Java org.springframework.dao.InvalidDataAccessResourceUsageException:无法准备SQL语句
我使用小型Spring boot应用程序,从控制器中获取由findAll方法触发的错误Java org.springframework.dao.InvalidDataAccessResourceUsageException:无法准备SQL语句,java,spring,hibernate,Java,Spring,Hibernate,我使用小型Spring boot应用程序,从控制器中获取由findAll方法触发的错误 @Controller @RequestMapping(value = "/") public class UserController { @Autowired private UserService userService; @GetMapping(value = "/") public String index() { return "redirect
@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;
}
}
@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.
}
这里有两行错误代码
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。我的简单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特权?尝试更改您的权限
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);
}
到
idk如果这能解决你的问题,
但正如我所看到的,您的userrepository设置为Entity ID type,但您的用户Entitiy ID使用Integer作为其类型。将其添加到application.yml文件中 春天: jpa: show-sql:true 生成ddl:true 冬眠:
ddl auto:CREATEDROP其他语句有效吗?如果不检查数据库的权限,用户是否有适当的访问权限。它的小应用程序带有HSQL数据库,没有定义的角色。我已经做了用户实体的Id,它不能解决问题。我仍然收到相同的错误。尝试将此添加到属性或yaml中,以避免失败:create table user spring.jpa.hibernate.ddl auto:create drop我在属性文件spring.jpa.hibernate.ddl auto=create中有此语句,也将其更改为create drop。这些对我都没有帮助。sry,但似乎我帮不了你,以前从未使用过HQL,但我希望这两个链接可以帮助你,虽然这段代码可能会解决这个问题,但如何以及为什么解决这个问题将真正有助于提高您的帖子质量,并可能导致更多的投票。请记住,你是在将来回答读者的问题,而不仅仅是现在提问的人。请在回答中添加解释,并说明适用的限制和假设。
@Configuration
@EnableJpaRepositories(basePackages = {
"com.boot.repository",
"com.boot.entity"
})
@EnableTransactionManagement
class PersistenceContext {
}
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);
}