JPA using @SQLInsert throws Parameter index out of range
@Data
@MappedSuperclass
public class BaseModel implements Serializable {
private static final Long serialVersionUID = -1442801573244745790L;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@JsonFormat(pattern = "yyyy-MM-dd HH:mm:ss")
@Convert(converter = LocalDateTimeConverter.class)
private LocalDateTime createAt = LocalDateTime.now();
@JsonFormat(pattern = "yyyy-MM-dd HH:mm:ss")
@Convert(converter = LocalDateTimeConverter.class)
private LocalDateTime updateAt = LocalDateTime.now();
}
@Data
@Entity
@Table(name = "tb_vip_code")
@SQLInsert(sql = "insert ignore into tb_vip_code (code, duration) values (?, ?)")
public class VipCode extends BaseModel {
private static final Long serialVersionUID = -4697221755301869573L;
private String code;
private Integer duration;
private Integer status = 0;
private Long userId;
public VipCode() {}
}
@Test
public void addOne() throws Exception {
VipCode vipCode = new VipCode();
vipCode.setCode("123456");
vipCode.setDuration(1);
service.addOne(vipCode);
}
create table tb_vip_code
(
id bigint auto_increment
primary key,
code varchar(20) not null,
status int default '0' not null,
user_id bigint null,
duration int not null,
create_at datetime default CURRENT_TIMESTAMP not null,
update_at timestamp default CURRENT_TIMESTAMP not null,
constraint tb_vip_code_code_uindex unique (code)
);
All code as above, I am trying to use a custom sql to save object, but it throws an exception:
Caused by: java.sql.SQLException: Parameter index out of range (3 > number of parameters, which is 2).
I only have tow parameters, why it says needed three?
java sql hibernate spring-data-jpa parameterbinding
edited Nov 15 '18 at 6:55
Jens Schauder
46.7k17115240
asked Nov 15 '18 at 2:40
林永福林永福
1
add a comment |
@Data
@MappedSuperclass
public class BaseModel implements Serializable {
private static final Long serialVersionUID = -1442801573244745790L;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@JsonFormat(pattern = "yyyy-MM-dd HH:mm:ss")
@Convert(converter = LocalDateTimeConverter.class)
private LocalDateTime createAt = LocalDateTime.now();
@JsonFormat(pattern = "yyyy-MM-dd HH:mm:ss")
@Convert(converter = LocalDateTimeConverter.class)
private LocalDateTime updateAt = LocalDateTime.now();
}
@Data
@Entity
@Table(name = "tb_vip_code")
@SQLInsert(sql = "insert ignore into tb_vip_code (code, duration) values (?, ?)")
public class VipCode extends BaseModel {
private static final Long serialVersionUID = -4697221755301869573L;
private String code;
private Integer duration;
private Integer status = 0;
private Long userId;
public VipCode() {}
}
@Test
public void addOne() throws Exception {
VipCode vipCode = new VipCode();
vipCode.setCode("123456");
vipCode.setDuration(1);
service.addOne(vipCode);
}
create table tb_vip_code
(
id bigint auto_increment
primary key,
code varchar(20) not null,
status int default '0' not null,
user_id bigint null,
duration int not null,
create_at datetime default CURRENT_TIMESTAMP not null,
update_at timestamp default CURRENT_TIMESTAMP not null,
constraint tb_vip_code_code_uindex unique (code)
);
All code as above, I am trying to use a custom sql to save object, but it throws an exception:
Caused by: java.sql.SQLException: Parameter index out of range (3 > number of parameters, which is 2).
I only have tow parameters, why it says needed three?
java sql hibernate spring-data-jpa parameterbinding
edited Nov 15 '18 at 6:55
Jens Schauder
46.7k17115240
asked Nov 15 '18 at 2:40
林永福林永福
1
add a comment |
@Data
@MappedSuperclass
public class BaseModel implements Serializable {
private static final Long serialVersionUID = -1442801573244745790L;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@JsonFormat(pattern = "yyyy-MM-dd HH:mm:ss")
@Convert(converter = LocalDateTimeConverter.class)
private LocalDateTime createAt = LocalDateTime.now();
@JsonFormat(pattern = "yyyy-MM-dd HH:mm:ss")
@Convert(converter = LocalDateTimeConverter.class)
private LocalDateTime updateAt = LocalDateTime.now();
}
@Data
@Entity
@Table(name = "tb_vip_code")
@SQLInsert(sql = "insert ignore into tb_vip_code (code, duration) values (?, ?)")
public class VipCode extends BaseModel {
private static final Long serialVersionUID = -4697221755301869573L;
private String code;
private Integer duration;
private Integer status = 0;
private Long userId;
public VipCode() {}
}
@Test
public void addOne() throws Exception {
VipCode vipCode = new VipCode();
vipCode.setCode("123456");
vipCode.setDuration(1);
service.addOne(vipCode);
}
create table tb_vip_code
(
id bigint auto_increment
primary key,
code varchar(20) not null,
status int default '0' not null,
user_id bigint null,
duration int not null,
create_at datetime default CURRENT_TIMESTAMP not null,
update_at timestamp default CURRENT_TIMESTAMP not null,
constraint tb_vip_code_code_uindex unique (code)
);
All code as above, I am trying to use a custom sql to save object, but it throws an exception:
Caused by: java.sql.SQLException: Parameter index out of range (3 > number of parameters, which is 2).
