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up vote 0 down vote favorite

I am trying to send a php array to ajax but it does not work. To be honest I do not know what am I doing wrong.

I am using json_encode() which returns null.

my php code:

$info = array();
$info['NEW YORK CITY'] = array(
'Name' => 'New York City'
);

$city = $_POST['city'];

if (strtoupper($city) === 'NEW YORK CITY') {

echo "Name: " . json_encode($info['NEW YORK CITY']['Name']) . ".<br>";
} else {
echo "error.";
}

my ajax code:

$('form.ajax').on('submit', function() {
var that = $(this),
    url = that.attr('action'),
    type = that.attr('method'),
    data = {};

    that.find('[name]').each(function(index, value) {
        var that = $(this),
            name = that.attr('name'),
            value = that.val();

            data[name] = value;

    });

    //console.log(data);

    $.ajax({
        url: url,
        type: type,
        data: data,
        success: function(response) {
            //console.log(response);

            $('form.ajax').html(response);
        }
    }).fail(function(jqXHR) {
        alert(jqXHR.statusText);
    });

return false;
});

Fixed it! I had the json_encode before the array. It works now that I put the array on top.

share|improve this question
    
Can i see the print_r($info);? – aldrin27 Sep 6 at 6:30
    
$info = array(); $info['NEW YORK CITY'] = array( 'Name' => 'New York City', 'Rank' => '1st, US', 'Population' => '8,491,079' ); – Rony Alvarez Sep 6 at 6:55

3 Answers 3

up vote 0 down vote

json_encode accepts an array as an argument

Such that it will take the following

array('key' => 'value')

This will be sent as properly formated json key:value. But a single value will not be handled correctly and may lead to unwanted results.

share|improve this answer
2  
json_encode's first argument isn't limited to arrays. A string or a number is a perfectly acceptable input. – curtis1000 Sep 6 at 6:02
    
Though correct, it's much more straight forward to keep up with the practice, and is usually advised. My mistake though. To the asking user or any other viewers. here is a link to the documentation for json_encode() php.net/manual/en/function.json-encode.php – Andrew Mata Sep 6 at 6:05
up vote 0 down vote

replace your php code with the following

$city = $_POST['city'];
 if (strtoupper($city) === 'NEW YORK CITY') {
     echo json_encode($info['NEW YORK CITY']);
 } else {
     echo json_encode(array("error."));
 }
share|improve this answer
    
I get an internal server error. – Rony Alvarez Sep 6 at 6:39
up vote 0 down vote

Try this if this works:

  $city = $_POST['city'];

   if (strtoupper($city) === 'NEW YORK CITY') {
    echo json_encode(['Name' => $info['NEW YORK CITY']['Name'] ]);
   } else {
     echo json_encode(['error']);
   }

Ajax:

 $('form.ajax').on('submit', function() {
  var that = $(this),
  url = that.attr('action'),
  type = that.attr('method'),
  data = {};

  that.find('[name]').each(function(index, value) {
    var that = $(this),
        name = that.attr('name'),
        value = that.val();

        data[name] = value;

   $.ajax({
    url: url,
    type: type,
    data: data,
    success: function(response) {
        //console.log(response);

        $('form.ajax').html(response);
     }
     }).fail(function(jqXHR) {
     alert(jqXHR.statusText);
   });

    return false;

  });

});

Your ajax should be after the on submit then it will trigger the AJAX functions.

share|improve this answer

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