c# - 从连接到远程 MySQL 数据库的 PHP 服务检索/获取数据以显示到我的 ASP.Net Web 应用程序

Question

所以我有一个远程(托管)MySQL 数据库，我通过 PHP 服务连接到该数据库。我需要我的 ASP.Net c# web 应用程序和我的 android 来与之通信。但是，我正在努力使用从服务中检索到的所有信息来填充我的 Web 应用程序模板。例如，我想填充用户的个人资料页面。

下面是我的 PHP 连接和与数据库的通信:

    `// Create connection
         $conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT * FROM Vendor Where VendorId = 2"; //this is just a test
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        echo . $row["id"]. . $row["BusinessName"]. . $row["Location"]. . $row["Email"]. .$row["Website"]. .$row["ProductType"]. .$row["Contact"]. .$row["PaymentOption"]. .$row["ProfileImg"]."<br>";
    }
} else {
    echo "0 results";
}
$conn->close();
` 

然后(不共享我的所有设置)这将是 asp.net c# 与我的 PHP 文件/服务通信的代码示例。

public void getUserInfo(int id)
        {


            string BusinessName = lblBusiness.Text.Trim();
            string email = lblEmail.Text.Trim();
            string Contact = lblPhone.Text.Trim();

            string location = lblLocation.Text.Trim();
            string Website = lblWebsite.Text.Trim();

            string payment = lblPayment.Text.Trim();

            //Variables to get information from the service
            Stream dataStream = null;
            WebResponse response = null;
            StreamReader reader = null;
            //Stores the result from the server
            string responseFromServer = null;
            try
            {
                string requestMethod = "GET";
                //Sending this data across the stream
                string postData = "&Email=" + email +   "&BusinessName=" + BusinessName + "&Website=" + Website + "&PaymentOption=" + payment + "&Location=" + location + "&ProductType=" + ProductType + "&Contact=" + Contact + "";
                byte[] byteArray = Encoding.UTF8.GetBytes(postData);
                string URL = "";// url of php service location
                string contenttype = "application/x-www-form-urlencoded";
                //Create link to web service
                WebRequest request = WebRequest.Create(URL);
                //Pass the request method
                request.Method = requestMethod;
                request.ContentType = contenttype;
                request.ContentLength = byteArray.Length;
                dataStream = request.GetRequestStream();
                dataStream.Write(byteArray, 0, byteArray.Length);
                //Get response from the server
                response = request.GetResponse();
                dataStream = response.GetResponseStream();
                reader = new StreamReader(dataStream);
                responseFromServer = reader.ReadToEnd();
            }
            catch (WebException ex)
            {
                Console.WriteLine(ex.ToString());
            }
            finally
            {
                if (dataStream != null && reader != null && response != null)
                {
                    dataStream.Close();
                    reader.Close();
                    response.Close();
                }
                //Getting the response from the service
                //string result = responseFromServer.ToString();

            }

        }

另外，我不确定从该函数返回什么。请帮忙。

Answer 1

您的 php 文件是我按的“API”。
你基本上需要从你的 php 文件中返回它。

$vendorArray = [
    "Id" => $row["id"],
    "BusinessName" => $row["BusinessName"],
    // ... this is just pseudo code, convert the array or take the array or something like that
];

header('Content-type: application/json');
echo json_encode($vendorArray);

然后在 asp.net 中你做:

var deserializedVendor = JsonConvert.DeserializeObject<Vendor>(responseFromServer);

您的供应商类必须与您的 jsonObject 匹配才能反序列化

public class Vendor {

    public string Id {get;set;}

    public string BusinessName {get;set;}

    ...
}

这取决于你的 jsonResponse...

您还可以直接反序列化为具有供应商列表的复杂项目，如下所示:

var allTheVendorsDeserialized = JsonConvert.DeserializeObject<AllTheVendors>(responseFromServer);


public class AllTheVendors {

    public bool Success {get;set}

    public List<Vendor> {get;set}

}

哪里 php:

$arr = ["Success" => true, $myArrayOfVendors];

header('Content-type: application/json');
echo json_encode($arr);