Spring MVC 传入List参数的接收问题
2016-06-21 17:56
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帖一个歪果仁的解决方案,测试没有问题。建议使用此方式。
Here is the code that works for me. The key is that you need a wrapper class.
A PersonWrapper class
My Controller methods
The request sent is json in POST
And the response is
Here is the code that works for me. The key is that you need a wrapper class.
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
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
@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
{"persons":[{"name":"shail1","age":"2"},{"name":"shail2","age":"3"}]}
And the response is
["Saved person: Person [name=shail1, age=2]","Saved person: Person [name=shail2, age=3]"]http://stackoverflow.com/questions/23012841/receiving-json-and-deserializing-as-list-of-object-at-spring-mvc-controller
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