我需要在WebClient中使用“HTTP Post”将一些数据发布到我拥有的特定URL。
现在,我知道这可以通过WebRequest完成,但出于某些原因我想使用WebClient。那可能吗?如果是这样,有人能告诉我一些例子还是指向正确的方向?
答案 0 :(得分:355)
我刚刚找到解决方案,是的,它比我想象的容易:)
所以这是解决方案:
string URI = "http://www.myurl.com/post.php";
string myParameters = "param1=value1¶m2=value2¶m3=value3";
using (WebClient wc = new WebClient())
{
wc.Headers[HttpRequestHeader.ContentType] = "application/x-www-form-urlencoded";
string HtmlResult = wc.UploadString(URI, myParameters);
}
它就像魅力一样:)
答案 1 :(得分:335)
有一个名为UploadValues的内置方法可以发送HTTP POST(或任何类型的HTTP方法)并处理请求体的构造(连接参数与“&”和通过url编码转义字符)以适当的格式数据格式:
using(WebClient client = new WebClient())
{
var reqparm = new System.Collections.Specialized.NameValueCollection();
reqparm.Add("param1", "<any> kinds & of = ? strings");
reqparm.Add("param2", "escaping is already handled");
byte[] responsebytes = client.UploadValues("http://localhost", "POST", reqparm);
string responsebody = Encoding.UTF8.GetString(responsebytes);
}
答案 2 :(得分:38)
使用WebClient.UploadString
或WebClient.UploadData
,您可以轻松地将数据发布到服务器。我将使用UploadData显示一个示例,因为UploadString的使用方式与DownloadString相同。
byte[] bret = client.UploadData("http://www.website.com/post.php", "POST",
System.Text.Encoding.ASCII.GetBytes("field1=value1&field2=value2") );
string sret = System.Text.Encoding.ASCII.GetString(bret);
答案 3 :(得分:18)
string URI = "site.com/mail.php";
using (WebClient client = new WebClient())
{
System.Collections.Specialized.NameValueCollection postData =
new System.Collections.Specialized.NameValueCollection()
{
{ "to", emailTo },
{ "subject", currentSubject },
{ "body", currentBody }
};
string pagesource = Encoding.UTF8.GetString(client.UploadValues(URI, postData));
}
答案 4 :(得分:17)
//Making a POST request using WebClient.
Function()
{
WebClient wc = new WebClient();
var URI = new Uri("http://your_uri_goes_here");
//If any encoding is needed.
wc.Headers["Content-Type"] = "application/x-www-form-urlencoded";
//Or any other encoding type.
//If any key needed
wc.Headers["KEY"] = "Your_Key_Goes_Here";
wc.UploadStringCompleted +=
new UploadStringCompletedEventHandler(wc_UploadStringCompleted);
wc.UploadStringAsync(URI,"POST","Data_To_Be_sent");
}
void wc__UploadStringCompleted(object sender, UploadStringCompletedEventArgs e)
{
try
{
MessageBox.Show(e.Result);
//e.result fetches you the response against your POST request.
}
catch(Exception exc)
{
MessageBox.Show(exc.ToString());
}
}
答案 5 :(得分:0)
使用简单的client.UploadString(adress, content);
通常可以正常工作,但是我想应该记住,如果未返回HTTP成功状态代码,则会抛出WebException
。我通常这样处理,以打印远程服务器返回的任何异常消息:
try
{
postResult = client.UploadString(address, content);
}
catch (WebException ex)
{
String responseFromServer = ex.Message.ToString() + " ";
if (ex.Response != null)
{
using (WebResponse response = ex.Response)
{
Stream dataRs = response.GetResponseStream();
using (StreamReader reader = new StreamReader(dataRs))
{
responseFromServer += reader.ReadToEnd();
_log.Error("Server Response: " + responseFromServer);
}
}
}
throw;
}
答案 6 :(得分:0)
在模型中使用webapiclient发送序列化json参数请求。
PostModel.cs
public string Id { get; set; }
public string Name { get; set; }
public string Surname { get; set; }
public int Age { get; set; }
WebApiClient.cs
internal class WebApiClient : IDisposable
{
private bool _isDispose;
public void Dispose()
{
Dispose(true);
GC.SuppressFinalize(this);
}
public void Dispose(bool disposing)
{
if (!_isDispose)
{
if (disposing)
{
}
}
_isDispose = true;
}
private void SetHeaderParameters(WebClient client)
{
client.Headers.Clear();
client.Headers.Add("Content-Type", "application/json");
client.Encoding = Encoding.UTF8;
}
public async Task<T> PostJsonWithModelAsync<T>(string address, string data,)
{
using (var client = new WebClient())
{
SetHeaderParameters(client);
string result = await client.UploadStringTaskAsync(address, data); // method:
//The HTTP method used to send the file to the resource. If null, the default is POST
return JsonConvert.DeserializeObject<T>(result);
}
}
}
业务呼叫者方法
public async Task<ResultDTO> GetResultAsync(PostModel model)
{
try
{
using (var client = new WebApiClient())
{
var serializeModel= JsonConvert.SerializeObject(model);// using Newtonsoft.Json;
var response = await client.PostJsonWithModelAsync<ResultDTO>("http://www.website.com/api/create", serializeModel);
return response;
}
}
catch (Exception ex)
{
throw new Exception(ex.Message);
}
}
答案 7 :(得分:0)
这是明确的答案:
public String sendSMS(String phone, String token) {
WebClient webClient = WebClient.create(smsServiceUrl);
SMSRequest smsRequest = new SMSRequest();
smsRequest.setMessage(token);
smsRequest.setPhoneNo(phone);
smsRequest.setTokenId(smsServiceTokenId);
Mono<String> response = webClient.post()
.uri(smsServiceEndpoint)
.header(HttpHeaders.CONTENT_TYPE, MediaType.APPLICATION_JSON_VALUE)
.body(Mono.just(smsRequest), SMSRequest.class)
.retrieve().bodyToMono(String.class);
String deliveryResponse = response.block();
if (deliveryResponse.equalsIgnoreCase("success")) {
return deliveryResponse;
}
return null;
}
答案 8 :(得分:-3)
string URI = "http://www.myurl.com/post.php";
string myParameters = "param1=value1¶m2=value2¶m3=value3"
可以简化为
http://www.myurl.com/post.php?param1=value1¶m2=value2¶m3=value3
这总是有效的。我发现原来的一个可以开启和关闭。