XML文件到datagridview C＃

Question

我正在使用一个API，该API返回以XML格式保存的XML文件中的文本。 每当我尝试将信息显示到datagridview时。 但是我收到一个错误消息，说该文件已经打开并正在使用。 这是接收文本，将其保存到XML并尝试将其显示到数据网格的代码。

     using (WebResponse response = request.GetResponse())
        {
            using (Stream stream = response.GetResponseStream())
            {
                using (StreamReader sr99 = new StreamReader(stream))
                {
                    responseContent = sr99.ReadToEnd();
                }
            }
        }
        XmlDocument doc = new XmlDocument();
        doc.LoadXml(responseContent);
        XmlWriterSettings settings = new XmlWriterSettings();
        settings.Indent = true;
        // Save the document to a file and auto-indent the output.
        XmlWriter writer = XmlWriter.Create("ResponseContent.xml", settings);
        doc.Save(writer);



        XmlReader xmlFile = XmlReader.Create(@"C:\Users\Tyler\Documents\Repo\New Trunk\WalmartSmiles\WalmartSmiles\bin\Debug\ResponseContent.xml", new XmlReaderSettings());
        DataSet dataSet = new DataSet();
        //Read xml to dataset
        dataSet.ReadXml("ResponseContent.xml");
        //Pass empdetails table to datagridview datasource
        dataGridView1.DataSource = dataSet.Tables["ns2:feed"];
        //Close xml reader
        xmlFile.Close();

Answer 1

XmlReader xmlFile ;
xmlFile = XmlReader.Create("Product.xml", new XmlReaderSettings());
DataSet ds = new DataSet();
ds.ReadXml(xmlFile);
dataGridView1.DataSource = ds.Tables[0];

将数据传递到dataGridView的简单方法

Answer 2

通常，发生这种情况时，请在代码中查找其他也已打开，读取，写入或对该文件执行任何操作的其他事物，然后将其关闭！

    // Save the document to a file and auto-indent the output.
    XmlWriter writer = XmlWriter.Create("ResponseContent.xml", settings);
    doc.Save(writer);
    writer.Close();