如何在REST Response中以JSON Arrray格式检索结果(How to retrieve the result as JSON Arrray format in REST Respons

如何在REST Response中以JSON Arrray格式检索结果(How to retrieve the result as JSON Arrray format in REST Response)

我试图从数据库中检索数据，并以JSON数组的形式返回'Response'。 但现在我在浏览器中得到如下结果，这不是正确的JSON数组格式。 我如何以JSON数组的形式接收数据？

{"{\n \"id\": 14,\n \"name\": \"Test Doom Post\",\n \"email\": \"test@test1.com\...

JDK 1.7 Jersey（jaxrs-ri-2.25.1）Gson

//以下是我的Get方法：

@Path("/register") public class JSONService { @GET @Path("/get") @Produces("application/json") @Consumes("application/json") public Response getRegisterInJSON() { JSONObject requestedJSON = new JSONObject(); try { Class.forName("com.mysql.jdbc.Driver"); SoccerUtils dbConnection = new SoccerUtils(); Connection conn = dbConnection.getWeekendDBConnection(); PreparedStatement stmt = conn.prepareStatement("SELECT ID, FIRST_NAME, EMAIL FROM mycoolmap.weekendsoccer_login"); ResultSet rs = stmt.executeQuery(); while(rs.next()) { RegisterPlayer playerObj = new RegisterPlayer(); playerObj.setId(rs.getInt("ID")); playerObj.setName(rs.getString("FIRST_NAME")); playerObj.setEmail(rs.getString("EMAIL")); Gson gson = new GsonBuilder().setPrettyPrinting().create(); String json1 = gson.toJson(playerObj); requestedJSON.put(json1, json1); System.out.println(requestedJSON); } } catch (Exception e) { e.printStackTrace(); } finally { } return Response.status(Status.OK).entity(requestedJSON.toString()).build(); }

//注册Player POJO类：

@XmlRootElement public class RegisterPlayer implements Serializable { private int id; private String name; private String email; public RegisterPlayer() { } public RegisterPlayer(int id, String name, String email) { super(); this.id =id; this.name = name; this.email = email; } public int getId() { return id; } public void setId(int id) { this.id =id; } public String getName() { return name; } public void setName(String name) { this.name = name; } public String getEmail() { return email; } public void setEmail(String email) { this.email = email; } @Override public String toString() { return "RegisterPlayer[id=" + id +", name=" + name +", email="+ email +"]"; } }

I am trying to retrieve the data from database and return them in 'Response' as JSON array. But now I am getting the result in browser as below, which is not the correct JSON array format. How can I receive the data as JSON Array ?

{"{\n \"id\": 14,\n \"name\": \"Test Doom Post\",\n \"email\": \"test@test1.com\...

JDK 1.7 Jersey (jaxrs-ri-2.25.1) Gson

//Following is my Get method below:

@Path("/register") public class JSONService { @GET @Path("/get") @Produces("application/json") @Consumes("application/json") public Response getRegisterInJSON() { JSONObject requestedJSON = new JSONObject(); try { Class.forName("com.mysql.jdbc.Driver"); SoccerUtils dbConnection = new SoccerUtils(); Connection conn = dbConnection.getWeekendDBConnection(); PreparedStatement stmt = conn.prepareStatement("SELECT ID, FIRST_NAME, EMAIL FROM mycoolmap.weekendsoccer_login"); ResultSet rs = stmt.executeQuery(); while(rs.next()) { RegisterPlayer playerObj = new RegisterPlayer(); playerObj.setId(rs.getInt("ID")); playerObj.setName(rs.getString("FIRST_NAME")); playerObj.setEmail(rs.getString("EMAIL")); Gson gson = new GsonBuilder().setPrettyPrinting().create(); String json1 = gson.toJson(playerObj); requestedJSON.put(json1, json1); System.out.println(requestedJSON); } } catch (Exception e) { e.printStackTrace(); } finally { } return Response.status(Status.OK).entity(requestedJSON.toString()).build(); }

// Register Player POJO class:

@XmlRootElement public class RegisterPlayer implements Serializable { private int id; private String name; private String email; public RegisterPlayer() { } public RegisterPlayer(int id, String name, String email) { super(); this.id =id; this.name = name; this.email = email; } public int getId() { return id; } public void setId(int id) { this.id =id; } public String getName() { return name; } public void setName(String name) { this.name = name; } public String getEmail() { return email; } public void setEmail(String email) { this.email = email; } @Override public String toString() { return "RegisterPlayer[id=" + id +", name=" + name +", email="+ email +"]"; } }

最满意答案

正如罗斯在上述评论中所建议的，我创建了一个列表，添加对象并返回列表。 它按预期工作。

/Created a 'registerPlayerList' List List<RegisterPlayer> registerPlayerList = new ArrayList<RegisterPlayer>(); // Intialiaze the RegisterPlayer class RegisterPlayer playerObj = new RegisterPlayer(); //set all the values into the object playerObj.setId(rs.getInt("ID")); playerObj.setName(rs.getString("FIRST_NAME")); playerObj.setEmail(rs.getString("EMAIL")); ...... //add the playerObj to the created registerPlayerList registerPlayerList.add(playerObj); // return the list return registerPlayerList ;

As advised by Roman in above comment, I have created a list, add the object and return the list. It worked as expected.

/Created a 'registerPlayerList' List List<RegisterPlayer> registerPlayerList = new ArrayList<RegisterPlayer>(); // Intialiaze the RegisterPlayer class RegisterPlayer playerObj = new RegisterPlayer(); //set all the values into the object playerObj.setId(rs.getInt("ID")); playerObj.setName(rs.getString("FIRST_NAME")); playerObj.setEmail(rs.getString("EMAIL")); ...... //add the playerObj to the created registerPlayerList registerPlayerList.add(playerObj); // return the list return registerPlayerList ;