在Xamarin中,无法从我的php数据库中获取特定值,我是否需要DeserializeObject?

时间:2016-03-01 00:08:14

标签: c# php xamarin

这是我获取数据的地方。

        $contacts = array(); 

        while ($row = mysqli_fetch_array($stmt)) 
        {

            $contact = array("ID" => $row['ProduktID'],
                             "Name" => $row['ProduktNamn'],
                             "Number" => $row['ProduktPris']);

            array_push($contacts, $contact);
        }

        echo json_encode(array('results' => $contacts), JSON_PRETTY_PRINT);

现在,当我在Xamarin的代码中使用它时,我可以在日志中写出整个内容,但是当我尝试写出某个表的某个“名称”时却没有。这是代码:

static public async Task<JObject> getOurMainInfo()
    {
        var httpClientRequest = new HttpClient ();

        var result = await httpClientRequest.GetAsync ("http://localhost/getmystuff.php");

        var resultString = await result.Content.ReadAsStringAsync ();

        var jsonResult = JObject.Parse (resultString);

    //  var jsonResult = JsonConvert.DeserializeObject<Contact>(resultString); //Should I implement this somehow to my JObject? Contact is my public class.

        System.Diagnostics.Debug.WriteLine (jsonResult["Name"]); //i get nothing in the log. if i remove ["Name"] i get the entire data in the log.

        return jsonResult;

    } 

我的公共课(如果我需要这个以某种方式连接和使用edatbase):

public class Contact
{
    public string ID { get; set; }
    public string Name { get; set; }
    public string Number { get; set; }
}

0 个答案:

没有答案