its NOT DUPLICATE.Link that has been provided is an OLD one."http client" has been removed in api23
我要发送json对象:
{"emailId":"ashish.bhatt@mobimedia.in","address":"Naya bans","city":"Noida","pincode":"201301","account_number":"91123546374208","bank_name":"Axis Bank","branch_name":"91123546374208","ifsc_code":"UTI0000879"}
到网址:
http://10digimr.mobimedia.in/api/mobile_retailer/update_profile How do i do it? via post method?
方法:
POST /api/mobile_retailer/update_profile
强制键:
{"emailId","address"}
请求 JSON:
{"emailId":"ashish.bhatt@mobimedia.in","address":"Naya bans","city":"Noida","pincode":"201301","account_number":"91123546374208","bank_name":"Axis Bank","branch_name":"91123546374208","ifsc_code":"UTI0000879"}
响应:
{"message":"Mail Send","data":true,"status":200}
最佳答案
定义一个类 AsyncT
并在 onCreate
方法中调用它:
AsyncT asyncT = new AsyncT();
asyncT.execute();
类定义:
class AsyncT extends AsyncTask<Void,Void,Void>{
@Override
protected Void doInBackground(Void... params) {
try {
URL url = new URL(""); //Enter URL here
HttpURLConnection httpURLConnection = (HttpURLConnection)url.openConnection();
httpURLConnection.setDoOutput(true);
httpURLConnection.setRequestMethod("POST"); // here you are telling that it is a POST request, which can be changed into "PUT", "GET", "DELETE" etc.
httpURLConnection.setRequestProperty("Content-Type", "application/json"); // here you are setting the `Content-Type` for the data you are sending which is `application/json`
httpURLConnection.connect();
JSONObject jsonObject = new JSONObject();
jsonObject.put("para_1", "arg_1");
DataOutputStream wr = new DataOutputStream(httpURLConnection.getOutputStream());
wr.writeBytes(jsonObject.toString());
wr.flush();
wr.close();
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} catch (JSONException e) {
e.printStackTrace();
}
return null;
}
}
关于android - 在android中通过http post方法发送json对象,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36833798/