如何在c#中的ajax post请求中传递json body(对于API) [英] How to pass json body in ajax post request in c# (for API )
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问题描述
我已经开发了快递跟踪系统,
我必须只插入跟踪号码并且它检测到快递,
之后我必须将json身体传递给后快递,
任何人都可以建议???
解决方案将非常值得一提并提前感谢..
以下是以下代码.. :(
I have developed courier tracking system,
in i have to insert only tracking number and it detects courier,
After that i have to pass json body to post courier ,
Can any one suggest regarding that???
Solutions will be highly appreciable and thanks in advance..
Here is the code below.. :(
if (TrackingNumber != "")
{
HttpWebRequest request = (HttpWebRequest)WebRequest.Create("https://api.aftership.com/v3/trackings);
request.ContentType = "application/json";
request.Method = "POST";
request.Headers.Add("aftership-api-key:*************************");
HttpWebResponse response = (HttpWebResponse)request.GetResponse();
if (response.StatusCode == HttpStatusCode.OK)
{
Stream receiveStream = response.GetResponseStream();
StreamReader readStream = null;
if (response.CharacterSet == "")
readStream = new StreamReader(receiveStream);
else
readStream = new StreamReader(receiveStream, Encoding.GetEncoding(response.CharacterSet));
// List<string> list = new List<string>();
string data = readStream.ReadToEnd();
.
.
.
.
.
我想通过这样的json身体...... :(
And i want to pass json body like this... :(
{
"tracking": {
"slug": "dhl",
"tracking_number": "123456789",
"title": "Title Name",
}
}
这里只有tracking_number是强制性,否则所有字段都是可选的.....:(
Here only tracking_number is mandatory ,else all fields are optional..... :(
推荐答案
我已经解决了我自己.. :)
i have solved it my self.. :)
try
{
var httpWebRequest = (HttpWebRequest)WebRequest.Create("https://api.aftership.com/v3/trackings");
httpWebRequest.Method = "POST";
httpWebRequest.Headers.Add("aftership-api-key:********fdbfd93980b8c5***");
httpWebRequest.ContentType = "application/json";
using (var streamWriter = new StreamWriter(httpWebRequest.GetRequestStream()))
{
string json = "{" + "\"tracking\": {" +"\"slug\":\""+Courier+"\","+"\"tracking_number\":\""+trackNumber+"\"}}";
streamWriter.Write(json);
streamWriter.Flush();
streamWriter.Close();
var httpResponse = (HttpWebResponse)httpWebRequest.GetResponse();
using (var streamReader = new StreamReader(httpResponse.GetResponseStream()))
{
var result = streamReader.ReadToEnd();
}
}
}
catch
{
}
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