Question

我知道这可能很简单，我用谷歌搜索但仍然无法看到我的错误。

我的基本代码是：

public static void postToImgur(string imagFilePath, string apiKey, string apiSecret, string title, string description)
{
    byte[] imageData;
    FileStream fileStream = File.OpenRead(imagFilePath);
    imageData = new byte[fileStream.Length];
    fileStream.Read(imageData, 0, imageData.Length);
    fileStream.Close();

    const int MAX_URI_LENGTH = 32766;
    string base64img = System.Convert.ToBase64String(imageData);
    StringBuilder sb = new StringBuilder();
    for (int i = 0; i < base64img.Length; i += MAX_URI_LENGTH)
    {
        sb.Append(Uri.EscapeDataString(base64img.Substring(i, Math.Min(MAX_URI_LENGTH, base64img.Length - i))));
    }

    string uploadRequestString = "client_id" + apiKey + "client_secret" + apiSecret + "&title=" + title + "&imageTitle" + title + "&description" + description + "&caption=" + title + "img" + "&image=" + sb.ToString();

    HttpWebRequest webRequest = (HttpWebRequest)WebRequest.Create("https://api.imgur.com/3/upload.xml");
    // needs a space between
    webRequest.Headers.Add("Authorization", "Client-ID " + apiKey);
    webRequest.Method = "POST";
    webRequest.ContentType = "application/x-www-form-urlencoded";
    webRequest.ServicePoint.Expect100Continue = false;

    StreamWriter streamWriter = new StreamWriter(webRequest.GetRequestStream());

    streamWriter.Write(uploadRequestString);
    streamWriter.Close();

    WebResponse response = webRequest.GetResponse();
    Stream responseStream = response.GetResponseStream();
    StreamReader responseReader = new StreamReader(responseStream);
    string responseString = responseReader.ReadToEnd();

    //received response is a xml file. the link to the uploaded file is in between the tag< link ></ link >
    //using regular expression to retrive the link to the image.
    Regex regex = new Regex("<link>(.*)</link>");
    var test = regex.Match(responseString).Groups[1].ToString();
    //return regex.Match(responseString).Groups[1].ToString();
    dgView.Rows.Add(test);
}

private void btnPostToImgur_Click(object sender, EventArgs e)
{
    // validation

    postToImgur(txtBoxImageLocation.Text, "ccc8d227d", "5b64c6b63908aa3e51840db3ce874137d7", txtBoxTitle.Text, txtBoxDescription.Text);
}

我收到了错误

访问非静态方法或者需要对象引用属性

当我尝试将数据添加到dgView我的datagrid变量名称时，我们将不胜感激。

Answer 1

您的方法postToImgur是一种静态方法。来自msdn：

静态方法和属性无法访问非静态字段和包含类型的事件，他们无法访问实例任何对象的变量，除非它在方法中显式传递参数。

您已尝试使用

dgView.Rows.Add(test);

但是，您没有将DataGridView对象作为参数传递，并且变量不是静态的 - 因此，您无法访问静态方法中的dgView变量。

您的两个选择是：

使方法postToImgur非静态
将dgView变量传递给方法

添加到数据网格时的对象引用

1 个答案: