向PHP web服务器发送到MySQL的Swift JSON请求

我是一个初学者，正在做一个学习项目。我想通过php文档将swift变量传输到我的web服务器。这意味着对PHP的快速请求

这是我的Swift代码：

  let myUrl = URL(string: "http://example.net/stock_service3.php");

        var request = URLRequest(url:myUrl!)

        request.httpMethod = "POST"// Compose a query string

        let postString = "firstName=51&lastName=6";

        request.httpBody = postString.data(using: String.Encoding.utf8);

        let task = URLSession.shared.dataTask(with: request) { (data: Data?, response: URLResponse?, error: Error?) in

            if error != nil
            {
                print("error=\(String(describing: error))")
                return
            }

            // You can print out response object
            print("response = \(String(describing: response))")

            //Let's convert response sent from a server side script to a NSDictionary object:
            do {
                let json = try JSONSerialization.jsonObject(with: data!, options: .mutableContainers) as? NSDictionary

                if let parseJSON = json {

                    // Now we can access value of First Name by its key
                    let firstNameValue = parseJSON["firstName"] as? String
                    print("firstNameValue: \(String(describing: firstNameValue))")




                }
            } catch {
                print(error)
            }
        }
        task.resume()

这是PHP和MySQL代码：

<?php

// Create connection
$con=mysqli_connect(".......".........".........");

// Check connection
if (mysqli_connect_errno())
{
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$firstName= $_REQUEST["firstName"];
$lastName = $_REQUEST["lastName"];

// Select all of our stocks from table 'stock_tracker'
$sql ="SELECT m.*


      , ( ACOS( COS( RADIANS( $firstName) ) 
              * COS( RADIANS( m.latitude ) )
              * COS( RADIANS( m.longitude ) - RADIANS( $lastName) )
              + SIN( RADIANS($firstName ) )
              * SIN( RADIANS( m.Latitude) )
          )
        * 6371
        ) AS distance_in_km

  FROM TankBilliger m
  HAVING distance_in_km <= 100
 ORDER BY distance_in_km ASC
 LIMIT 100";

// Confirm there are results
if ($result = mysqli_query($con, $sql))
{
    // We have results, create an array to hold the results
        // and an array to hold the data
    $resultArray = array();
    $tempArray = array();

    // Loop through each result
    while($row = $result->fetch_object())
    {
        // Add each result into the results array
        $tempArray = $row;
        array_push($resultArray, $tempArray);
    }

    // Encode the array to JSON and output the results

    echo json_encode($resultArray);
}

// Close connections
mysqli_close($con);
?>

---

不知何故，它不起作用，我也不知道为什么。

“不知何故，它不起作用，我也不知道为什么”什么是“它”？哪一行？您需要更改它，这样，如果它无法连接或运行查询，它仍然会返回有效的JSON。目前，您只是重复一个错误，因此无法将其解析为JSON。请阅读-总结是，这不是解决志愿者问题的理想方式，可能会对获得答案产生反作用。请不要将此添加到您的问题中。