java - JPA/Spring 从 2 个 MySQL 表返回 JSON 作为嵌套对象

Question

JPA/Spring 从 2 个 MySQL 表返回 JSON 作为嵌套对象

我是 JPA/Spring 的新手，所以如果你至少能指出我正确的方向，那将会非常有帮助!干杯!

这是我得到的:

[{"conId":1,"phone1":"test","tblPeople":   {"pepId":1,"pepAccountIdFk":1,"firstName":"TestFirst","lastName":"Last","title":"Title","notes":"Notes"}}]

我想要的是:

 [{"conId":1,"phone1":"test",  "pepId":1,"pepAccountIdFk":1,"firstName":"TestFirst","lastName":"Last","title":"Title","notes":"Notes"}]

Java 类 tblPeopleContactInfo:

package CMS;




@Entity
public class TblPeopleContactInfo implements Serializable{

private static final long serialVersionUID = 1L;

@Id
private int conId;

@Column(name="con_phone_1")
private String phone1;



@OneToOne(optional=false)
@JoinColumn( name="con_person_id_fk")
private TblPeople tblPeople;


public String getphone1() {
    return phone1;
}

public void setphone1(String phone1) {
    this.phone1 = phone1;
}

public TblPeople getTblPeople() {
    return tblPeople;
}

public void setTblPeople(TblPeople tblPeople) {
    this.tblPeople = tblPeople;
} 

public int getConId() {
    return conId;
}

public void setConId(int conId) {
    this.conId = conId;
}




public TblPeopleContactInfo(){

}
}

tbl人:

@Entity
public class TblPeople{
@Id
private int pepId;
private int pepAccountIdFk;
@Column(name="pep_first_name")
private String firstName;
@Column(name="pep_last_name")
private String lastName;
@Column(name="pep_title")
private String title;
@Column(name="pep_notes")
private String notes;

@OneToOne(mappedBy="tblPeople")
@JsonIgnore
private TblPeopleContactInfo tblPeopleContactInfo;


public TblPeopleContactInfo getTblPeopleContactInfo() {
    return tblPeopleContactInfo;
}


public void setTblPeopleContactInfo(TblPeopleContactInfo    tblPeopleContactInfo) {
    this.tblPeopleContactInfo = tblPeopleContactInfo;
} 







public int getPepId() {
    return pepId;
}


public void setPepId(int pepId) {
    this.pepId = pepId;
}


public int getPepAccountIdFk() {
    return pepAccountIdFk;
}


public void setPepAccountIdFk(int pepAccountIdFk) {
    this.pepAccountIdFk = pepAccountIdFk;
}


public String getFirstName() {
    return firstName;
}


public void setFirstName(String firstName) {
    this.firstName = firstName;
}


public String getLastName() {
    return lastName;
}


public void setLastName(String lastName) {
    this.lastName = lastName;
}


public String getTitle() {
    return title;
}


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


public String getNotes() {
    return notes;
}


public void setNotes(String notes) {
    this.notes = notes;
}


public TblPeople() {
}

Answer 1

而不是这个

public class TblPeopleContactInfo implements Serializable{
       private int conId;
       private String phone1;
       private TblPeople tblPeople;
}

你需要这个类结构(以获取你想要的 JSON)

 public class TblPeopleContactInfo implements Serializable{
           private int conId;
           private String phone1;
           private int pepId;
           private int pepAccountIdFk;
           private String firstName;
           private String lastName;
           private String title;
           private String notes;
    }

这意味着，您需要创建一个单独的类结构，并将所有属性包含在 Single 类中以获得您想要的 JSON。而且和JPA没有任何关系。