在JAVA中读取XML文件并创建数据库

Question

我在从 XML 文件中读取“特征”并在数据库中填充表时遇到问题。我成功阅读了教师和学生名单，但尝试阅读特征时出错。对于解析，我使用的是 jackson-dataformat-xml。任何帮助将不胜感激。

我得到的错误：

JSON parse error: Cannot deserialize value of type `java.util.ArrayList<com.Homework.Structure.Trait>` from Object value (token `JsonToken.START_OBJECT`); nested exception is com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot deserialize value of type `java.util.ArrayList<com.Better.Homework.Structure.Trait>` from Object value (token `JsonToken.START_OBJECT`)\n 

这是我的 XML 文件：

<teacher id="100" class="English">
    <students>
        <student>
            <id>1</id>
            <first_name>Alice</first_name>
            <last_name>Wild</last_name>
            <traits>
                <trait>nice</trait>
                <trait>good_grades</trait>
            </traits>
        </student>
        <student>
            <id>2</id>
            <first_name>John</first_name>
            <last_name>Doe</last_name>
            <traits>
                <trait>kind</trait>
                <trait>likes_to_help</trait>
            </traits>
        </student>
    </students>
</teacher>

@Data
@Entity
@Table
public class Teacher{

    @SequenceGenerator(
            name = "teacher_sequence",
            sequenceName = "teacher_sequence",
            allocationSize = 1
    )

    @JacksonXmlProperty(isAttribute = true, localName="id") private Integer id;
    @JacksonXmlProperty(isAttribute = true, localName = "class") private String class;

    @OneToMany (cascade = CascadeType.ALL)
    private List<Student> students;

    public Integer getId() {
        return id;
    }

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

@Data
@Entity
@Table
public class Student{
    @SequenceGenerator(
            name = "student_sequence",
            sequenceName = "student_sequence",
            allocationSize = 1
    )

    @Id
    private Integer id;
    private String first_name;
    private String last_name;

    @OneToMany (cascade = CascadeType.ALL)
    private List <Trait> traits;

//    @ElementCollection
//    private List <String> traits;
}

@Table
@Data
@Entity
public class Trait {
    @SequenceGenerator(
            name = "trait_sequence",
            sequenceName = "trait_sequence",
            allocationSize = 1
    )
    @GeneratedValue(
            strategy = GenerationType.SEQUENCE,
            generator = "trait_sequence"
    )

    @Id
    private String traits;


//    @JsonCreator
//    public Trait(@JsonProperty("traits") String trait) {
//        this.trait = trait;
//    }

    public Trait() {

    }
}

Answer 1

你需要创建一个Traits类，是这样的……

class Traits {
  List<Trait> traits
}

然后学生有一个Traits 类的实例组成其中。我认为这将允许您反序列化。

这能解决您的问题吗？在 cmets 中告诉我。