如何将JSON中的details转为ArrayList<Details>集合？

方案一：让Spring自动完成反序列化（推荐） #

直接把控制器的入参改成NewQuote，Spring自带的Jackson会自动把前端传的JSON转换成实体对象，包括嵌套的Details列表，完全不用手动处理JSON：

修改后的控制器代码

@PostMapping(value = "/save-quote", produces = MediaType.APPLICATION_JSON_VALUE)
public void saveQuote(@RequestBody NewQuote newQuote) {
    newQuoteService.saveQuote(newQuote);
}

修改服务层代码

public void saveQuote(NewQuote newQuote) {
    newQuoteRepository.save(newQuote);
}

注意：实体类需要补充必要结构
JPA和Jackson反序列化都要求有无参构造，同时要给所有字段添加getter/setter：

@Entity
public class NewQuote {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;
    private String description;
    @ElementCollection(targetClass = Details.class)
    private List<Details> details = new ArrayList<>();

    // 必须添加无参构造
    public NewQuote() {}

    public NewQuote(String description, ArrayList<Details> details) {
        this.description = description;
        this.details = details;
    }

    // 所有字段的getter和setter
    public Long getId() { return id; }
    public void setId(Long id) { this.id = id; }
    public String getDescription() { return description; }
    public void setDescription(String description) { this.description = description; }
    public List<Details> getDetails() { return details; }
    public void setDetails(List<Details> details) { this.details = details; }
}

Details类补充getter/setter即可：

@Embeddable
public class Details {
    private String description;
    private String label;

    public Details(String description, String label) {
        this.description = description;
        this.label = label;
    }

    public Details() {}

    // getter和setter
    public String getDescription() { return description; }
    public void setDescription(String description) { this.description = description; }
    public String getLabel() { return label; }
    public void setLabel(String label) { this.label = label; }
}

方案二：如果必须用Map接收（不推荐） #

要是因为特殊场景只能用Map<String, String>接收，就用Jackson的ObjectMapper把details对应的JSON字符串转成List<Details>：

修改后的服务层代码

import com.fasterxml.jackson.core.type.TypeReference;
import com.fasterxml.jackson.databind.ObjectMapper;
import java.util.List;
import java.util.ArrayList;

// 注入Spring自动配置的ObjectMapper
private final ObjectMapper objectMapper;

public YourServiceClass(ObjectMapper objectMapper) {
    this.objectMapper = objectMapper;
}

public void saveQuote(Map<String, String> data) {
    String description = data.get("description");
    String detailsJson = data.get("details");
    
    try {
        // 将JSON字符串转成List<Details>
        List<Details> details = objectMapper.readValue(detailsJson, new TypeReference<List<Details>>() {});
        NewQuote newQuote = new NewQuote(description, new ArrayList<>(details));
        newQuoteRepository.save(newQuote);
    } catch (Exception e) {
        // 处理JSON解析异常，比如抛自定义异常或打日志
        throw new RuntimeException("解析details失败", e);
    }
}

内容的提问来源于stack exchange，提问作者xRay