我将请求正文传递给 postman 的POST请求,类似于以下内容:
"name":"Mars",
"artifacts":[
{
"elements":[
{
"name":"carbon",
"amount":0.5,
"measurement":"g"
}
],
"typeName":"typeA"
},
{
"elements":[
{
"name":"hydrogen",
"amount":0.2,
"measurement":"g"
}
],
"typeName":"typeB"
}
]
rest Controller 中的create方法如下所示。
@RequestMapping("/create")
public Planet create(@RequestBody Planet data) {
Planet mars = planetService.create(data.getName(),data.getArtifacts());
return mars;
Planet及其所有嵌套对象都具有默认构造函数,例如:
public Planet() {}
但是,由于缺少默认构造函数,我无法创建新的Planet对象。请帮忙!
编辑:
行星类
public class Planet {
@JsonProperty("name")
private String name;
@Field("artifacts")
private List<Artifact> artifacts;
public Planet() {}
public Planet(String name, List<Artifact> artifacts)
{
this.name = name;
this.artifacts = artifacts;
}
//setters and getters
}
Artifact 类:
public class Artifact() {
@Field("elements")
private List<Element> elements;
@JsonProperty("typeName")
private String typeName;
public Artifact() {}
public Artifact(String typeName, List<Element> elements)
{
this.typeName = typeName;
this.elements = elements;
}
}
元素类:
public class Element() {
@JsonProperty("elementName")
private String name;
@JsonProperty("amount")
private double amount;
@JsonProperty("measurement")
private String measurement;
public Element() {}
public Element(String name, double amount, String measurement)
{
//assignments
}
}
最佳答案
我不明白您面临的问题是什么,但是我可以立即看到一个错误,因此我猜测这就是您面临的问题,我将为您提供解决方案。
创建一个与您的json数据结构相匹配的类,如下所示:
Class PlanetData {
private String name;
private List<Planet> artifacts;
public PlanetData(String name, List<Planet> artifacts){
name = name;
artifacts = artifacts;
}
// include rest of getters and setters here.
}
然后,您的 Controller 应如下所示。基本上,您需要将
@RequestBody
放入要从请求JSON接收的所有参数。之前,您只将@RequestBody
放在名称参数上,而不是工件参数上,并且由于请求正文只能被使用一次,因此您需要一个包装器类来使用单个@RequestBody
批注接收完整的请求正文。@RequestMapping("/create")
public String create(@RequestBody PlanetData data) {
Planet mars = planetService.create(data.getName(),data.getArtifacts());
return mars.toString();
}
编辑:看着Planet类,它也需要一些修改
public class Planet {
private String typeName; // key in json should match variable name for proper deserialization or you need to use some jackson annotation to map your json key to your variable name.
private List<Element> elements;
public Planet() {}
public Planet(String typeName, List<Element> elements)
{
this.typeName = typeName;
this.elements = elements;
}
//setters and getters. Remember to change your setters and getter from name to typeName.
}
希望这能解决您的问题。