AJAX没有从php返回变量

Question

I know there is a few questions like this on here. but I have done a lot of researching and bug fixing all day to try work out why my ajax does not return a response from the php file. All I want is for it to tell me a user has been registered so I can let the user move on with the signing up process. And I just need someones wise guidance to tell me what I am doing wrong!!

so I wont bore you with the validation part of the js file just the ajax

  if(ValidationComplete == true){
   var that = $(this),
url = that.attr('action'),
type = that.attr('method'),
data = {};

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

        data[name] = value; 
});

    $.ajax({
    url:url, 
    type:type,
    data: data,
    dataType: 'json', 
    success: function(result){
        alert(result.status);
        console.log(result.data);

    },
error: function(xhr, textStatus, error){
  console.log(xhr.statusText);
  console.log(textStatus);
  console.log(error);
   }        

});

return false;
} else {
    return false;
}

currently with this, if I remove the dataType bit the alert bit happens but currently with it there nothing does.

again I will just skip to the chase on the php file

    $query = "INSERT INTO person 
   VALUES('','$first_Name','$surname','$email','$dob','$password',
  '1','','0','1','','','','$emailCode')";
  if($query_run =mysql_query($query)) {
      echo json_encode(array("response"='true'));

Any help would be amazing!!!!!

updated code:

  <?php    

if( isset($_POST['firstname']) && 
isset($_POST['surname']) && 
isset($_POST['email']) && 
isset($_POST['day']) && 
isset($_POST['month']) && 
isset($_POST['year']) && 
isset($_POST['password']) && 
isset($_POST['re_type_password'])){

  $first_Name = $_POST['firstname'];
  $surname = $_POST['surname'];
  $email = $_POST['email'];
$password = $_POST['password'];
$day = $_POST['day'];
$month = $_POST['month'];
$year = $_POST['year'];
$re_type_password = $_POST['re_type_password'];
$emailCode = md5($_POST['$first_Name'] + microtime());

if(!empty($first_Name)&& 
!empty($surname)&& 
!empty($email)&& 
!empty($day) && 
!empty($month) && 
!empty($year) && 
!empty($password)&&                                   
!empty($re_type_password)){



  if(strlen($firstname)>30 || strlen($surname)>30 || strlen($email)>50){
  echo 'the data enetered is to long';
} else {
  if($password != $re_type_password){
  echo 'passwords do not match, please try again.';
} else{ 
$query = "SELECT email FROM person WHERE email ='$email'";
$query_run = mysql_query($query);
if(mysql_num_rows($query_run)==1){
  echo 'Email address already on databse';         
} else{
  if($day>31 || $month>12){
    echo 'date of birth wrong';
  } else{

    $dob= $year.'-'.$day.'-'.$month;
  $query = "INSERT INTO person 
  VALUES('','$first_Name','$surname','$email','$dob','$password'
   ,'1','','0','1','','','','$emailCode')";
   if($query_run =mysql_query($query)) {
      email($email, 'Email Confirmation', "hello ". $first_Name." ,
    \n\n you need to   activate your account so click the link  ");
    $return_data['status'] = 'success';
     echo json_encode($return_data);
  } else {
      echo @mysql_error();
  }

}

 }

  } 
  }

  } else {
      echo "<p id='error'> All fields are required. Please try again.</p>";

  }
      }

   ?>

  <?php
  } else if (loggedIn()) {

echo 'you are already registed and logged in';

   }


  ?>

   </body>
  </html>

Answer 1

the last line it should be

echo json_encode(array("response"=>'true'));

see the added > in the array declaration, that is used to assign arrays with keys.

also in general you should put a error capture in your ajax statement, see this answer for more info

EDIT: Ok wow, that's some spaghetti code you have there, but after a little clean-up your problem is too many closing braces } you have to remove the } just before the following line also get rid of the closing and opening tags around this line, they serve no use.

} // <------- THIS ONE!

} else if (loggedIn()) {

    echo 'you are already registed and logged in';

}

I should also mention two other issues with your code

1. You are accepting input from the user without cleaning it up and testing it properly. This is no no read here to find out more
2. You are using mysl_ functions, these are old and depreciated they are also security risks. Check out PDO instead

EDIT:

Add ini_set('error_reporting',1); to the top of your php script.