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Home/ Questions/Q 6175097
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T23:51:57+00:00 2026-05-23T23:51:57+00:00

http://jsfiddle.net/MQHkA/2/ $(document).ready(function() { var mystring=fusioncharts,om,bdutt; var arr = mystring.split(‘,’); //array returned for(var i =

  • 0

http://jsfiddle.net/MQHkA/2/

$(document).ready(function() {
    var mystring="fusioncharts,om,bdutt";
    var arr = mystring.split(','); //array returned
    for(var i = 0; i < arr.length; i++) {
      alert(arr[i]);
    }
}

Would the above code work ?

EDIT—

Well the real code block is this :

handle1 = getUrlVars();
if(handle1 == '') {
    $("input#handle1").val('barackobama');
    $("input#handle2").val('aplusk');
    $("input#handle3").val('charliesheen');
    handle1 = 'barackobama,aplusk,charliesheen';
} else {
    alert(handle1);       // this says fusioncharts,om,bdutt
    var queryvals = [];
    queryvals = handle1.split(',');
    alert('length'+queryvals.length);         // *** this says nothing ***
    for(var i = 0; i < queryvals.length; i++) {
        alert(queryvals[i]);                   // *** nothing here too.. ****
    }
}

And the entire block is in a $(document).ready()…

Must be some simple error which I’m unable to spot..

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-23T23:51:59+00:00Added an answer on May 23, 2026 at 11:51 pm

    you are missing the closing parentheses other than that it works fine

    $(document).ready(function() {
        var mystring="fusioncharts,om,bdutt";
        var arr = mystring.split(','); //array returned
        for(var i = 0; i < arr.length; i++) {
          alert(arr[i]);
        }
    }); // this one is missing on yours
    
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