如何使用Spring Boot将数据从一个html表单传递到多个表

时间:2018-09-10 11:48:04

标签: java hibernate spring-mvc spring-boot jpa

我已经编码了三个类(UserUserCredentialAddress),我想在其中使用映射将数据存储到表中。我正在使用JSON将数据存储到表中。

当我存储数据时,数据存储在所有表中,但是在用户ID中,它显示1,在UserCredential ID中,它显示3,在地址ID中,它显示2条目的所有ID应该为1

类是

package com.spring.demo.model;

import java.util.Date;

import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.Lob;
import javax.persistence.OneToOne;
import javax.persistence.Table;
import javax.persistence.Transient;

@Entity
@Table(name="user")
public class User {

    @Id
    @Column(name="user_id")
    @GeneratedValue(strategy=GenerationType.AUTO)
    private int id;

    private String fName;

    private String lName;

    @Column(unique=true,nullable=true)
    private String email;

    @Column(unique=true,nullable=true)
    private long mobile;

    private Date dob;

    @Lob
    private byte[] image;   

    @Transient
    private String base64Image;

    @OneToOne(cascade=CascadeType.ALL,fetch =FetchType.EAGER)
    @JoinColumn(name="userCredential_id")
    private UserCredential userCredential;

    @OneToOne(cascade=CascadeType.ALL,fetch =FetchType.EAGER)
    @JoinColumn(name="add_id")
    private Address address;



    public Address getAddress() {
        return address;
    }

    public void setAddress(Address address) {
        this.address = address;
    }

    public int getId() {
        return id;
    }

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

    public String getfName() {
        return fName;
    }

    public void setfName(String fName) {
        this.fName = fName;
    }

    public String getlName() {
        return lName;
    }

    public void setlName(String lName) {
        this.lName = lName;
    }

    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

    public long getMobile() {
        return mobile;
    }

    public void setMobile(long mobile) {
        this.mobile = mobile;
    }

    public Date getDob() {
        return dob;
    }

    public void setDob(Date dob) {
        this.dob = dob;
    }

    public byte[] getImage() {
        return image;
    }

    public void setImage(byte[] image) {
        this.image = image;
    }

    public UserCredential getUserCredential() {
        return userCredential;
    }

    public void setUserCredential(UserCredential userCredential) {
        this.userCreenter code heredential = userCredential;
    }



}

UserCredential.java

package com.spring.demo.model;

import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.OneToOne;

import com.fasterxml.jackson.annotation.JsonIgnore;

     @Entity
public class UserCredential {

    @Id
    @Column(name="credential_id")
    @GeneratedValue(strategy=GenerationType.AUTO)
    private int id;

    @Column(unique=true,nullable=true)
    private String username;

    private String password;
    private String cnfrmpassword;



    @JsonIgnore
    @OneToOne(cascade=CascadeType.ALL)
    @JoinColumn(name="user_id")
    private User user;

    public int getId() {
        return id;
    }
    public void setId(int id) {
        this.id = id;
    }
    public String getUsername() {
        return username;
    }
    public void setUsername(String username) {
        this.username = username;
    }
    public String getPassword() {
        return password;
    }
    public void setPassword(String password) {
        this.password = password;
    }
    public String getCnfrmpassword() {
        return cnfrmpassword;
    }
    public void setCnfrmpassword(String cnfrmpassword) {
        this.cnfrmpassword = cnfrmpassword;
    }

    public User getUser() {
        return user;
    }
    public void setUser(User user) {
        this.user = user;
    }

}

Address.java

package com.spring.demo.model;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.ManyToOne;
import javax.persistence.Table;

@Entity
@Table(name="address")
public class Address {

    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    @Column(name="add_id")
    private int id;

    @Column(name="city")
    private String city;

    @Column(name="state")
    private String state;

    @Column(name="house_no")
    private String h_no;

    @ManyToOne
    @JoinColumn(name="user_id", nullable=true)
    private User user;

    public int getId() {
        return id;
    }

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

    public String getCity() {
        return city;
    }

    public void setCity(String city) {
        this.city = city;
    }

    public String getState() {
        return state;
    }

    public void setState(String state) {
        this.state = state;
    }

    public String getH_no() {
        return h_no;
    }

    public void setH_no(String h_no) {
        this.h_no = h_no;
    }

    public User getUser() {
        return user;
    }

    public void setUser(User user) {
        this.user = user;
    }

}

我真的不明白如何克服这个问题。

JSON格式以存储数据

{

    "fName":"suresh kumst",
    "lName":"dingh",
    "mobile":4595498366,
    "email":"ksuraj1sd00@gmail.com",
    "dob":"2012-04-23T18:25:43.511Z",
    "address":{
        "city":"noida",
        "state":"up",
        "h_no":"123"
    },
    "userCredential":{

        "username":"ksuraj1asd002",
        "password":"12345",
        "cnfrmpassword":"12345"
    }
}

以及具有不同id的响应,尽管它们在第一项中应为1,并且用户id应具有值

    {
        "id": 1,
        "fName": "suresh kumst",
        "lName": "dingh",
        "email": "ksuraj1sd00@gmail.com",
        "mobile": 4595498366,
        "dob": "2012-04-23T18:25:43.511+0000",
        "image": null,
        "userCredential": {
            "id": 3,
            "username": "ksuraj1asd002",
            "password": "12345",
            "cnfrmpassword": "12345"
        },
        "address": {
            "id": 2,
            "city": "noida",
            "state": "up",
            "h_no": "123",
            "user": null
        }
    }

3 个答案:

答案 0 :(得分:1)

在这里,您正在使用@GeneratedValue(strategy=GenerationType.AUTO)。 一些数据库将使用公共序列来生成并分配序列号。 Hibernate将创建一个表hibernate_sequence,所有实体表都将引用该表以获取下一个序列号。因此,主键将分散在实体之间。 要使每个实体的主键都从1开始,请对每个实体使用@GeneratedValue(strategy=GenerationType.IDENTITY)

答案 1 :(得分:0)

请使用数据库序列,以便您更好地控制生成的ID。在这里,我仅假设后端DB为Oracle/mysql

 @Id
   @GeneratedValue(strategy=GenerationType.AUTO, generator = "employee_sequence")
   @SequenceGenerator(name = "employee_sequence", sequenceName = "EMP_SN")
    private Long empNo;

这里EMP_SN是数据库序列。 希望这能解决您的问题。

答案 2 :(得分:0)

当您使用@GeneratedValue(strategy=GenerationType.AUTO)时,底层的ORM框架将使用标识列,序列或表,具体取决于底层的DB。在您的情况下,我猜它正在使用一个序列(用于为所有表生成ID的单个序列)。在这种情况下,为了使每个表ID都独立于其他表,必须为每个实体赋予不同的序列名称。

如果您使用的是Hibernate,请考虑使用@GenericGenerator。来自其他SO answer的示例:

@GenericGenerator(
    name = "wikiSequenceGenerator",
    strategy = "org.hibernate.id.enhanced.SequenceStyleGenerator",
    parameters = {
            @Parameter(name = "sequence_name", value = "WIKI_SEQUENCE"),
            @Parameter(name = "initial_value", value = "1000"),
            @Parameter(name = "increment_size", value = "1")
    }
)
@Id
@GeneratedValue(generator = "wikiSequenceGenerator")