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java - org.springframework.dao.InvalidDataAccessResourceUsageException : could not prepare SQL statement

转载 作者:行者123 更新时间:2023-12-02 12:11:12 26 4
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我使用小型 Spring boot 应用程序,并收到由 Controller 中的方法 findAll() 触发的错误,

@Controller
@RequestMapping(value = "/")
public class UserController {

@Autowired
private UserService userService;

@GetMapping(value = "/")
public String index() {
return "redirect:/users";
}


@GetMapping(value = "/users")
public String showAllUsers(Model model) {

model.addAttribute("users", userService.findAll());
return "list";
}
}

错误堆栈的重要部分是,

org.springframework.dao.InvalidDataAccessResourceUsageException: could not prepare statement; SQL [select user0_.id as id1_1_, user0_.address as address2_1_, user0_.confirm_password as confirm_3_1_, user0_.country as country4_1_, user0_.email as email5_1_, user0_.name as name6_1_, user0_.newsletter as newslett7_1_, user0_.number as number8_1_, user0_.password as password9_1_, user0_.sex as sex10_1_ from user user0_]; nested exception is org.hibernate.exception.SQLGrammarException: could not prepare statement

Caused by: java.sql.SQLSyntaxErrorException: user lacks privilege or object not found: USER0_.ADDRESS in statement [select user0_.id as id1_1_, user0_.address as address2_1_, user0_.confirm_password as confirm_3_1_, user0_.country as country4_1_, user0_.email as email5_1_, user0_.name as name6_1_, user0_.newsletter as newslett7_1_, user0_.number as number8_1_, user0_.password as password9_1_, user0_.sex as sex10_1_ from user user0_]

Caused by: org.hsqldb.HsqlException: user lacks privilege or object not found: USER0_.ADDRESS

提供了实体类,

@Entity
public class User {

// form:hidden - hidden value
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
Integer id;

// form:input - textbox
@Column(name = "name", columnDefinition = "VARCHAR(30)", nullable = false)
String name;

// form:input - textbox
@Column(name = "email", columnDefinition = "VARCHAR(50)", nullable = false)
String email;

// form:textarea - textarea
@Column(name = "address", columnDefinition = "VARCHAR(255)", nullable = true)
String address;

// form:input - password
@Column(name = "password", columnDefinition = "VARCHAR(20)", nullable = false)
String password;

// form:input - password
String confirmPassword;

// form:checkbox - single checkbox
@Column(name = "newsletter", nullable = true)
boolean newsletter;

// form:checkboxes - multiple checkboxes
// @Column(columnDefinition = "VARCHAR(500)", nullable = false)
@ElementCollection
List<String> framework;

// form:radiobutton - radio button
@Column(name = "sex", columnDefinition = "VARCHAR(1)", nullable = true)
String sex;

// form:radiobuttons - radio button
@Column(name = "number", nullable = true)
Integer number;

// form:select - form:option - dropdown - single select
@Column(name = "", columnDefinition = "VARCHAR(10)", nullable = true)
String country;

// form:select - multiple=true - dropdown - multiple select
// @Column(columnDefinition = "VARCHAR(500)", nullable = true)
@ElementCollection
List<String> skill;

//Check if this is for New of Update
public boolean isNew() {
return (this.id == null);
}


public Integer getId() {
return id;
}

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

public String getName() {
return name;
}

public void setName(String name) {
this.name = name;
}

public String getEmail() {
return email;
}

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

public String getAddress() {
return address;
}

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

public String getPassword() {
return password;
}

public void setPassword(String password) {
this.password = password;
}

public String getConfirmPassword() {
return confirmPassword;
}

public void setConfirmPassword(String confirmPassword) {
this.confirmPassword = confirmPassword;
}

public boolean isNewsletter() {
return newsletter;
}

public void setNewsletter(boolean newsletter) {
this.newsletter = newsletter;
}

public List<String> getFramework() {
return framework;
}

public void setFramework(List<String> framework) {
this.framework = framework;
}

public String getSex() {
return sex;
}

public void setSex(String sex) {
this.sex = sex;
}

public Integer getNumber() {
return number;
}

public void setNumber(Integer number) {
this.number = number;
}

public String getCountry() {
return country;
}

public void setCountry(String country) {
this.country = country;
}

public List<String> getSkill() {
return skill;
}

public void setSkill(List<String> skill) {
this.skill = skill;
}

@Override
public boolean equals(Object o) {
if (this == o) return true;
if (!(o instanceof User)) return false;

User user = (User) o;

if (isNewsletter() != user.isNewsletter()) return false;
if (!getId().equals(user.getId())) return false;
if (!getName().equals(user.getName())) return false;
if (!getEmail().equals(user.getEmail())) return false;
if (getAddress() != null ? !getAddress().equals(user.getAddress()) : user.getAddress() != null) return false;
if (!getPassword().equals(user.getPassword())) return false;
if (getConfirmPassword() != null ? !getConfirmPassword().equals(user.getConfirmPassword()) : user.getConfirmPassword() != null)
return false;
if (!getFramework().equals(user.getFramework())) return false;
if (getSex() != null ? !getSex().equals(user.getSex()) : user.getSex() != null) return false;
if (getNumber() != null ? !getNumber().equals(user.getNumber()) : user.getNumber() != null) return false;
if (getCountry() != null ? !getCountry().equals(user.getCountry()) : user.getCountry() != null) return false;
return getSkill() != null ? getSkill().equals(user.getSkill()) : user.getSkill() == null;
}

