Using AJAX and JSON to display HTML content from PHP

Question

I am trying to retrieve information from my php file which contains the following code. My code contains the colums Email, FirstName, LastName, and State.

$query = 'SELECT * FROM users WHERE LOWER(Email) = :email';
$stmt = $dbh->prepare($query);
$stmt->bindValue(':email', $email);
$stmt->execute();

if ($stmt->rowCount() == 1) {
    $row = $stmt->fetch(PDO::FETCH_ASSOC);
    $firstName = $row['FirstName'];
    $lastName = $row['LastName'];
    $state = $row['State'];     
} echo json_encode($row);

My Ajax code is:

$.ajax({
    datatype: 'json',
    type: "POST",
    url: 'json-data.php',         
      success: function(data) {
        //called when successful
        $('#firstname').append(data.FirstName);     
      },
      error: function(e) {
        //called when there is an error
        //console.log(e.message);
      }

});

When I type $('#firstname').append(data);, it shows me the following output:

{"FirstName":"Foo","LastName":"Bar","State":"Florida","Email":"[email protected]"}

How do I make it so I can get only the first name and append it to a div?

It doesn't display when I append using $('#firstname').append(data.FirstName); but it does when I use $('#firstname').append(data); but this method outputs the entire array when I am only looking for one element. — Bhavik P.
– Bhavik P., Commented Jan 30, 2013 at 23:18
I found my error, instead of dataType, I was using datatype... >.< — Bhavik P.
– Bhavik P., Commented Jan 31, 2013 at 4:52

vincenzodb · Accepted Answer · 2013-01-30 22:29:26Z

1

try with:

var obj = jQuery.parseJSON(data);
$('#firstname').append(OBJ.FirstName);

answered Jan 30, 2013 at 22:29

vincenzodb

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2 Comments

Bhavik P. Over a year ago

Had to make an edit to have it work but it did. Thank you very much. $('#firstname').append(obj.FirstName); Is there a way to do it without parsing it again like Sedz said?

Bhavik P. Over a year ago

I found my error, instead of 'dataType', I was using 'datatype'

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