如何将文件发布到远程网址?

发布于 2024-12-06 06:19:13 字数 4673 浏览 0 评论 0原文

我关注了有关将文件发布到远程网址的另一个问题: 上传带有 HTTPWebrequest (multipart/form-data) 的文件

它使我的 HttpHandler 中的远程 url 崩溃:“无法转换类型的对象‘System.String’来输入‘System.Web.HttpPostedFile’。”该方法是否发布了正确的文件数据,或者只是发布了一个字符串?

这是用于将文件发布到远程服务器的方法:

   public void HttpUploadFile(string url, string file, string paramName, string contentType, NameValueCollection nvc)
    {
        Response.Write(string.Format("Uploading {0} to {1}", file, url));
        string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
        byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");

        var wr = (HttpWebRequest)WebRequest.Create(url);
        wr.ContentType = "multipart/form-data; boundary=" + boundary;
        wr.Method = "POST";
        wr.KeepAlive = true;
        wr.Credentials = System.Net.CredentialCache.DefaultCredentials;

        var rs = wr.GetRequestStream();

        string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}";
        foreach (string key in nvc.Keys)
        {
            rs.Write(boundarybytes, 0, boundarybytes.Length);
            string formitem = string.Format(formdataTemplate, key, nvc[key]);
            byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
            rs.Write(formitembytes, 0, formitembytes.Length);
        }
        rs.Write(boundarybytes, 0, boundarybytes.Length);

        string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n";
        string header = string.Format(headerTemplate, paramName, file, contentType);
        byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
        rs.Write(headerbytes, 0, headerbytes.Length);

        var fileStream = new FileStream(file, FileMode.Open, FileAccess.Read);
        byte[] buffer = new byte[4096];
        int bytesRead = 0;
        while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
        {
            rs.Write(buffer, 0, bytesRead);
        }
        fileStream.Close();

        byte[] trailer = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "--\r\n");
        rs.Write(trailer, 0, trailer.Length);
        rs.Close();

        WebResponse wresp = null;
        try
        {
            wresp = wr.GetResponse();
            var stream2 = wresp.GetResponseStream();
            var reader2 = new StreamReader(stream2);
            Response.Write(string.Format("File uploaded, server response is: {0}", reader2.ReadToEnd()));
        }
        catch (Exception ex)
        {
            Response.Write("Error uploading file: " + ex.Message);
            if (wresp != null)
            {
                wresp.Close();
                wresp = null;
            }
        }
        finally
        {
            wr = null;
        }
    }

这是远程服务器上的 HttpHandler:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Collections.Specialized;

namespace SitefinityWebApp.Widgets.Files
{
    public class UploadFileHandler : IHttpHandler
    {
        public void ProcessRequest(HttpContext context)
        {
            context.Response.ContentType = "text/plain";

            //VALIDATE FILES IN REQUEST
            if (context.Request.Files.Count > 0)
            {
                try
                {
                    //HANDLE EACH FILE IN THE REQUEST
                    foreach (HttpPostedFile item in context.Request.Files)
                    {
                        item.SaveAs(context.Server.MapPath("~/Temp/" + item.FileName));
                        context.Response.Write("File uploaded");
                    }
                }
                catch (Exception ex)
                {
                    //NO FILES IN REQUEST TO HANDLE
                    context.Response.Write("Error: " + ex.Message);
                }
            }
            else
            {
                //NO FILES IN REQUEST TO HANDLE
                context.Response.Write("No file uploaded");
            }
        }

        public bool IsReusable
        {
            get
            {
                return false;
            }
        }
    }
}

这是使用 HttpUploadFile 方法的方法:

    var nvc = new NameValueCollection();
    nvc.Add("user", userName);
    nvc.Add("password", password);
    nvc.Add("library", libraryName);

    HttpUploadFile(destinationUrl, uploadFile, "file", "image/png", nvc);

I followed another question about posting a file to a remote url: Upload files with HTTPWebrequest (multipart/form-data)

It is making the remote url crash in my HttpHandler with: "Unable to cast object of type 'System.String' to type 'System.Web.HttpPostedFile'." Is the method posting the right file data, or is it posting just a string?

