java - JSON 解析错误 : Cannot deserialize instance of. 。超过 START_ARRAY token

Question

我知道 stackoverflow 上有一些关于这个问题的问题。但我花了几个小时试图解决这个错误，但没有成功。

我使用 mysql 数据库来存储值。

我不断收到来自com.example.springboot.Recipe 文件。

这是 springboot 配方文件

package com.example.springboot;
import com.example.springboot.Recipe;
import javax.persistence.*;
import javax.validation.constraints.NotBlank;
import  javax.validation.constraints.NotNull;

@Entity // This tells Hibernate to make a table out of this class

public class Recipe {

    public Recipe(){

    }

    public Recipe(Integer id, String name, String description, String type, Integer preptime, Integer cooktime, String content, Integer difficulty){
        this.id = id;
        this.name = name;
        this.description = description;
        this.type = type;
        this.preptime = preptimee;
        this.cooktime = cooktime;
        this.content = content;
        this.difficulty = difficulty;
    }

    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    private Integer id;


    private String name;


    private String description;


    private String type;


    private Integer preptime;


    private Integer cooktime;


    @Column(columnDefinition = "TEXT")
    private String content;

    private Integer difficulty;






    public Integer getId() {
        return id;
    }

    public void setId(Integer id) {
        this.id = id;
    }

    public String getTitle() {
        return name;
    }

    public void setTitle(String name) {
        this.name = name;
    }

    public String getDescription() {
        return description;
    }

    public void setDescription(String description) {
        this.description = description;
    }

    public String getType() {
        return type;
    }

    public void setType(String type) {
        this.type = type;
    }

    public String getContent() {
        return content;
    }

    public void setContent(String content) {
        this.content = content;
    }

    public Integer getDifficulty() {
        return difficulty;
    }

    public void setDifficulty(Integer difficulty) {
        this.difficulty = difficulty;
    }

    public Integer getCookingtime() {
        return cooktime;
    }

    public void setCookingtimeime(Integer cooktime) {
        this.cooktime = cooktime;
    }

    public Integer getPreparationtime() {
        return preptime;
    }

    public void setPreparationtime(Integer preptime) {
        this.preptime = preptime;
    }
}

Main Controller:
@PutMapping("/recipes/edit/{id}")
    void updateRecipe2(@PathVariable int id, @RequestBody Recipe recipe ) {

        Recipe recipe_ = recipeRepository.findById(id).get();
        recipe_.setTitle(recipe.getTitle());


        System.out.println("sss " + recipe.getname());



        System.out.println("change");
        recipeRepository.save(recipe_);
    }

服务.ts:

updateRecipe2 (id: number, recipe: any): Observable<any > {
    const url = `${this.usersUrl}/edit/${id}`;
  return this.http.put(url ,recipe);
}

调用 updateRecipe2 的位置:

 save(): void {
    const id = +this.route.snapshot.paramMap.get('name');
     this.recipeService.updateRecipe2(id, this.recipes)
       .subscribe(() => this.gotoUserList());

   }

一旦用户单击“保存”，此功能就会保存所做的更改。我希望我提供的代码片段足以帮助解决问题。预先感谢您。

我正在使用 Spring Boot 构建一个 REST API，并且使用 AngularJS 作为前端。我对网络开发还很陌生。

Answer 1

您正在向需要单个配方对象的 API 端点发送配方列表。

您的选择是:

• 一次仅发送一个配方对象，例如:

  this.recipeService.updateRecipe2(id, this.recipes[0])
  

• 或者:创建一个新的 API 端点来接受配方列表，并“批量”编辑它们

  @PutMapping("/recipes/edit")
  void updateRecipes(@RequestBody List<Recipe> recipe ) {