spring 在spring mvc控制器接收json并反序列化为对象列表
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receiving json and deserializing as List of object at spring mvc controller
提问by HKumar
My code is as below:
我的代码如下：
controller
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
TestS pojo
public class TestS implements Serializable {
private String name;
private String age;
//getter setter
}
Json request
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'
});
It's giving this error
它给出了这个错误
java.lang.ClassCastException: java.util.LinkedHashMap cannot be cast to com.air.cidb.entities.TestS
java.lang.ClassCastException：java.util.LinkedHashMap 无法转换为 com.air.cidb.entities.TestS
I am using spring 3.1.2and Hymanson 2.0.4
我正在使用spring 3.1.2和Hymanson 2.0.4
Edit:I want to receive list of TestS objects at my controller method, and process them. I am not able to find if I am sending wrong json or my method signature is wrong.
编辑：我想在我的控制器方法中接收 TestS 对象列表，并处理它们。我无法确定是我发送了错误的 json 还是我的方法签名错误。
回答by Shailendra
Here is the code that works for me. The key is that you need a wrapper class.
这是对我有用的代码。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 + "]";
}
A PersonWrapper class
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;
}
}
My Controller methods
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;
}
The request sent is json in POST
发送的请求是 POST 中的 json
{"persons":[{"name":"shail1","age":"2"},{"name":"shail2","age":"3"}]}
And the response is
回应是
["Saved person: Person [name=shail1, age=2]","Saved person: Person [name=shail2, age=3]"]
回答by Dirk Lachowski
This is not possible the way you are trying it. The Hymanson unmarshalling works on the compiled java code aftertype erasure. So your
这在您尝试的方式中是不可能的。Hymanson 解组在类型擦除后对已编译的 Java 代码起作用。所以你的
public @ResponseBody ModelMap setTest(@RequestBody List<TestS> refunds, ModelMap map)
is really only
真的只是
public @ResponseBody ModelMap setTest(@RequestBody List refunds, ModelMap map)
(no generics in the list arg).
（列表 arg 中没有泛型）。
The default type Hymanson creates when unmarshalling a Listis a LinkedHashMap.
Hymanson 在解组 a 时创建的默认类型List是 a LinkedHashMap。
As mentioned by @Saint you can circumvent this by creating your own type for the list like so:
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一点，如下所示：
class TestSList extends ArrayList<TestS> { }
and then modifying your controller signature to
然后将您的控制器签名修改为
public @ResponseBody ModelMap setTest(@RequestBody TestSList refunds, ModelMap map) {
回答by Deepak
@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;
}
We can use Array as shown above.
我们可以使用如上所示的 Array。
回答by Jis Mathew
I believe this will solve the issue
我相信这将解决问题
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'
});
回答by Pratik Goenka
For me below code worked, first sending json string with proper headers
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);
}
});
And then on Spring side -
然后在春天方面 -
@RequestMapping(value = "/save",
method = RequestMethod.POST,
consumes="application/json")
public @ResponseBody String save(@RequestBody ArrayList<KeyValue> keyValList) {
//Saving call goes here
return "";
}
Here KeyValue is simple pojo that corresponds to your JSON structure also you can add produces as you wish, I am simply returning string.
这里 KeyValue 是一个简单的 pojo，对应于您的 JSON 结构，您也可以根据需要添加产品，我只是返回字符串。
My json object is like this -
我的 json 对象是这样的 -
[{"storedKey":"vc","storedValue":"1","clientId":"1","locationId":"1"},
{"storedKey":"vr","storedValue":"","clientId":"1","locationId":"1"}]
回答by AmitG
Solution works very well,
解决方案效果很好，
public List<String> savePerson(@RequestBody Person[] personArray)
For this signature you can pass Personarray from postman like
对于此签名，您可以Person从邮递员传递数组，例如
[
{
"empId": "10001",
"tier": "Single",
"someting": 6,
"anything": 0,
"frequency": "Quaterly"
}, {
"empId": "10001",
"tier": "Single",
"someting": 6,
"anything": 0,
"frequency": "Quaterly"
}
]
Don't forget to add consumestag:
不要忘记添加consumes标签：
@RequestMapping(value = "/getEmployeeList", method = RequestMethod.POST, consumes="application/json", produces = "application/json")
public List<Employee> getEmployeeDataList(@RequestBody Employee[] employeearray) { ... }
