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Home/ Questions/Q 6473339
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T06:27:18+00:00 2026-05-25T06:27:18+00:00

I have the following json { id: 0001, type: donut, name: Cake, ppu: 0.55,

  • 0

I have the following json

{
    "id": "0001",
    "type": "donut",
    "name": "Cake",
    "ppu": 0.55,
    "batters": {
        "batter": [
                { "id": "1001", "type": "Regular" },
                { "id": "1002", "type": "Chocolate" },
                { "id": "1003", "type": "Blueberry" },
                { "id": "1004", "type": "Devil's Food" }
            ]
    },
    "topping": [
        { "id": "5001", "type": "None" },
        { "id": "5002", "type": "Glazed" },
        { "id": "5005", "type": "Sugar" },
        { "id": "5007", "type": "Powdered Sugar" },
        { "id": "5006", "type": "Chocolate with Sprinkles" },
        { "id": "5003", "type": "Chocolate" },
        { "id": "5004", "type": "Maple" }
    ]
}

I am trying to pass an xpath as a variable.

$(document).ready(function(){
    var json_offset = 'topping.types'
    ...
    $.getJSON('json-data.php', function(data) {
        var json_pointer = data.json_offset;
        ...
    });
});

Which dosn’t work. Can anyone help?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-25T06:27:18+00:00Added an answer on May 25, 2026 at 6:27 am

    Something like that should work (I didn’t actually test it, tho):

    Object.getPath = function(obj, path) {
        var parts = path.split('.');
        while (parts.length && obj = obj[parts.shift()]);
        return obj;
    }
    // Or in CoffeeScript:
    // Object.getPath = (obj, path) -> obj=obj[part] for part in path.split '.'; obj
    

    Than, use it like that:

    Object.getPath(data, json_offset)
    

    However, unless the path is dynamic and you can’t, you should simply use data.topping.types. Also, you referred to that path as “XPath”, but XPath is a very different thing that has nothing to do with you’re trying to do.

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