在spring mvc控制器接收json并反序列化为对象列表

Question

我的代码如下：

控制器

@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 无法转换为 com.air.cidb.entities.TestS

我正在使用 spring 3.1.2 和 jackson 2.0.4

编辑：我想在我的控制器方法中接收 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

我相信这会解决问题

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

Answer 2

解决方案效果很好，

public List<String> savePerson(@RequestBody Person[] personArray)  

对于这个签名，你可以像邮递员一样传递Person数组

[
{
  "empId": "10001",
  "tier": "Single",
  "someting": 6,
  "anything": 0,
  "frequency": "Quaterly"
}, {
  "empId": "10001",
  "tier": "Single",
  "someting": 6,
  "anything": 0,
  "frequency": "Quaterly"
}
]

别忘了加consumes标签：

@RequestMapping(value = "/getEmployeeList", method = RequestMethod.POST, consumes="application/json", produces = "application/json")
public List<Employee> getEmployeeDataList(@RequestBody Employee[] employeearray) { ... }

Answer 3

对我来说，下面的代码有效，首先发送带有正确标题的 json 字符串

$.ajax({
        type: "POST",
        url : 'save',
        data : JSON.stringify(valObject),
        contentType:"application/json; charset=utf-8",
        dataType:"json",
        success : function(resp){
            console.log(resp);
        },
        error : function(resp){
            console.log(resp);
        }
    });

然后在 Spring 方面 -

@RequestMapping(value = "/save", 
                method = RequestMethod.POST,
                consumes="application/json")
public @ResponseBody String  save(@RequestBody ArrayList<KeyValue> keyValList) {
    //Saving call goes here
    return "";
}

这里的 KeyValue 是简单的 pojo，对应于你的 JSON 结构，你也可以根据需要添加产品，我只是返回字符串。

我的json对象是这样的-

[{"storedKey":"vc","storedValue":"1","clientId":"1","locationId":"1"},
 {"storedKey":"vr","storedValue":"","clientId":"1","locationId":"1"}]

Answer 4

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

我们可以使用如上所示的数组。

Answer 5

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;
    }
}

我的控制器方法

@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]"]

Answer 6

这在您尝试的方式中是不可能的。 Jackson 解组处理在 类型擦除之后编译的java 代码。所以你的

public @ResponseBody ModelMap setTest(@RequestBody List<TestS> refunds, ModelMap map) 

真的只有

public @ResponseBody ModelMap setTest(@RequestBody List refunds, ModelMap map) 

（列表 arg 中没有泛型）。

Jackson 在解组List 时创建的默认类型是LinkedHashMap。

正如@Saint 所提到的，您可以通过为列表创建自己的类型来规避这种情况，如下所示：

class TestSList extends ArrayList<TestS> { }

然后将你的控制器签名修改为

public @ResponseBody ModelMap setTest(@RequestBody TestSList refunds, ModelMap map) {