Question

如何从Java servlet返回JSON对象。

以前使用servlet进行AJAX时，我返回了一个字符串。是否存在需要使用的JSON对象类型，或者只返回看起来像JSON对象的String，例如

String objectToReturn = "{ key1: 'value1', key2: 'value2' }";

Answer 1

将JSON对象写入响应对象的输出流。

您还应该按如下方式设置内容类型，该类型将指定您要返回的内容：

response.setContentType("application/json");
// Get the printwriter object from response to write the required json object to the output stream      
PrintWriter out = response.getWriter();
// Assuming your json object is **jsonObject**, perform the following, it will return your json object  
out.print(jsonObject);
out.flush();

Answer 2

首先将JSON对象转换为String。然后将其写入响应编写器以及内容类型application/json和UTF-8的字符编码。

以下是假设您使用Google Gson将Java对象转换为JSON字符串的示例：

protected void doXxx(HttpServletRequest request, HttpServletResponse response) {
    // ...

    String json = new Gson().toJson(someObject);
    response.setContentType("application/json");
    response.setCharacterEncoding("UTF-8");
    response.getWriter().write(json);
}

就是这样。

另见：

Answer 3

我完全按照你的建议（返回String）。

您可以考虑设置MIME类型以指示您正在返回JSON（根据this other stackoverflow post它是“application / json”）。

Answer 4

如何从Java Servlet返回JSON对象

response.setContentType("application/json");
response.setCharacterEncoding("utf-8");
PrintWriter out = response.getWriter();

  //create Json Object
  JsonObject json = new JsonObject();

    // put some value pairs into the JSON object .
    json.put("Mobile", 9999988888);
    json.put("Name", "ManojSarnaik");

    // finally output the json string       
    out.print(json.toString());

Answer 5

只需在输出流中写一个字符串即可。如果您感觉有帮助，可以将MIME类型设置为text/javascript（编辑：application/json显然是官方的）。（有一个小但非零的机会，它会让某些东西不会弄乱它，这是一个很好的做法。）

Answer 6

Gson对此非常有用。更容易均匀。这是我的例子：

public class Bean {
private String nombre="juan";
private String apellido="machado";
private List<InnerBean> datosCriticos;

class InnerBean
{
    private int edad=12;

}
public Bean() {
    datosCriticos = new ArrayList<>();
    datosCriticos.add(new InnerBean());
}

}

    Bean bean = new Bean();
    Gson gson = new Gson();
    String json =gson.toJson(bean);

的out.print（JSON）;

{＆＃34; NOMBRE＆＃34;：＆＃34;胡安＆＃34;＆＃34; apellido＆＃34;：＆＃34;马查多＆＃34;＆＃34; datosCriticos＆＃34;：[ {＆＃34; EDAD＆＃34; 12}]}

不得不说人们如果使用gson时你的vars是空的它不会为你构建json。只是

{}

Answer 7

可能有一个JSON对象可以方便Java编码。但最后数据结构将序列化为字符串。设置正确的MIME类型会很不错。

我建议来自JSON Java的json.org。

Answer 8

我使用Jackson将Java Object转换为JSON字符串并发送如下。

PrintWriter out = response.getWriter();
ObjectMapper objectMapper= new ObjectMapper();
String jsonString = objectMapper.writeValueAsString(MyObject);
response.setContentType("application/json");
response.setCharacterEncoding("UTF-8");
out.print(jsonString);
out.flush();

Answer 9

JsonObject是抽象的，具体取决于Java版本（或JDK，SDK，JRE ...，对Java生态系统来说是新的）。因此，这是一个新的实现：

import javax.json.Json;
import javax.json.JsonObject;

...

try (PrintWriter out = response.getWriter()) {
    response.setContentType("application/json");       
    response.setCharacterEncoding("UTF-8");

    JsonObject json = Json.createObjectBuilder().add("foo", "bar").build();

    out.print(json.toString());
}

Answer 10

response.setContentType（ “文本/ JSON”）;

