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Home/ Questions/Q 8016679
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T20:36:27+00:00 2026-06-04T20:36:27+00:00

I need to load database content according to the following scheme into the variable

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I need to load database content according to the following scheme into the variable data.

var data = {
    "62": {
        sku: "62",
        section: "bodyImage",
        img: "images/diy-images/config-images/62.png",
        label: "plain red",
        price: "100"
    },
    "63": {
        sku: "63",
        section: "bodyImage",
        img: "images/diy-images/config-images/63.png",
        label: "plain pink",
        price: "110"
    },
    "360": {
        sku: "360",
        section: "bodyImage",
        img: "images/diy-images/config-images/360.png",
        label: "plain gray",
        price: "120"
    },
};​

I tried to achieve that with the following function but it doesn’t work out. What am I missing?

var data = (function() {
    $.ajax({
        url: 'get_data.php',
        data: "",
        dataType: 'json',
        success: function(rows) {
            for (var i in rows) {
                var row = rows[i];

                var id = row[0];
                var section = row[1];
                var img = row[2];
                var label = row[3];
                var price = row[4];
            }
        }
    });
});​
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-04T20:36:29+00:00Added an answer on June 4, 2026 at 8:36 pm
    var data = {}; // data variable for future use
    $.ajax({
     url: '',
     data: '',
     dataType: 'json',
     success: function(rows) {
       $.each(rows, function(i, val) {
          var id = val.sku,
              section = val.section,
              img = val.img,
              label = val.label,
              price = val.price;
       });
       // if you want to update data variable then
       data[sku] = {sku: sku, img: img, section: section, label: label, price: price};
      // to set data to a content
       $('#output_div').html(JSON.stringify(data));
     }
    })
    
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