I only have tow parameters, why it says needed three?
java sql hibernate spring-data-jpa parameterbinding
edited Nov 15 '18 at 6:55
Jens Schauder
46.7k17115240
asked Nov 15 '18 at 2:40
林永福林永福
1
@Data
@MappedSuperclass
public class BaseModel implements Serializable {
private static final Long serialVersionUID = -1442801573244745790L;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@JsonFormat(pattern = "yyyy-MM-dd HH:mm:ss")
@Convert(converter = LocalDateTimeConverter.class)
private LocalDateTime createAt = LocalDateTime.now();
@JsonFormat(pattern = "yyyy-MM-dd HH:mm:ss")
@Convert(converter = LocalDateTimeConverter.class)
private LocalDateTime updateAt = LocalDateTime.now();
}
@Data
@Entity
@Table(name = "tb_vip_code")
@SQLInsert(sql = "insert ignore into tb_vip_code (code, duration) values (?, ?)")
public class VipCode extends BaseModel {
private static final Long serialVersionUID = -4697221755301869573L;
private String code;
private Integer duration;
private Integer status = 0;
private Long userId;
public VipCode() {}
}
@Test
public void addOne() throws Exception {
VipCode vipCode = new VipCode();
vipCode.setCode("123456");
vipCode.setDuration(1);
service.addOne(vipCode);
}
create table tb_vip_code
(
id bigint auto_increment
primary key,
code varchar(20) not null,
status int default '0' not null,
user_id bigint null,
duration int not null,
create_at datetime default CURRENT_TIMESTAMP not null,
update_at timestamp default CURRENT_TIMESTAMP not null,
constraint tb_vip_code_code_uindex unique (code)
);
All code as above, I am trying to use a custom sql to save object, but it throws an exception:
Caused by: java.sql.SQLException: Parameter index out of range (3 > number of parameters, which is 2).
I only have tow parameters, why it says needed three?
java sql hibernate spring-data-jpa parameterbinding
java sql hibernate spring-data-jpa parameterbinding
edited Nov 15 '18 at 6:55
Jens Schauder
46.7k17115240
asked Nov 15 '18 at 2:40
林永福林永福
1
edited Nov 15 '18 at 6:55
Jens Schauder
46.7k17115240
asked Nov 15 '18 at 2:40
林永福林永福
1
edited Nov 15 '18 at 6:55
Jens Schauder
46.7k17115240
edited Nov 15 '18 at 6:55
Jens Schauder
46.7k17115240
edited Nov 15 '18 at 6:55
Jens Schauder
46.7k17115240
46.7k17115240
asked Nov 15 '18 at 2:40
林永福林永福
1
asked Nov 15 '18 at 2:40
林永福林永福
1
asked Nov 15 '18 at 2:40
林永福林永福
1
1
add a comment |
add a comment |
1 Answer
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The SQL statement specified by @SQLInsert gets the fields of your entity bound as parameters.
You have at least 4 attributes in your VipCode entity directly, and I suspect even more from the BaseModel class it inherits from. But your SQL statment expects only two parameters.
To fix the problem remove the superfluous attributes from the entity, add parameters for them to the query or mark them as @Transient.
answered Nov 15 '18 at 6:54
Jens SchauderJens Schauder
46.7k17115240
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
The SQL statement specified by @SQLInsert gets the fields of your entity bound as parameters.
You have at least 4 attributes in your VipCode entity directly, and I suspect even more from the BaseModel class it inherits from. But your SQL statment expects only two parameters.
To fix the problem remove the superfluous attributes from the entity, add parameters for them to the query or mark them as @Transient.
answered Nov 15 '18 at 6:54
Jens SchauderJens Schauder
46.7k17115240
add a comment |
The SQL statement specified by @SQLInsert gets the fields of your entity bound as parameters.
You have at least 4 attributes in your VipCode entity directly, and I suspect even more from the BaseModel class it inherits from. But your SQL statment expects only two parameters.
To fix the problem remove the superfluous attributes from the entity, add parameters for them to the query or mark them as @Transient.
answered Nov 15 '18 at 6:54
Jens SchauderJens Schauder
46.7k17115240
add a comment |
The SQL statement specified by @SQLInsert gets the fields of your entity bound as parameters.
You have at least 4 attributes in your VipCode entity directly, and I suspect even more from the BaseModel class it inherits from. But your SQL statment expects only two parameters.
To fix the problem remove the superfluous attributes from the entity, add parameters for them to the query or mark them as @Transient.
answered Nov 15 '18 at 6:54
Jens SchauderJens Schauder
46.7k17115240
The SQL statement specified by @SQLInsert gets the fields of your entity bound as parameters.
You have at least 4 attributes in your VipCode entity directly, and I suspect even more from the BaseModel class it inherits from. But your SQL statment expects only two parameters.
To fix the problem remove the superfluous attributes from the entity, add parameters for them to the query or mark them as @Transient.
answered Nov 15 '18 at 6:54
Jens SchauderJens Schauder
46.7k17115240
answered Nov 15 '18 at 6:54
Jens SchauderJens Schauder
46.7k17115240
answered Nov 15 '18 at 6:54
Jens SchauderJens Schauder
46.7k17115240
answered Nov 15 '18 at 6:54
Jens SchauderJens Schauder
46.7k17115240
46.7k17115240
add a comment |
add a comment |
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