JSON中的返回值为Javascript格式PHP

Question

I have a form that submit data on a click via ajax it runs a php script. This bit is working what I want is for when the php runs it return a value for success or not but it seems to be failing and I can't see why.

This is the form

<div id="container"></div>
<form id="availability-check" class="appnitro"  method="post"  onsubmit="return false">
  <label class="description" for="element_1">Post Code </label>
  <input id="postcode" name="postcode" type="text" maxlength="10" value="" placeholder=" e.g. WC1  1AA"/>

  <input id="check-availability" class="art-button check-availability" type="submit" name="submit" value="Submit" />
</form>

This is the javascript

$(document).ready(function(){
  $("#availability-check").submit(function(){

  var getField = $("#postcode").val();
 $.ajax({
             url: '/index.php/postcodeupdate' , 

             type: 'POST',
             data: '{ postcode: getField }',

             dataType: "json",

             success: function(data){ 

               $('#container').append(data)      

             }
          });   
  });
});

And this is the php

<?php

if(!empty($_POST['postcode'])){

$postcode=$_POST["postcode"];

$postcode= preg_replace('/\s+/', '', $postcode);

$db = JFactory::getDBO();

$query = "SELECT * FROM rex71_postcodes WHERE postcode='". $postcode . "'";
$db->setQuery($query);
$reply = $db->query();
$rowsnum = $db->getNumRows();

if($rowsnum>0){

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

echo json_encode(array('result' => '1'));

}

else{

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

echo json_encode(array('result' => '0'));

}

}
?>

Answer 1

data: '{ postcode: getField }',

This makes no sense. You want this:

data: { postcode: getField },

Then you will receive the value as a POST value and assuming your PHP code is correct everything will work.

Oh, and please read about SQL injection. Right now your code is very vulnerable.