java - 如何在Java Spring中将对象转换为字符串

我想通过bedType搜索春季数据jpa。但是，bedType不是字符串。不是String bedType，而是BedType bedType（Object）。这是我的资料库

    @Query("select a,b.bedType,b.roomCategory from RoomDetail a left outer join RoomMaster b on a.roomId = b.id where lower(b.bedType) like %:bedType%")
<Page>RoomDetail findByBedType(
        @Param("bedType") BedType bedType,
        Pageable paging);

FindRoomStatus.Java

public class FindRoomStatus {

private BedType bedType;

public BedType getBedType() {
    return bedType;
}

public void setBedType(BedType bedType) {
    this.bedType = bedType;
}

在我的控制器中，我遇到了错误


  字符串不能转换为BedType

data = roomDetailRepository.findByBedType(findRoomStatus.getBedType().toString(), paging);

这是BedType.Java

public class BedType implements Serializable{
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY, generator = "bed_type_seq")
@SequenceGenerator(name = "bed_type_seq", sequenceName = "bed_type_id_bed_type_seq", allocationSize = 1, initialValue = 1)
private int id;

@Column(name = "bed_type_name", length = 20)
private String bedTypeName;

@JsonIgnore
@Column(name = "status", length = 10)
private String status;

public int getId() {
    return id;
}

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

public String getBedTypeName() {
    return bedTypeName;
}

public void setBedTypeName(String bedTypeName) {
    this.bedTypeName = bedTypeName;
}

public String getStatus() {
    return status;
}

public void setStatus(String status) {
    this.status = status;
}

列出房间状态，在此列表中，我想按BedType查找

 {
            "id": 105,
            "roomId": 43,
            "floor": "1",
            "roomNumber": "001",
            "description": "Normal",
            "status": "Vacant Clean"
        },
        {
            "id": 11,
            "bedTypeName": "King size"
        },
        {
            "id": 39,
            "categoryName": "President Suite"
        }

最佳答案

您在查询中使用LIKE关键字，但在查询中接受BedType对象作为参数。您正在从控制器发送String作为参数。这就是问题。 toString方法将给出对象的字符串表示形式。

您可以做的是，将参数更改为String，它接受bedTypeName，如下所示：

@Query("select a,b.bedType,b.roomCategory from RoomDetail a left outer
join RoomMaster b on a.roomId = b.id where lower(b.bedType.bedTypeName) like
%:bedTypeName%")
<Page>RoomDetail findByBedType(
    @Param("bedTypeName") String bedTypeName,
    Pageable paging);

从控制器

data = roomDetailRepository.findByBedType(findRoomStatus.getBedType().getBedTypeName(),
paging);