使用 System.Net.WebClient 发送 HTTP POST
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Sending HTTP POST with System.Net.WebClient
提问by Endy Tjahjono
Is it possible to send HTTP POST with some form data with System.Net.WebClient?
是否可以使用 System.Net.WebClient 发送带有某些表单数据的 HTTP POST？
If not, is there another library like WebClient that can do HTTP POST? I know I can use System.Net.HttpWebRequest, but I'm looking for something that is not as verbose.
如果没有，是否有另一个像 WebClient 这样可以执行 HTTP POST 的库？我知道我可以使用 System.Net.HttpWebRequest，但我正在寻找不那么冗长的东西。
Hopefully it will look like this:
希望它看起来像这样：
Using client As New TheHTTPLib
client.FormData("parm1") = "somevalue"
result = client.DownloadString(someurl, Method.POST)
End Using
回答by Endy Tjahjono
Based on @carlosfigueira 's answer, I looked further into WebClient's methods and found UploadValues, which is exactly what I want:
根据@carlosfigueira 的回答，我进一步研究了 WebClient 的方法并找到了UploadValues，这正是我想要的：
Using client As New Net.WebClient
Dim reqparm As New Specialized.NameValueCollection
reqparm.Add("param1", "somevalue")
reqparm.Add("param2", "othervalue")
Dim responsebytes = client.UploadValues(someurl, "POST", reqparm)
Dim responsebody = (New Text.UTF8Encoding).GetString(responsebytes)
End Using
The key part is this:
关键部分是这样的：
client.UploadValues(someurl, "POST", reqparm)
It sends whatever verb I type in, and it also helps me create a properly url encoded form data, I just have to supply the parameters as a namevaluecollection.
它发送我输入的任何动词，它还帮助我创建一个正确的 url 编码表单数据，我只需要提供参数作为 namevaluecollection。
回答by carlosfigueira
WebClientdoesn't have a direct support for form data, but you can send a HTTP post by using the UploadString method:
WebClient不直接支持表单数据，但您可以使用 UploadString 方法发送 HTTP 帖子：
Using client as new WebClient
result = client.UploadString(someurl, "param1=somevalue¶m2=othervalue")
End Using
回答by faester
As far as the http verb is concerned the WebRequestmight be easier. You could go for something like:
就 http 动词而言，这WebRequest可能更容易。你可以去这样的事情：
WebRequest r = WebRequest.Create("http://some.url");
r.Method = "POST";
using (var s = r.GetResponse().GetResponseStream())
{
using (var reader = new StreamReader(r, FileMode.Open))
{
var content = reader.ReadToEnd();
}
}
Obviously this lacks exception handling and writing the request body (for which you can use r.GetRequestStream()and write it like a regular stream, but I hope it may be of some help.
显然，这缺少异常处理和编写请求正文（您可以r.GetRequestStream()像常规流一样使用和编写它，但我希望它可能有所帮助。
