Question

I have a problem with handling json variables in javascript.

The problem appears when I try to post data to a php class and then it should return 3 variables using JSON, but Firebug + FirePHP told me that the JSON variable returned correctly. Then when when I try to store each of them in different variables (3 returned json variables in 3 javascript variables) I just get the first parameter and the other two are NULL.

You can see in next javascript code:

function UpdateContact(ID)
{
    // get current Contact data
    var full_name,email,mobile;
    $.post("/Cards/index.php/Cont/GetContactInfo",{'id' : ID},
       function(data){
                full_name = data.full_name;
                email = data.email;
                mobile = data.mobile;
                   $( "#dialog2-form" ).html(
    '<p class="validateTips">All form fields are required.</p>'+
    '<form>'+
    '<fieldset>'+
        '<label for="name">Name</label>'+
        '<input type="text" name="Updatefull_name" id="Updatefull_name" class="text ui-widget-content ui-corner-all" value="'+full_name+'" /><br />'+
        '<label for="email">Email</label>'+
        '<input type="text" name="Updateemail" id="Updateemail" value="" class="text ui-widget-content ui-corner-all" value="'+data.email+'"/><br />'+
                '<label for="email">Mobile</label>'+
        '<input type="text" name="Updatemobile" id="Updatemobile" value="" class="text ui-widget-content ui-corner-all" value="'+data.mobile+'"/><br />'+
    '<input type="hidden" id="UpdateContactID" value="'+ID+'"></fieldset>'+
    '</form>'
);
   $( "#dialog2-form" ).dialog( "open" );
 }, "json");
}

Just the full_name variable is getting a value, but email and mobile variables do not get any value.

The PHP function is:

function GetContactInfo()
{
    $Contact = $this->Contacts->GetContacByID($this->input->post('id'));
    $data['full_name'] = $Contact->full_name;
    $data['email'] = $Contact->email;
    $data['mobile'] = $Contact->mobile;
    echo json_encode($data);
}

Note: I use CodeIgniter PHP Framework on my XAMPP localhost server

Answer 1

You need to check that the fields are present before using them:

var field_to_read = some_default_value;
if (typeof data['name_of_field'] !== 'undefined') {
   field_to_read = data['name_of_field'];
}

I suggest you use a construct similar to the above for reading your JSON data.