从XML文件到datagridview c#

Question

我使用的api返回XML中的文本，并将其保存到XML文件中。每当我试图将信息显示到数据视图时。但是，我收到一个错误，说该文件已经打开并正在使用。以下是接收文本的代码，将其保存到XML并尝试将其显示给数据存储。

     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];

将数据传递给dataGridView的简单方法

Answer 2

通常，当这种情况发生时，在代码中查找其他也打开、读取、写入或处理该文件的内容，并关闭它们！

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