//创建JSON字符串，我建议使用一些框架。

字符串your_string;

out.write（your_string.getBytes（ “UTF-8”））;

Answer 11

您可以像下面这样使用波纹管。

如果要使用json数组：

下载 json-simple-1.1.1.jar 并添加到您的项目类路径中

创建一个名为 Model 的类，例如波纹管

public class Model {

 private String id = "";
 private String name = "";

 //getter sertter here
}

在sevlet getMethod中，您可以像下面这样使用

@Override
protected void doGet(HttpServletRequest request, HttpServletResponse response)
    throws ServletException, IOException {

  //begin get data from databse or other source
  List<Model> list = new ArrayList<>();
  Model model = new Model();
  model.setId("101");
  model.setName("Enamul Haque");
  list.add(model);

  Model model1 = new Model();
  model1.setId("102");
  model1.setName("Md Mohsin");
  list.add(model1);
  //End get data from databse or other source
try {

    JSONArray ja = new JSONArray();
    for (Model m : list) {
        JSONObject jSONObject = new JSONObject();
        jSONObject.put("id", m.getId());
        jSONObject.put("name", m.getName());
        ja.add(jSONObject);
    }
    System.out.println(" json ja = " + ja);
    response.addHeader("Access-Control-Allow-Origin", "*");
    response.setContentType("application/json");
    response.setCharacterEncoding("UTF-8");
    response.getWriter().print(ja.toString());
    response.getWriter().flush();
   } catch (Exception e) {
     e.printStackTrace();
  }

 }

输出：

    [{"name":"Enamul Haque","id":"101"},{"name":"Md Mohsin","id":"102"}]

我想让json对象像这样使用：

@Override
protected void doGet(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException {
    try {

        JSONObject json = new JSONObject();
        json.put("id", "108");
        json.put("name", "Enamul Haque");
        System.out.println(" json JSONObject= " + json);
        response.addHeader("Access-Control-Allow-Origin", "*");
        response.setContentType("application/json");
        response.setCharacterEncoding("UTF-8");
        response.getWriter().print(json.toString());
        response.getWriter().flush();
        // System.out.println("Response Completed... ");
    } catch (Exception e) {
        e.printStackTrace();
    }

}

以上功能输出：

{"name":"Enamul Haque","id":"108"}

完整源代码提供给GitHub：https://github.com/enamul95/ServeletJson.git

Answer 12

使用Google Gson lib以4个简单的行接近 BalusC 答案。将这些行添加到servlet方法：

User objToSerialize = new User("Bill", "Gates");    
ServletOutputStream outputStream = response.getOutputStream();

response.setContentType("application/json;charset=UTF-8");
outputStream.print(new Gson().toJson(objToSerialize));

祝你好运！

Answer 13

通过使用Gson，您可以发送json响应，请参见以下代码

您可以看到此代码

@WebServlet(urlPatterns = {"/jsonResponse"})
public class JsonResponse extends HttpServlet {

@Override
protected void doGet(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException {
    response.setContentType("application/json");
    response.setCharacterEncoding("utf-8");
    Student student = new Student(12, "Ram Kumar", "Male", "1234565678");
    Subject subject1 = new Subject(1, "Computer Fundamentals");
    Subject subject2 = new Subject(2, "Computer Graphics");
    Subject subject3 = new Subject(3, "Data Structures");
    Set subjects = new HashSet();
    subjects.add(subject1);
    subjects.add(subject2);
    subjects.add(subject3);
    student.setSubjects(subjects);
    Address address = new Address(1, "Street 23 NN West ", "Bhilai", "Chhattisgarh", "India");
    student.setAddress(address);
    Gson gson = new Gson();
    String jsonData = gson.toJson(student);
    PrintWriter out = response.getWriter();
    try {
        out.println(jsonData);
    } finally {
        out.close();
    }

  }
}

对json response from servlet in java有帮助

如何从Java Servlet返回JSON对象

13 个答案:

另见：