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Home/ Questions/Q 3844548
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Editorial Team
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Editorial Team
Asked: May 19, 20262026-05-19T16:06:24+00:00 2026-05-19T16:06:24+00:00

I think I’m missing something here: Using AjAX I get some data from a

  • 0

I think I’m missing something here:

Using AjAX I get some data from a database and send it back in JSON format
$jsondata = array();

while ($Row = mysql_fetch_array($params))
{

    $jsondata[]= array('cat_id'=>$Row["cat_id"], 
                          'category'=>$Row["category"], 
                     'category_desc'=>$Row["category_desc"],
                     'cat_bgd_col'=>$Row["cat_bgd_col"]);
};

echo("{\"Categories\": ".json_encode($jsondata)."};");

No problem so far I think.

On the cleint side I receive back the above into

ajaxRequest.responseText

and if I do this

var categoriesObject = ajaxRequest.responseText; 
alert(categoriesObject);

I see what I expect to see ie the entire array in the alert.

Where it all goes wrong is trying to access the response. The error I get is that the “categoriesObject” is not an object – if not what is it? what’s bugginh me is that I can’t even access it like this:

document.write(categoriesObject.Categories[0].category);

so what am I doing wrong?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-19T16:06:25+00:00Added an answer on May 19, 2026 at 4:06 pm
    1. You should not create JSON manually. Use:

      echo json_encode(array('Categories' => $jsondata));
      

      or just

      echo json_encode($jsondata);
      

      I don’t see a reason to add Categories.

    2. You have to decode the JSON on the client side, using JSON.parse (available in most browsers, but also available as script):

      var data = JSON.parse(ajaxRequest.responseText);
      
    3. If you want to be very correct, add

      header('Content-type: application/json');
      

      to your PHP script.

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