WebClient上传带有POST值的UploadFile

Question

我想使用WebClient类将文件上传到主机。我还想传递一些应该在服务器部分（PHP）的$ _POST数组中显示的值。我想一口气做

我用过下面的代码

using (WebClient wc = new WebClient())
{
    wc.Encoding = Encoding.UTF8;
    NameValueCollection values = new NameValueCollection();
    values.Add("client", "VIP");
    values.Add("name", "John Doe"); 
    wc.QueryString = values; // this displayes in $_GET
    byte[] ans= wc.UploadFile(address, dumpPath);
}

如果我使用QueryString属性，则$ _GET数组中显示的值。但我想通过post方法发送它

Answer 1

没有内置功能可让您执行此操作。我已经在博客中发布了可以使用的扩展程序。以下是相关的类：

public class UploadFile
{
    public UploadFile()
    {
        ContentType = "application/octet-stream";
    }
    public string Name { get; set; }
    public string Filename { get; set; }
    public string ContentType { get; set; }
    public Stream Stream { get; set; }
}

public byte[] UploadFiles(string address, IEnumerable<UploadFile> files, NameValueCollection values)
{
    var request = WebRequest.Create(address);
    request.Method = "POST";
    var boundary = "---------------------------" + DateTime.Now.Ticks.ToString("x", NumberFormatInfo.InvariantInfo);
    request.ContentType = "multipart/form-data; boundary=" + boundary;
    boundary = "--" + boundary;

    using (var requestStream = request.GetRequestStream())
    {
        // Write the values
        foreach (string name in values.Keys)
        {
            var buffer = Encoding.ASCII.GetBytes(boundary + Environment.NewLine);
            requestStream.Write(buffer, 0, buffer.Length);
            buffer = Encoding.ASCII.GetBytes(string.Format("Content-Disposition: form-data; name=\"{0}\"{1}{1}", name, Environment.NewLine));
            requestStream.Write(buffer, 0, buffer.Length);
            buffer = Encoding.UTF8.GetBytes(values[name] + Environment.NewLine);
            requestStream.Write(buffer, 0, buffer.Length);
        }

        // Write the files
        foreach (var file in files)
        {
            var buffer = Encoding.ASCII.GetBytes(boundary + Environment.NewLine);
            requestStream.Write(buffer, 0, buffer.Length);
            buffer = Encoding.UTF8.GetBytes(string.Format("Content-Disposition: form-data; name=\"{0}\"; filename=\"{1}\"{2}", file.Name, file.Filename, Environment.NewLine));
            requestStream.Write(buffer, 0, buffer.Length);
            buffer = Encoding.ASCII.GetBytes(string.Format("Content-Type: {0}{1}{1}", file.ContentType, Environment.NewLine));
            requestStream.Write(buffer, 0, buffer.Length);
            file.Stream.CopyTo(requestStream);
            buffer = Encoding.ASCII.GetBytes(Environment.NewLine);
            requestStream.Write(buffer, 0, buffer.Length);
        }

        var boundaryBuffer = Encoding.ASCII.GetBytes(boundary + "--");
        requestStream.Write(boundaryBuffer, 0, boundaryBuffer.Length);
    }

    using (var response = request.GetResponse())
    using (var responseStream = response.GetResponseStream())
    using (var stream = new MemoryStream())
    {
        responseStream.CopyTo(stream);
        return stream.ToArray();
    }
}

现在您可以在应用程序中使用它：

using (var stream = File.Open(dumpPath, FileMode.Open))
{
    var files = new[] 
    {
        new UploadFile
        {
            Name = "file",
            Filename = Path.GetFileName(dumpPath),
            ContentType = "text/plain",
            Stream = stream
        }
    };

    var values = new NameValueCollection
    {
        { "client", "VIP" },
        { "name", "John Doe" },
    };

    byte[] result = UploadFiles(address, files, values);
}

现在，在你的PHP脚本，你可以使用$_POST["client"]，$_POST["name"]和$_FILES["file"]。