Parsing a JSON from HTTP Response in Java

Question

Hi I am using Client Http (apache), and json-simple.

I want to access the attributes of the json response, and then use them.

Any idea how to do this? I read a post and did not work as it but me.

This is my answer json:

{"Name":"myname","Lastname":"mylastname","Age":19}

This is my code java:

DefaultHttpClient httpClient = new DefaultHttpClient();

HttpGet getRequest = new HttpGet(
    "http://localhost:8000/responsejava");
getRequest.addHeader("accept", "application/json");

HttpResponse response = httpClient.execute(getRequest);

if (response.getStatusLine().getStatusCode() != 200) {
    throw new RuntimeException("Failed : HTTP error code : "
             + response.getStatusLine().getStatusCode());
}

BufferedReader br = new BufferedReader(
    new InputStreamReader( 
        (response.getEntity().getContent())
    )
);

StringBuilder content = new StringBuilder();
String line;
while (null != (line = br.readLine())) {
    content.append(line);
}

Object obj=JSONValue.parse(content.toString());
JSONObject finalResult=(JSONObject)obj;
System.out.println(finalResult);

httpClient.getConnectionManager().shutdown();

I printed null, What am I doing wrong?

A better approach to do this is using Gson library, take a look at this github.com/google/gson — Ignacio Chiazzo
– Ignacio Chiazzo, Commented Feb 21, 2016 at 0:40

armysheng · Accepted Answer · 2016-02-21 01:36:36Z

4

Better and easier to use Gson

Gson gson = new Gson;
NameBean name = gson.fromJson(content.toString(),NameBean.class)

NameBean is the object where you persist the json string.

public class NameBean implements Serializable{
public String name;
public String lastname;
public Int age;

public String getName() {
    return name;
}

public void setName(String name) {
    this.name = name;
}

public String getLastname() {
    return lastname;
}

public void setLastname(String lastname) {
    this.lastname = lastname;
}

public Int getAge() {
    return age;
}

public void setAge(Int age) {
    this.age = age;
}

}

edited Feb 21, 2016 at 1:36

answered Feb 21, 2016 at 1:12

armysheng

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2 Comments

jojemapa Over a year ago

How I define Name Bean? ;(

armysheng Over a year ago

NameBean should define with fields and getter setter matches with the json String. Check the Gson user guide for more, it's easy and won't take long minutes

David Sousa · Accepted Answer · 2016-02-21 01:10:02Z

4

instead of

Object obj=JSONValue.parse(content.toString());
JSONObject finalResult=(JSONObject)obj;
System.out.println(finalResult);

try this:

JSONObject jsonObject = new JSONObject(content.toString());
System.out.println(jsonObject.getString("Name") + " " jsonObject.getString("Lastname") + " " + jsonObject.getInt("Age"));

answered Feb 21, 2016 at 1:10

David Sousa

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4 Comments

jojemapa Over a year ago

JSONObject does not accept the string argument. The method getString not exist. Thanks for your time ;(

David Sousa Over a year ago

I think you are using a native library, because i used an external json jar and worked fine, I hope you can solve your problem ^^

jojemapa Over a year ago

Thank you. Apparently you use json javax . It is not the same as Gson , or single - json . And can resolve my problem. Thank you.

Roman Vottner Over a year ago

@VipPunkJoshersDroopy org.json.JSONObject does have a getString(String) method. Though, always make sure to check with has(String) beforehand if the JSON actually contained this field on the respective object

Beno · Accepted Answer · 2017-09-21 20:54:52Z

0

I higly recomend http-request built on apache http api.

HttpRequest<Data> httpRequest = HttpRequestBuilder.createGet(yourUri, Data.class)
    .addDefaultHeader("accept", "application/json")
    .build();

public void send(){
   ResponseHandler<Data> responseHandler = httpRequest.execute();
   Data data = responseHandler.orElseThrow(); // returns the data or throws ResponseException If response code is not success
}

Data class which you get as response.

public Data{
   private String Name;
   private String Lastname;
   private int Age;

   // getters and setters
}

I also recommend watching my answer here If you want get response as String

answered Sep 21, 2017 at 20:54

Beno

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