@Override
public int hashCode() {

int result = getId().hashCode();

result = 31 * result + getName().hashCode();
result = 31 * result + getEmail().hashCode();
result = 31 * result + (getAddress() != null ? getAddress().hashCode() : 0);
result = 31 * result + getPassword().hashCode();
result = 31 * result + (getConfirmPassword() != null ? getConfirmPassword().hashCode() : 0);
result = 31 * result + (isNewsletter() ? 1 : 0);
result = 31 * result + getFramework().hashCode();
result = 31 * result + (getSex() != null ? getSex().hashCode() : 0);
result = 31 * result + (getNumber() != null ? getNumber().hashCode() : 0);
result = 31 * result + (getCountry() != null ? getCountry().hashCode() : 0);
result = 31 * result + (getSkill() != null ? getSkill().hashCode() : 0);
return result;
}
}

提供的repositoy目录下的接口(interface),

@NoRepositoryBean
public interface CrudRepository<T, ID extends Serializable>
extends Repository<T, ID> {

<S extends T> S save(S entity);

T findOne(ID primaryKey);

Iterable<T> findAll();

Long count();

void delete(T entity);
void delete(ID idx);

boolean exists(ID primaryKey);

// … more functionality omitted.
}

以及界面的扩展,

public interface UserRepository extends CrudRepository<User, Long>{

User save(User user);

@Query("SELECT t.name FROM User t where t.id = :id")
String findNameById(@Param("id") Long id);

@Query("UPDATE User SET NAME=:name, EMAIL=:email, ADDRESS=:address, PASSWORD=:password, NEWSLETTER=:newsletter, FRAMEWORK=:framework, SEX=:sex, NUMBER=:number, COUNTRY=:country, SKILL=:skill WHERE id=:id")
User update (@Param("id") Long id);
}

用户 Controller 类已在前面提供。如果我关闭 address 字段,它会向另一个字段显示错误。如果需要,我可以提供更多信息。这里有什么问题吗?

我注释掉了除 ID 和名称之外的所有内容,但仍然出现相同类型的错误,

Caused by: java.sql.SQLSyntaxErrorException: user lacks privilege or object not found: USER0_.NAME in statement [select user0_.id as id1_1_, user0_.name as name2_1_ from user user0_]

....
....
Caused by: org.hsqldb.HsqlException: user lacks privilege or object not found: USER0_.NAME

这里有 2 行更多的错误,

2017-10-03 12:50:52.537 ERROR 8116 --- [on(2)-127.0.0.1] org.hibernate.tool.hbm2ddl.SchemaExport  : HHH000389: Unsuccessful: create table user (id integer generated by default as identity (start with 1), name VARCHAR(30) not null, primary key (id))
2017-10-03 12:50:52.538 ERROR 8116 --- [on(2)-127.0.0.1] org.hibernate.tool.hbm2ddl.SchemaExport : object name already exists: USER in statement [create table user (id integer generated by default as identity (start with 1), name VARCHAR(30) not null, primary key (id))]

我认为问题在于缺乏特权。我有一个简单的配置文件,现在是空的,

@Configuration
@EnableJpaRepositories(basePackages = {
"com.boot.repository",
"com.boot.entity"
})
@EnableTransactionManagement
class PersistenceContext {

}

如何在应用程序中为用户启用权限?

最佳答案

尝试改变自己

public interface UserRepository extends CrudRepository<User, Long>{

User save(User user);

@Query("SELECT t.name FROM User t where t.id = :id")
String findNameById(@Param("id") Long id);

@Query("UPDATE User SET NAME=:name, EMAIL=:email, ADDRESS=:address, PASSWORD=:password, NEWSLETTER=:newsletter, FRAMEWORK=:framework, SEX=:sex, NUMBER=:number, COUNTRY=:country, SKILL=:skill WHERE id=:id")
User update (@Param("id") Long id);
}

public interface UserRepository extends CrudRepository<User, Integer>{

User save(User user);

@Query("SELECT t.name FROM User t where t.id = :id")
String findNameById(@Param("id") Integer id);

@Query("UPDATE User SET NAME=:name, EMAIL=:email, ADDRESS=:address, PASSWORD=:password, NEWSLETTER=:newsletter, FRAMEWORK=:framework, SEX=:sex, NUMBER=:number, COUNTRY=:country, SKILL=:skill WHERE id=:id")
User update (@Param("id") Integer id);
}

不知道这是否能解决您的问题,但正如我所见,您的用户存储库将 Long 设置为实体 ID 类型,但您的用户实体 Id 使用 Integer 作为其类型。

关于java - org.springframework.dao.InvalidDataAccessResourceUsageException : could not prepare SQL statement,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46538176/

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