Pri*_*him 72 c# windows-phone-8.1
我正在开发一个wp8.1应用程序C#,我已经设法通过从textbox.texts创建一个json对象(bm)来在json中执行一个POST方法到我的api.这是我的代码如下.我如何使用相同的textbox.text并将它们作为内容类型= application/x-www-form-urlencoded进行POST.那是什么代码?
Profile bm = new Profile();
bm.first_name = Names.Text;
bm.surname = surname.Text;
string json = JsonConvert.SerializeObject(bm);
MessageDialog messageDialog = new MessageDialog(json);//Text should not be empty
await messageDialog.ShowAsync();
HttpClient client = new HttpClient();
client.DefaultRequestHeaders.Clear();
client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json"));
client.DefaultRequestHeaders.TryAddWithoutValidation("Content-Type", "application/json");
byte[] messageBytes = Encoding.UTF8.GetBytes(json);
var content = new ByteArrayContent(messageBytes);
content.Headers.ContentType = new MediaTypeHeaderValue("application/json");
var response = client.PostAsync("myapiurl", content).Result;
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Raw*_*aew 122
var nvc = new List<KeyValuePair<string, string>>();
nvc.Add(new KeyValuePair<string, string>("Input1", "TEST2"));
nvc.Add(new KeyValuePair<string, string>("Input2", "TEST2"));
var client = new HttpClient();
var req = new HttpRequestMessage(HttpMethod.Post, url) { Content = new FormUrlEncodedContent(nvc) };
var res = await client.SendAsync(req);
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要么
var dict = new Dictionary<string, string>();
dict.Add("Input1", "TEST2");
dict.Add("Input2", "TEST2");
var client = new HttpClient();
var req = new HttpRequestMessage(HttpMethod.Post, url) { Content = new FormUrlEncodedContent(dict) };
var res = await client.SendAsync(req);
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小智 14
对我来说最好的解决方案是:
// Add key/value
var dict = new Dictionary<string, string>();
dict.Add("Content-Type", "application/x-www-form-urlencoded");
// Execute post method
using (var response = httpClient.PostAsync(path, new FormUrlEncodedContent(dict))){}
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小智 9
var params= new Dictionary<string, string>();
var url ="Please enter URLhere";
params.Add("key1", "value1");
params.Add("key2", "value2");
using (HttpClient client = new HttpClient())
{
client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json"));
HttpResponseMessage response = client.PostAsync(url, new FormUrlEncodedContent(dict)).Result;
var tokne= response.Content.ReadAsStringAsync().Result;
}
//Get response as expected
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