java - Spring Boot REST Controller 没有将给定的 Json 参数解析为 Java 对象

Question

我想使用 Spring Boot 构建一个 Rest Controller 。但是 Spring 无法自动将 Json 转换为 Java 对象。是的:

• 我有空的构造函数
• 我尝试了@RequestParam、@RequestBody、@RequestPart
• 我的 JSON 参数是有效的(我尝试用 Gson 解析，它成功了)

这是我的 REST Controller :

@RestController
public class HelloController {

     @RequestMapping(value= "/abc", method=RequestMethod.POST)
     @ResponseStatus(HttpStatus.OK)
     public @ResponseBody MyResponse handlePurchase(@RequestParam A request){
         return new MyResponse("Simply the best");
     }

}

这是 A.java:

public class A implements Serializable {

    private B something;

    public B getSomething() {
        return Something;
    }

    public void setSomething(B something) {
        this.Something = something;
    }

    @Override
    public String toString() {
        return "A [Something=" + something + "]";
    }

    public A(B something) {
        super();
        this.Something = something;
    }

    public A() {
        super();
    }


}

这是 A.java 使用的 B.java:

public class B implements Serializable {
    private String something;
    private int catsNumber;
    private int dogsNumber;

    public String getSomething() {
        return something;
    }

    public void setSomething(String something) {
        this.something = something;
    }

    public int getCatsNumber() {
        return catsNumber;
    }

    public void setCatsNumber(int catsNumber) {
        this.catsNumber = catsNumber;
    }

    public int getDogsNumber() {
        return dogsNumber;
    }

    public void setDogsNumber(int dogsNumber) {
        this.dogsNumber = dogsNumber;
    }

    @Override
    public String toString() {
        return "B [something=" + something + ", catsNumber=" + catsNumber + ", dogsNumber=" + dogsNumber + "]";
    }

    public B() {
        super();
    }


}

希望有人能帮我解决这个问题，但都没有成功:

• Spring.io
• 其他 Stackoverflow 问答

Answer 1

@RequestParam表示一个url参数http://foo.com?parameter=x。使用post时，需要定义一个@RequestBody

@RequestMapping(value= "/abc", method=RequestMethod.POST)
@ResponseStatus(HttpStatus.OK)
public @ResponseBody MyResponse handlePurchase(@RequestBody A a){
     return new MyResponse("Simply the best");
}