Load database content into variable

Question

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];
            }
        }
    });
});

score 1 · Answer 1 · edited Jun 02 '12 at 11:33

You can get json object value by '.' operator

var data = (function() {

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

                var id = row.sku;          
                var section = row.section;
                var img = row.img;
                var label = row.label;
                var price = row.price;

            }
        }
    });

});

score 0 · Answer 2 · answered Jun 02 '12 at 10:59

You can address the properties by name:

for (var i in rows) {
    var row = rows[i];          

    var id = row.sku;          // or var id = i;
    var section = row.section;
    var img = row.img;
    var label = row.label;
    var price = row.price;
    ...
}

thecodeparadox · Accepted Answer · 2012-06-02T12:18:25.127

0

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));
 }
})

edited Jun 02 '12 at 12:18

answered Jun 02 '12 at 11:01

thecodeparadox

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Thanks, however I still have problems with the output of the data. If I output them into a div I only get undefined values. I need to have the output exactly in the same way as in the example at the top of the page. – Stefan Schneider Jun 02 '12 at 11:53
The only output I get is {}. Maybe something wrong with the mysql query: $result = mysql_query("SELECT * FROM table"); $data = array(); while ( $row = mysql_fetch_row($result) ) { $data[] = $row; } echo json_encode( $data ); – Stefan Schneider Jun 02 '12 at 14:17

Load database content into variable

3 Answers3