can use Array.prototype.forEach( ), through call( ), and passing the string as an argument.

    And, for comparison, use charCodeAt( ), which returns the ASCII code of the character at that position.

    function countDigits( str ) {
  var acu = 0;

  Array.prototype.forEach.call( str, function( val ) {
    acu += ( val.charCodeAt( 0 ) > 47 ) && ( val.charCodeAt( 0 ) < 58 ) ? 1 : 0;
  } );

  return acu;
}

console.log( countDigits( 'abcde123fgh' ) );
console.log( countDigits( 'abcdefgh' ) );

    Lastly, if you want it all in one call, you can return an object:

    function examineString( str ) {
  var acu = 0,
      withoutSpaces = str.replace( /\s/g, '' ).length;

  Array.prototype.forEach.call( str, function( val ) {
    acu += ( val.charCodeAt( 0 ) > 47 ) && ( val.charCodeAt( 0 ) < 58 ) ? 1 : 0;
  } );

  return {
    withoutSpaces: withoutSpaces,
    digitsCount: acu,
    otherChars: withoutSpaces - acu
  };
}

console.log( examineString( 'Pues vaya, después de 10 o 15 intentos, lo encontré :-)' ) );

    EDIT

    To count the digits, it can also be used Array.prototype.reduce( )... it's even simpler:

    function countDigits( str ) {
  return Array.prototype.reduce.call( str, function( acu, val ) {
    return ( val.charCodeAt( 0 ) > 47 ) && ( val.charCodeAt( 0 ) < 58 ) ? acu + 1 : acu;
  }, 0 );
}

console.log( countDigits( 'abcde012fgh' ) );
console.log( countDigits( 'abcdefgh' ) );

    • 16

  2. A variation on @Trauma's answer , making reduce more elegant :

    1. We create constant of the values ​​that can be position 0 is letter while position 1 is number
    2. We use the Trauma formula to find those that are letters and multiply by 1 (this is to make the change to integer)
    3. We add the accumulator and voila!

    const values = ["letras", "numeros", "espacios"];
function countDigits( str ) {

  return Array.prototype.reduce.call( str, function( acu, val ) {
  
    let pos =  val.charCodeAt( 0 ) == 32 ? 2 : 
             (val.charCodeAt( 0 ) > 47 ) && ( val.charCodeAt( 0 ) < 58) * 1;
    acu[values[pos]]++
    return acu;
  }, {letras : 0, numeros : 0, espacios : 0} );
}

console.log( countDigits( 'abcde012fgh ' ) );
console.log( countDigits( 'abcdefgh' ) );
console.log( countDigits( "asdjn kajsd5 46546 54aaa asdjn kajsd5 46546 " ) );

    Long live Functional Programming!

    • 6

  3. According to what you say, you need:

    1. how many characters there are without counting spaces or numbers.
    2. send it to print. Characters without counting spaces or numbers?
    3. if there are numbers and how many, the same without spaces.

    I would use the callback of replace(), passing it 2 regular expressions, [^\s^\d]and \drespectively, in order to get everything you need together at once:

    const str = "asdjn kajsd5 46546 54aaa";

const c = { noNumeros: "", cantNoNumeros: 0, numeros: "", cantNumeros: 0, }

str.replace(/([^\s^\d])|(\d)/g, (s, p1, p2, ) => {
    if (p1) {
        c.noNumeros += p1;
        c.cantNoNumeros++;
    }
    if (p2) {
        c.numeros += p2;
        c.cantNumeros++;
    }
});

console.log(c);

    If you just want the amounts, you can use reduce()and test():

    const str = "asdjn kajsd5 46546 54aaa asdjn kajsd5 46546 ";

const res = [...str].reduce((p,c)=>{
    (/[^\s^\d]/.test(c))?p.cantNoNumeros++:
    (/\d/.test(c))?p.cantNumeros++:null;
    return p;
},{ cantNoNumeros: 0, cantNumeros: 0, });

console.log(res)

    • 5

  4. I think that the problem can also be solved in a simple way with more common functions like split(), join()and filter():

    function contarNum(cadena) {
    ncad=cadena.split(' ').filter(String).join('').split(''); //convertir la cadena en array de caracteres eliminando los espacios
    nnum=ncad.filter(num => num >= 0 && num <= 9);            //para obtener los numeros solo basta con esta comparacion  
    
    return {
      numeros: nnum.length,                        //total de numeros
      espacios: cadena.split(' ').length-1,        //total de espacios
      otroscaracteres: ncad.length-nnum.length     //el numero de letras es el total de caracteres - el de numeros
    };
  }

//ejemplos

console.log(contarNum('00 6789687 yuhyugi 7tyjk 00'));

console.log(contarNum('asdjn kajsd5 46546 54aaa asdjn kajsd5 46546 '));

console.log(contarNum('abcde123fgh '));

console.log(contarNum('Pues vaya, después de 10 o 15 intentos, lo encontré :-)'));

    You can compare the results of the examples with those of other answers and you can see that they are the same.

    I hope I have contributed to solve the problem, greetings.

    • 1