php - Unexpected end of JSON input - Uncaught SyntaxError -

everybody. have ajax call returns me error mentioned in title. believe line causes error: var obj = jquery.parsejson(data); maybe wrongly constructed userdata.

this jquery:

var userdata = 'email=' + email + '&password=' + password;   $.ajax({     type: 'post',     url: './api/getinfo.php',     data: userdata,     success: function(data){         var obj = jquery.parsejson(data);           $('#name').html(obj.firstname + ' ' + obj.lastname);         ...     },     error: function(){         alert('error');     } }); 

and getinfo.php:

if($_server['request_method'] == 'post') {      $email = prepareinput($_post['email']);      $password = prepareinput($_post['password']);       $stmt = $connection->conn->prepare('select firstname,lastname,... tb_users email = ? , password = ?');      $stmt->bind_param('ss',$email,$password);       $stmt->execute();       $result = $stmt->get_result();       $obj = $result->fetch_assoc();       echo json_encode($obj); } 

can tell me if i'm doing wrong?

update

function prepareinput($data) {     $data = trim($data);     $data = stripslashes($data);     $data = htmlspecialchars($data);      return $data; } 

data passed php empty if $obj contains values (i checked echoing them) must problem echo json_encode($obj); statement.

solution

i found answer - link. encoding problem. if strings not utf-8 json_encode() return empty string deal need convert these strings utf-8.

since mention: success function not receive data

it seems server response empty, because (a) credentials wrong or (b) there error in php , code never reaches echo line.

add error handling / confirmation in php , make sure $obj has value before return it.

try {   // ... sql code    if ( $obj && is_array( $obj ) ) {     echo json_encode( $obj );   } else {     echo json_encode( array( 'error' => 'wrong username or password' ) );   } } catch (exception $ex) {   echo json_encode( array( 'error' => $ex->getmessage() ) ); } 

update; notes debugging sql

to test sql:

1. as first line in condition (method == post) add echo json_encode(array('resp'=>'test')); exit; , run ajax code. console.log() should display {resp:test} json object. way know ajax call reaches part sql. if still not output else wrong in code...

2. first use hardcoded sql: enter email/password in know give result. not use ->bind_param() or $_post data, execute plain sql statement , see if ajax returns value.

3. if ajax works now, modify hardcoded sql move static email/password string bind_param(). still not using $_post data now, check if bind_param method works. again check if ajax still working.

4. if still working, use direkt $_post data in bind_param() call, ->bind_param('ss', $_post['email'], $_post['password']) - if working, then know there's problem prepareinput() function.

when testing each step, should find out, part not working.