从 PHP Web 服务未正确返回编码响应

Question

我编写了简单的 PHP 代码来从我的 iphone 应用程序中注册用户详细信息。它工作正常并返回 JSON 输出。我在下面添加了该代码

header('Content-type: application/json');
include 'connection.php';
$response = array();

$username = $_POST['username'];
$password = $_POST['password'];
$email = $_POST['email'];

if($username == NULL || $password == NULL || $email == NULL ){

    $response["success"] = 0;
    $response["message"] = "Something Empty";

}else{

    $sql = "INSERT INTO User (username, password, email)
VALUES ('".$username."', '".$password."', '".$email."')";

if ($conn->query($sql) === TRUE) {

    $response["success"] = 1;
    $response["message"] = "Done";

} else {

    $response["success"] = 0;
    $response["message"] = "Error";
}
}

echo json_encode($response);
$conn->close();

但是当我尝试在添加到表之前检查用户名已经存在时。我从我的 Xcode 日志中收到错误消息。

JSON text did not start with array or object and option to allow fragments not set.

if($username == NULL || $password == NULL || $email == NULL ){

    $response["success"] = 0;
    $response["message"] = "Something Empty";

}else{

   $query = mysql_query("SELECT * FROM User WHERE username='".$username."'");


    if (mysql_num_rows($query) != 0)
  {
      $response["success"] = 0;
      $response["message"] = "Username Already Exists";

  }else{

      $response["success"] = 1;
      $response["message"] = "That Name Fine";
  }


}

Answer 1

很可能是 php 错误（可能只是通知或警告，具体取决于您的 error_reporting 设置）污染了 json 输出。

您需要调试问题，并且可以考虑将error_reporting 级别用于生产。