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Home/ Questions/Q 7634051
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Editorial Team
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Editorial Team
Asked: May 31, 20262026-05-31T07:04:04+00:00 2026-05-31T07:04:04+00:00

I have a fiddle code: //create array with values thisArray= [”,[]]; thisArray[0] = [‘numbers’,[‘one’,’two’,’three’,’four’,’five’]];

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I have a fiddle

code:

//create array with values
thisArray= ['',[]];
thisArray[0] = ['numbers',['one','two','three','four','five']];
// have a clean clone of input
var newElem = $('#input').clone();

//search through aray and print the nested array inside the
//   value of the cloned input
for(i = 0; i<thisArray[0][1].length; i++){

    $(newElem).val(thisArray[0][1][i]);
    $('input.input').last().after(newElem);
}

Conditions: I cannot change the HTML or layout of the array.

Question: how can i edit the for loop to display the numbers in the array one after another inside the cloned input field? so the final output will be six input fields with a number in each except the first one.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-31T07:04:05+00:00Added an answer on May 31, 2026 at 7:04 am

    You need to make five copies in order to insert five new inputs:

    //create array with values
    thisArray= ['',[]];
    thisArray[0] = ['numbers',['one','two','three','four','five']];
    
    var newElem;
    
    //search through aray and print the nested array inside the value of the cloned input
    for(var i = 0; i<thisArray[0][1].length; i++){
        newElem = $('.input').eq(0).clone();    // <-- cloning moved here!
        newElem.val(thisArray[0][1][i]);
        $('input.input').last().after(newElem);
    }
    

    See http://jsfiddle.net/7NHXf/4/
    ​

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