问题

我正在使用Spring Boot创建一个Web应用程序。其中一个端点期望一个具有一个属性的JSON对象，即```studentId```。我像我的其他函数一样使用DTO来捕获有效载荷。

```java
@PostMapping("/courses/{id}/students")
public SuccessResponse<Void> addEnrolls(@PathVariable Long id, @RequestBody StudentIdPayload payload) throws HandledException {
    courseService.addEnrolls(id, payload.getStudentId());
    return success(HttpStatus.OK);
}

@Data
@AllArgsConstructor
public class StudentIdPayload {
    private Long studentId;
}

但是，当我尝试使用JSON主体{"studentId":1}发布端点时，我收到以下错误：

com.fasterxml.jackson.databind.exc.MismatchedInputException: 无法构造`org.bimoadityar.univms.dto.input.StudentIdPayload`的实例（尽管至少存在一个创建者）：无法从对象值反序列化（无委托或基于属性的创建者）

而如果我只使用值1进行发布，则可以正常工作。

如何使它能够处理对象有效载荷？

有趣的是，当我向StudentIdPayload添加另一个属性，比如String placeholder时，它按预期工作，尽管这个解决方案看起来有些不正规。

<details>
<summary>英文:</summary>

I am using Spring Boot to create a web application. One of the endpoints expect a json object having one property, i.e. ```studentId```. I am using DTO like my other functions to capture the payload.

```java
@PostMapping(&quot;/courses/{id}/students&quot;)
public SuccessResponse&lt;Void&gt; addEnrolls(@PathVariable Long id, @RequestBody StudentIdPayload payload) throws HandledException {
    courseService.addEnrolls(id, payload.getStudentId());
    return success(HttpStatus.OK);
}

@Data
@AllArgsConstructor
public class StudentIdPayload {
    private Long studentId;
}

But when I tried to post the endpoint with json body {"studentId":1}, I got the following error :

com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot construct instance of `org.bimoadityar.univms.dto.input.StudentIdPayload` (although at least one Creator exists): cannot deserialize from Object value (no delegate- or property-based Creator)

While it works if I post using just the value 1.

How can I get it to work with the object payload?

Interestingly, when I add another property to the StudentIdPayload, such as String placeholder, it works as intended, although this solution feels hacky.

答案1

得分: 1

考虑到 https://github.com/FasterXML/jackson-databind/issues/1498，似乎这是预期的行为。

对于我的特定情况，我满意地向我的构造函数添加了 @JsonCreator。

@Data
@AllArgsConstructor(onConstructor = @__(@JsonCreator))
public class StudentIdPayload {
    private Long studentId;
}

@Data
@AllArgsConstructor(onConstructor = @__(@JsonCreator))
public class StudentIdPayload {
    private Long studentId;
}

答案2

得分: -2

默认情况下，反序列化需要无参构造函数，因此添加 @NoArgsConstructor：

@Data
@AllArgsConstructor
@NoArgsConstructor
public class StudentIdPayload {
    private Long studentId;
}

另请参阅：

• Annotations: using custom constructor

@Data
@AllArgsConstructor
@NoArgsConstructor
public class StudentIdPayload {
    private Long studentId;
}