对象提取中的rest API json数组

Question

我收到一个错误：com.fasterxml.jackson.databind.exc.MismatchedInputException: 无法反序列化 `java.util.ArrayList

可能是因为我在 Main 中的 ObjectMapper 或 Entity 中的 ArrayList 期间的数据类型错误... 我不知道如何修复它，因为我似乎不明白错误具体在哪里，或者我是否应该以不同的方式解决它（即：不在 main 中执行业务逻辑并创建一个 serviceImpl 和也许是一个 dto 包。

我猜这个问题有两个方面：（1）如何解决这个错误。 （2）在restAPIs中标记json数据的标准化最佳方法是什么，然后我们可以通过弄乱url进行排序。即 localhost:8080/recipes?name=scrambledEggs

1) JSON 文件

    "recipes": [
      {
        "name": "scrambledEggs",
        "ingredients": [
          "1 tsp oil",
          "2 eggs",
          "salt"
        ],
        "instructions": [
          "Beat eggs with salt",
          "Heat oil in pan",
          "Add eggs to pan when hot",
          "Gather eggs into curds, remove when cooked",
          "Salt to taste and enjoy"
        ]
      },
      {
        "name": "garlicPasta",
        "ingredients": [
          "500mL water",
          "100g spaghetti",
          "25mL olive oil",
          "4 cloves garlic",
          "Salt"
        ],
        "instructions": [
          "Heat garlic in olive oil",
          "Boil water in pot",
          "Add pasta to boiling water",
          "Remove pasta from water and mix with garlic olive oil",
          "Salt to taste and enjoy"
        ]
      

2) 实体

public class RecipesInfo {

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

    private ArrayList<RecipesFile> recipes;

    //get + set

    public class RecipesFile {
        private String name;
        private ArrayList<String> ingredients;
        private ArrayList<String> instructions;

        //get + set

3) 主文件

ublic class ApiApplication {

    public static void main(String[] args) {
        SpringApplication.run(ApiApplication.class, args);
    }

    @Bean
    CommandLineRunner runner(RecipesService recipesService) {
        return args -> {
            ObjectMapper mapper = new ObjectMapper();
            TypeReference<ArrayList<RecipesInfo>> typeReference = new TypeReference<ArrayList<RecipesInfo>>() {};
            InputStream inputStream =TypeReference.class.getResourceAsStream("/json/data.json");
            try {
                ArrayList<RecipesInfo> recipes = mapper.readValue(inputStream, typeReference);
                recipesService.save(recipes);
                System.out.println("data saved: recipes");

            } catch (IOException e) {
                System.out.println("unable to save data: recipes");
                System.out.println(e);
            }
        };

Answer 1

JSON 应该包含一个数组或 JSON 对象。在您的情况下，根据您的实体，您应该在 recipes 数组周围添加大括号

{
    "recipes": [
      {
        "name": "scrambledEggs",
        "ingredients": [
          "1 tsp oil",
          "2 eggs",
          "salt"
        ],
        "instructions": [
          "Beat eggs with salt",
          "Heat oil in pan",
          "Add eggs to pan when hot",
          "Gather eggs into curds, remove when cooked",
          "Salt to taste and enjoy"
        ]
      },
      {
        "name": "garlicPasta",
        "ingredients": [
          "500mL water",
          "100g spaghetti",
          "25mL olive oil",
          "4 cloves garlic",
          "Salt"
        ],
        "instructions": [
          "Heat garlic in olive oil",
          "Boil water in pot",
          "Add pasta to boiling water",
          "Remove pasta from water and mix with garlic olive oil",
          "Salt to taste and enjoy"
        ]
    ]
}
      

之后，您可以简单地反序列化单个 RecipesInfo 而不是它的集合

RecipesInfo recipes = mapper.readValue(inputStream, RecipesInfo.class);

另外，我注意到你的内部 RecipesFile 类不是 static。没有它，Object Mapper 将无法创建 RecipesFile 类的实例

public class RecipesInfo {

    ...
    public static class RecipesFile {
        private String name;
        private ArrayList<String> ingredients;
        private ArrayList<String> instructions;

        //get + set
    ...

关于第二个问题，可以查看以下Stackoverflow article。