This is the method that is used to post the file to the remote server:

   public void HttpUploadFile(string url, string file, string paramName, string contentType, NameValueCollection nvc)
    {
        Response.Write(string.Format("Uploading {0} to {1}", file, url));
        string boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x");
        byte[] boundarybytes = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "\r\n");

        var wr = (HttpWebRequest)WebRequest.Create(url);
        wr.ContentType = "multipart/form-data; boundary=" + boundary;
        wr.Method = "POST";
        wr.KeepAlive = true;
        wr.Credentials = System.Net.CredentialCache.DefaultCredentials;

        var rs = wr.GetRequestStream();

        string formdataTemplate = "Content-Disposition: form-data; name=\"{0}\"\r\n\r\n{1}";
        foreach (string key in nvc.Keys)
        {
            rs.Write(boundarybytes, 0, boundarybytes.Length);
            string formitem = string.Format(formdataTemplate, key, nvc[key]);
            byte[] formitembytes = System.Text.Encoding.UTF8.GetBytes(formitem);
            rs.Write(formitembytes, 0, formitembytes.Length);
        }
        rs.Write(boundarybytes, 0, boundarybytes.Length);

        string headerTemplate = "Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"\r\nContent-Type: {2}\r\n\r\n";
        string header = string.Format(headerTemplate, paramName, file, contentType);
        byte[] headerbytes = System.Text.Encoding.UTF8.GetBytes(header);
        rs.Write(headerbytes, 0, headerbytes.Length);

        var fileStream = new FileStream(file, FileMode.Open, FileAccess.Read);
        byte[] buffer = new byte[4096];
        int bytesRead = 0;
        while ((bytesRead = fileStream.Read(buffer, 0, buffer.Length)) != 0)
        {
            rs.Write(buffer, 0, bytesRead);
        }
        fileStream.Close();

        byte[] trailer = System.Text.Encoding.ASCII.GetBytes("\r\n--" + boundary + "--\r\n");
        rs.Write(trailer, 0, trailer.Length);
        rs.Close();

        WebResponse wresp = null;
        try
        {
            wresp = wr.GetResponse();
            var stream2 = wresp.GetResponseStream();
            var reader2 = new StreamReader(stream2);
            Response.Write(string.Format("File uploaded, server response is: {0}", reader2.ReadToEnd()));
        }
        catch (Exception ex)
        {
            Response.Write("Error uploading file: " + ex.Message);
            if (wresp != null)
            {
                wresp.Close();
                wresp = null;
            }
        }
        finally
        {
            wr = null;
        }
    }

This is the HttpHandler on the remote server:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Collections.Specialized;

namespace SitefinityWebApp.Widgets.Files
{
    public class UploadFileHandler : IHttpHandler
    {
        public void ProcessRequest(HttpContext context)
        {
            context.Response.ContentType = "text/plain";

            //VALIDATE FILES IN REQUEST
            if (context.Request.Files.Count > 0)
            {
                try
                {
                    //HANDLE EACH FILE IN THE REQUEST
                    foreach (HttpPostedFile item in context.Request.Files)
                    {
                        item.SaveAs(context.Server.MapPath("~/Temp/" + item.FileName));
                        context.Response.Write("File uploaded");
                    }
                }
                catch (Exception ex)
                {
                    //NO FILES IN REQUEST TO HANDLE
                    context.Response.Write("Error: " + ex.Message);
                }
            }
            else
            {
                //NO FILES IN REQUEST TO HANDLE
                context.Response.Write("No file uploaded");
            }
        }

        public bool IsReusable
        {
            get
            {
                return false;
            }
        }
    }
}

And this is how you use the HttpUploadFile method:

    var nvc = new NameValueCollection();
    nvc.Add("user", userName);
    nvc.Add("password", password);
    nvc.Add("library", libraryName);

    HttpUploadFile(destinationUrl, uploadFile, "file", "image/png", nvc);

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两相知 2024-12-13 06:19:13

迭代HttpFileCollection将仅返回文件键。因此,您必须按如下方式更改 HTTP 处理程序:

//HANDLE EACH FILE IN THE REQUEST
foreach (string key in context.Request.Files)
{
  HttpPostedFile item = context.Request.Files[key];
  item.SaveAs(context.Server.MapPath("~/Temp/" + item.FileName));
  context.Response.Write("File uploaded");
}

希望这会有所帮助。

Iterating through the HttpFileCollection will return only the file keys. So you have to change your HTTP handler as follow:

//HANDLE EACH FILE IN THE REQUEST
foreach (string key in context.Request.Files)
{
  HttpPostedFile item = context.Request.Files[key];
  item.SaveAs(context.Server.MapPath("~/Temp/" + item.FileName));
  context.Response.Write("File uploaded");
}

Hope, this helps.

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