spring - 在spring mvc Controller 接收json并反序列化为对象列表

Question

我的代码如下:

Controller

@RequestMapping(value="/setTest", method=RequestMethod.POST, consumes="application/json")
public @ResponseBody ModelMap setTest(@RequestBody List<TestS> refunds, ModelMap map) {
    for(TestS r : refunds) {
        System.out.println(r.getName());
    }
    // other codes
}

TestS pojo

public class TestS implements Serializable {
    private String name;
    private String age;
    //getter setter
}

Json 请求

var z = '[{"name":"1","age":"2"},{"name":"1","age":"3"}]';
$.ajax({
    url: "/setTest",
    data: z,
    type: "POST",
    dataType:"json",
    contentType:'application/json'               
});

它给出了这个错误

java.lang.ClassCastException: java.util.LinkedHashMap cannot be cast to com.air.cidb.entities.TestS

我正在使用 spring 3.1.2 和 jackson 2.0.4

编辑:我想在我的 Controller 方法中接收 TestS 对象的列表，并处理它们。我无法找到我是否发送了错误的 json 或我的方法签名错误。

Answer 1

这是适合我的代码。关键是你需要一个包装类。

public class Person {

    private String name;
    private Integer age;

    public String getName() {
        return name;
    }

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

    public Integer getAge() {
        return age;
    }

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

    @Override
    public String toString() {
        return "Person [name=" + name + ", age=" + age + "]";
    }

一个 PersonWrapper 类

public class PersonWrapper {

    private List<Person> persons;

    /**
     * @return the persons
     */
    public List<Person> getPersons() {
        return persons;
    }

    /**
     * @param persons the persons to set
     */
    public void setPersons(List<Person> persons) {
        this.persons = persons;
    }
}

我的 Controller 方法

@RequestMapping(value="person", method=RequestMethod.POST,consumes="application/json",produces="application/json")
    @ResponseBody
    public List<String> savePerson(@RequestBody PersonWrapper wrapper) {
        List<String> response = new ArrayList<String>();
        for (Person person: wrapper.getPersons()){
        personService.save(person);
         response.add("Saved person: " + person.toString());
    }
        return response;
    }

发送的请求是 POST 中的 json

{"persons":[{"name":"shail1","age":"2"},{"name":"shail2","age":"3"}]}

响应是

["Saved person: Person [name=shail1, age=2]","Saved person: Person [name=shail2, age=3]"]