-3
中的POST请求,并在将JSON字符串发送到POST请求时遇到问题。如何将json字符串作为输入参数发送到android
这是我的网址:http://172.25.183.183:8080/JIRAservice/rest/runquery
键:查询
值:
{ "jql": "project=<projectkey>",
"startAt": 0,
"maxResults": 100,
"fields": [
"summary",
"customfield_10006",
"status",
"description"
]
}
哪里<projectkey>
是存储在共享偏好的价值,请帮助
这是我的代码
try{
TextView op=(TextView) findViewById(R.id.resp);
URL url=new URL("http://172.25.183.183:8080/JIRAservice/rest/runquery");
HttpsURLConnection conn=(HttpsURLConnection)url.openConnection();
conn.setRequestMethod("POST");
String projectKey=Home.savedid;
JSONObject jsonParam = new JSONObject();
jsonParam.put("query", " "{ \"jql\": \"project=" + projectKey + "\", \"startAt\": 0, \"maxResults\": 100, \"fields\": [\"summary\",\"customfield_10006\", \"status\", \"description\"] }"");
how to send the parameters??
conn.setDoOutput(true);
DataOutputStream dbstrm=new DataOutputStream(conn.getOutputStream());
dbstrm.flush();
dbstrm.close();
int respnse=conn.getResponseCode();
String output="Request URl"+url;
output+=System.getProperty("line.separator");
output+=System.getProperty("line.separator")+"Response Code"+respnse;
BufferedReader br=new BufferedReader(new InputStreamReader(conn.getInputStream()));
String line= "";
StringBuilder respop=new StringBuilder();
while((line=br.readLine())!=null){
respop.append(line);
}
br.close();
output +=System.getProperty("line.separator")+respop.toString();
op.setText(output);
}catch(MalformedURLException ae){
ae.printStackTrace();
}catch (IOException e){
e.printStackTrace();
}
哪一部分的代码,你已经? – devnull69
这很大程度上取决于服务器端的api ...您可以构建json对象并执行异步任务来发布您的数据......但是执行此操作的方法真的取决于您的服务器 –
关键和值应该同时发送 – Sireesha