调用rest api客户端

Question

我在eclipse中创建了一个休息api作为maven项目。 用于rest api的MobileAnalyticsModel类是

package org.subhayya.amazonws.mobileanalytics;

import java.util.Date;

import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement
public class MobileAnalyticsModel {

    private String name;
    private Date created;
    private String location;
    private String prize;
    private String requirement;

    public MobileAnalyticsModel() {

    }
    public MobileAnalyticsModel(String name, String location, String prize, String requirement) {

        this.name = name;
        this.location = location;
        this.prize = prize;
        this.requirement = requirement;
        this.created = new Date();
    }

    public String getName() {
        return name;
    }

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

    public Date getCreated() {
        return created;
    }

    public void setCreated(Date created) {
        this.created = created;
    }

    public String getLocation() {
        return location;
    }

    public void setLocation(String location) {
        this.location = location;
    }

    public String getPrize() {
        return prize;
    }

    public void setPrize(String prize) {
        this.prize = prize;
    }

    public String getRequirement() {
        return requirement;
    }

    public void setRequirement(String requirement) {
        this.requirement = requirement;
    }
}

这是创建的api的json响应：

和 这是我调用rest api的示例测试代码：

package org.subhayya.example;
import javax.ws.rs.client.Client;
import javax.ws.rs.client.ClientBuilder;
import javax.ws.rs.core.MediaType;

public class SampleTestREstClient {

    public static void main(String[] args) {


        Client client = ClientBuilder.newClient( );
        String reponse = client.target("http://localhost:8080/AgentBasedCloudServiceCompositionFramework/webapi/mobileanalytics/mobileanalyticsjson")
                .request(MediaType.APPLICATION_JSON)
                .get(String.class);

        System.out.println(reponse);
    }}

然后我得到了完整的json响应..作为

 [{"created":"2017-03-30T14:36:58.56","location":"http://api.server.com","name":"Mobile Analytics","prize":"$1.00 per 1,000,000 Amazon Mobile Analytics events per month thereafter","requirement":"PutEvents"}]

但我希望将单个参数作为我的输出，例如，名称，位置或要求。我也在同一个maven项目中创建客户端调用代码。所以我编写了如下代码

Client client = ClientBuilder.newClient( );
MobileAnalyticsModel reponse = 
        client.target("http://localhost:8080/AgentBasedCloudServiceCompositionFramework/webapi/mobileanalytics/mobileanalyticsjson")
                    .request(MediaType.APPLICATION_JSON)
                    .get(MobileAnalyticsModel.class);

System.out.println(reponse.getName());

但我得到例外，所以我将其改为System.out.println(reponse); 获得至少JSON响应，然后也得到错误。

如何从JSON响应中获取单个名称参数？我是新来的这个休息api ..请帮助我尽快解决这个问题。谢谢提前

Answer 1

您的回复是一个字符串。访问JSON响应元素的最简单方法是将共振转换为Json-Object。然后，您可以通过名称轻松访问这些字段。 看一下： How to parse JSON in Java

Answer 2

您还可以查看以下链接，将json转换为对象。

Parse a JSON response as an object

Answer 3

此代码适用于我..

 String url = "http://localhost:8080/AgentBasedCloudServiceCompositionFramework/webapi/mobileanalytics/";
            String city = "mobileanalyticsjson";
            Client client = ClientBuilder.newClient();
            WebTarget webTarget = client.register(JsonProcessingFeature.class).target(url);
            JsonArray jsonArray = webTarget.path(city)
                .request(MediaType.APPLICATION_JSON_TYPE).get(JsonArray.class);
            for (JsonObject jsonObject : jsonArray.getValuesAs(JsonObject.class)) {
                   System.out.println(jsonObject.getString("name")); 
                   System.out.println(jsonObject.getString("location")); }