Obtain data from a [closed] chain

1

How can I get each and every data from a chain? in Javascript, for example:

var cadena = "13msi2";

and send an alert with the values of that chain separately.

    
asked by José MN 09.12.2016 в 00:08
source

3 answers

3

You can:

  • Convert the string to an array
  • Do a for cycle to traverse the array
  • Send alert with each array value
  • var cadena = "13msi2";
        var cadenaSeparada = cadena.split("");
        for (i = 0; i < cadenaSeparada.length; i++) { 
        	alert(cadenaSeparada[i]);
    	}

    You can review the w3schools example: link

    A small modification for your example: link

        
    answered by 09.12.2016 / 00:27
    source
    1
    var cadena = "13msi2";
    var cadenaSeparada = cadena.split("");
    for (var i in cadenaSeparada) {
        document.write(cadenaSeparada[i]);
        document.write("<br/>");
    }
    
    // Output: 
    // 1
    // 3
    // m
    // s
    // i
    // 2
    

    The substr () method returns the characters of a string that start at a specified location and according to the number of characters that are specified.

    Example: Using substr

    var cadena = "abcdefghij";
    
    console.log("(1,2): "    + cadena.substr(1,2));   // '(1, 2): bc'
    console.log("(-3,2): "   + cadena.substr(-3,2));  // '(-3, 2): hi'
    console.log("(-3): "     + cadena.substr(-3));    // '(-3): hij'
    console.log("(1): "      + cadena.substr(1));     // '(1): bcdefghij'
    console.log("(-20, 2): " + cadena.substr(-20,2)); // '(-20, 2): ab'
    console.log("(20, 2): "  + cadena.substr(20,2));  // '(20, 2): '
    

    This script shows:

    (1,2): bc (-3,2): hi (-3): hij (1): bcdefghij (-20, 2): ab (20, 2):

        
    answered by 09.12.2016 в 01:13
    0

    If you are using jQuery you can do the following:

    $(document).ready(function(){
    	var cadena = "13msi2";
    	var array = cadena.split("");
    
    	$.each(array,function(i){
    	   alert(array[i]);
    	});
    
    });
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
        
    answered by 09.12.2016 в 01:29