Проблема с передачей переменной в PHP SQL скрипт из Swift

Question

Что я должен добавить к своему коду Swift, чтобы передать переменную в следующий скрипт PHP?Мой PHP-скрипт ожидает значение customerName в $ID = $_POST['customerName'];

Как я могу жестко закодировать значение в своем коде Swift для отправки значения?

Код Swift:

import Foundation

protocol FeedDetailProtocol: class {
    func itemsDownloaded(items: NSArray)
}


class FeedDetail: NSObject, URLSessionDataDelegate {



    weak var delegate: FeedDetailProtocol!



func downloadItems() {


    let url = URL(string: "https://www.example.com/test/test1.php")!
    let defaultSession = Foundation.URLSession(configuration: URLSessionConfiguration.default)
    var request = URLRequest(url: url)
    request.setValue("application/x-www-form-urlencoded", forHTTPHeaderField: "Content-Type")
    request.httpMethod = "POST"**strong text**
    let parameters: [String: Any] = ["customerName": "John"]

    request.httpBody = parameters.percentEscaped().data(using: .utf8)

    let task = defaultSession.dataTask(with: url) { (data, response, error) in

        if error != nil {
            print("Error")
        }else {
            print("details downloaded")
            self.parseJSON(data!)
        }

    }

    task.resume()
}

    func parseJSON(_ data:Data) {

        var jsonResult = NSArray()

        do{
            jsonResult = try JSONSerialization.jsonObject(with: data, options:JSONSerialization.ReadingOptions.allowFragments) as! NSArray

        } catch let error as NSError {
            print(error)

        }

        var jsonElement = NSDictionary()
        let stocks = NSMutableArray()

        for i in 0 ..< jsonResult.count
        {

            jsonElement = jsonResult[i] as! NSDictionary

            let stock = DetailModel()

            //the following insures none of the JsonElement values are nil through optional binding
            if let rma = jsonElement["rma"] as? String,
                let customer = jsonElement["customer"] as? String,
                let manufacturer = jsonElement["manufacturer"] as? String,
                let model = jsonElement["model"] as? String

            {

                stock.rma = rma
                stock.manufacturer = manufacturer
                stock.model = model
                stock.customer = customer

            }

            stocks.add(stock)

        }

        DispatchQueue.main.async(execute: { () -> Void in

            self.delegate.itemsDownloaded(items: stocks)

        })
    }
}

PHP скрипт:

$con=mysqli_connect("localhost","username","password","dbName");

// Check connection
if (mysqli_connect_errno())
{
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

// This SQL statement selects ALL from the table 'Equipment'

$ID = $_POST['customerName'];

$sql = "SELECT customer, rma, manufacturer, model, status FROM Equipment 
        WHERE customer = '$ID' ";

// Check if there are results
if ($result = mysqli_query($con, $sql))
{
    // Create temporary connection
    $resultArray = array();
    $tempArray = array();

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

    // Finally, encode the array to JSON and output the results
    echo json_encode($resultArray);
}

mysqli_close($con);

Answer 1

Ваш PHP-код принимает значение POST ($ ID = $ _POST ['customerName']), поэтому вам нужно создать запрос POST.

Примерно так:

let url = URL(string: "https://www.example.com/test1/test1.php")!
var request = URLRequest(url: url)
request.setValue("application/x-www-form-urlencoded", forHTTPHeaderField: "Content-Type")
request.httpMethod = "POST"
let parameters: [String: Any] = ["customerName": "John"]`

HTTP-запрос в Swift с методом POST