问题描述
新到Android,我需要使用下面的参数在我的Web服务调用。据我所知,参数实际上是JSON对象。
New to Android, I need to use below parameters in my webservices call. I understand that parameters are actually JSON objects.
下面code返回XML与标题:坏请求时,它应该返回登录的用户信息。 logcat的是显示值 - > JSON:{查询:com.androidatc.customviewindrawer.Query@f1eb09f,includeUserMiscInfo:真正}意味着我的参数是不正确的。如何正确地传递呢?
Below code returns XML with "title: Bad request" when it should return logged in user info. logcat is showing value as --> json: {"Query":"com.androidatc.customviewindrawer.Query@f1eb09f","includeUserMiscInfo":true} means my parameter is not correct. How to pass it correctly?
protected void sendJson(final String email, final String pwd) {
Thread t = new Thread() {
public void run() {
Looper.prepare(); //For Preparing Message Pool for the child Thread
HttpClient client = new DefaultHttpClient();
HttpConnectionParams.setConnectionTimeout(client.getParams(), 10000); //Timeout Limit
HttpResponse response;
JSONObject json = new JSONObject();
try {
HttpPost post = new HttpPost("http://192.168.0.102/SDService_SAFTI/ServiceSD.svc/LoginUser");
Query queryObj = new Query();
queryObj.setLogin("WT");
queryObj.setPassword("3");
json.put("Query", queryObj);
// json.put("email", email);
// json.put("password", pwd);
json.put("includeUserMiscInfo", true);
StringEntity se = new StringEntity( json.toString());
se.setContentType(new BasicHeader(HTTP.CONTENT_TYPE, "application/json"));
post.setEntity(se);
response = client.execute(post);
/*Checking response */
if(response!=null){
InputStream in = response.getEntity().getContent(); //Get the data in the entity
Toast.makeText(getActivity().getApplicationContext(), "Response:" + convertStreamToString(in),Toast.LENGTH_LONG).show();
}
} catch(Exception e) {
e.printStackTrace();
// getActivity().createDialog("Error", "Cannot Estabilish Connection");
}
Looper.loop(); //Loop in the message queue
}
};
t.start();
}
private static String convertStreamToString(InputStream is) {
BufferedReader reader = new BufferedReader(new InputStreamReader(is));
StringBuilder sb = new StringBuilder();
String line = null;
try {
while ((line = reader.readLine()) != null) {
sb.append((line + "\n"));
}
} catch (IOException e) {
e.printStackTrace();
} finally {
try {
is.close();
} catch (IOException e) {
e.printStackTrace();
}
}
return sb.toString();
}
Query.java
Query.java
public class Query {
String login;
public void setPassword(String password) {
this.password = password;
}
public void setLogin(String login) {
this.login = login;
}
String password;
public String getPassword() {
return password;
}
public String getLogin() {
return login;
}
}
任何建议都AP preciated。提前很多感谢。
Any suggestions are appreciated. Many thanks in advance.
推荐答案
由于您的JSON类型,你可能需要做这样的,
As your Json Type, you may have to do like this,
JSONArray array = new JSONArray();
array.put("login=WT&password=3");
json.put("Query", array);
json.put("includeUserMiscInfo", "true");
我想建议你去尝试更换这个太行,
I want to suggest you to try replacing this line too,
json.put("includeUserMiscInfo", true);
与
json.put("includeUserMiscInfo", "true");
这篇关于传递JSON作为OBJ参数web服务 - 收益"坏请求"响应的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!