title: Difficulty with jQuery's AJAX response from PHP
body: <p>I am fairly new to AJAX. I am trying to get a simple login script to work. This is using jQuery 1.6.4. In the AJAX below, when the user clicks a button, it sends the email address and password to login.php. 
  That all seems to work fine. The trouble is with the success function. When the email and password are correct, it should return true. It should return false when it does not work. Using Firebug, I see that it works with console.log. It also works correctly if I write alert(response);. However, the conditional always evaluates to false even when response is equal to true. I've tried both     <code>if(response=="true")</code> and <code>if(response==="true")</code>, putting the variable outside the function, and a few other things without success. Would anyone have any ideas on how to fix this?</p>
  
  <p>Thank you for any help or ideas,
  Jason.</p>
  
  <p>AJAX:</p>
  
  <pre><code>$("#firstpage").live('pageinit', function (evt) {
  $('#button').click(function(){       
  var $form = $('#login'),
  $inputs = $form.find("input"),
  serializedData = $form.serialize();
  $.ajax({
    type: 'POST',
    url: 'php/login.php',
    data: serializedData,
    success: function(response){
      console.log("Response: "+response);
      if(response=="true") 
      {
  $('#firstpage #wrong').text("Login script is working");
  } else {
  $('#firstpage #wrong').text("Your email and password combination did not match.");
  }
  
      },      
    dataType: 'json'
  });
  });  
  });
  </code></pre>
  
  <p>If it helps, this is my login.php script.</p>
  
  <pre><code>$email = $_POST['email'];
  $password = $_POST['password'];
  require_once("DB.php");
  $connection = mysql_connect($host, $user, $pass) or die ("Unable to connect!"); 
  mysql_select_db($db) or die ("Unable to select database!"); 
  
  $query = "SELECT * FROM member WHERE email='$email' AND password='".md5($_POST['password'])."'";
  $result = mysql_query($query) or die ("Error in query: $query. ".mysql_error()); 
  
  $num_rows = mysql_num_rows($result);
  if($num_rows&gt;0){
  $output = true;
  } else {
  $output = false;
  }
  echo json_encode($output);
  </code></pre>