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Home/ Questions/Q 7066229
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T05:03:30+00:00 2026-05-28T05:03:30+00:00

I want to save Json into two variables so that I can manipulate one,

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I want to save Json into two variables so that I can manipulate one, and save the original when I need to restore and reset the data to the original.

The Json has 4 items. I have my two variables, they both share the same data initially, and I can see they are working in the console. However, when I splice the “Current” var the “Original” var somehow also gets spliced too. I just want to splice, pop, and push on the Current var.

My goal is to have two objects and only to manipulate one. I can’t use cookies or the server.

<script type="text/javascript" src="http://code.jquery.com/jquery-latest.pack.js"></script>
<script type="text/javascript">

        var jsonOriginal;//used for the original json object
        var jsonCurrent;//used for the filtered json object that gets manipulated

        $.ajax({
          url: "sources/json.txt",
          dataType: 'json',
          success: (function(json) 
        { 
            //save the JSON into two variables for later use
             jsonOriginal = json;
             jsonCurrent= json;
             doSomething();
         })
        });


        function doSomething(){

            console.log(jsonOriginal);//has 4 items
            console.log(jsonCurrent);//has 4 items

            //Splice ONLY CURRENT
            jsonCurrent.items.splice(2, 3);//remove 2 items from jsonCurrent

            console.log(jsonOriginal);//has 2 items -- WHAT????
            console.log(jsonCurrent);//has 2 items as expected

            //reset Current to the Original
            jsonCurrent=jsonOriginal;//should go back to the 4 items

        }

</script>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-28T05:03:31+00:00Added an answer on May 28, 2026 at 5:03 am

    You need to make a copy of the JSON, otherwise jsonOriginal and jsonCurrent are just references to the same object. Use

    var jsonOriginal = jQuery.extend(true, {}, json);
    

    instead of

    jsonOriginal = json;
    

    It would probably be a good idea to use the same method to copy jsonOriginal back when you want it back.

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