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This is tge error i am getting atthe time of compilation. I am new at spring and not able to understand the error.Plz help me to find out.

org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'userRepo' defined in com.example.jpa.repository.UserRepo defined in @EnableJpaRepositories declared on JpaRepositoriesRegistrar.EnableJpaRepositoriesConfiguration: Invocation of init method failed; nested exception is org.springframework.data.repository.query.QueryCreationException: Could not create query for public abstract java.util.List com.example.jpa.repository.UserRepo.findNameLike(java.lang.String)! Reason: Failed to create query for method public abstract java.util.List com.example.jpa.repository.UserRepo.findNameLike(java.lang.String)! No property findName found for type User!; nested exception is java.lang.IllegalArgumentException: Failed to create query for method public abstract java.util.List com.example.jpa.repository.UserRepo.findNameLike(java.lang.String)! No property findName found for type User!
This is my Repository class
According to that  the error in the @Query But i am getting how to fix and wher is my mistake.
i have created a tavle called user1 in mysql.
package com.example.jpa.repository;
import java.util.List;
import javax.persistence.criteria.CriteriaBuilder.In;
import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.data.jpa.repository.Modifying;
import org.springframework.data.jpa.repository.Query;
import org.springframework.data.repository.CrudRepository;
import org.springframework.stereotype.Repository;
import com.example.jpa.entity.User;
@Repository
public interface UserRepo extends JpaRepository<User, Integer> {
    public List<User> findByAgeGreaterThanEqual(int age) ;
    public List<User> findByAgeIn(int age);
    public List<User> findByNameOrderByName(String name);
    @Modifying
    @Query (value="select * from user1",nativeQuery=true)
    public List<User> getAllUser();
This is my Entity class
package com.example.jpa.entity;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Table;
@Entity
//@Table(name = "user")
public class User {
    @GeneratedValue(strategy = GenerationType.AUTO)
    private int id;
    private String name;
    private String city;
    private String status;
    public User() {
        super();
    public User(int id, String name, String city, String status) {
        super();
        this.id = id;
        this.name = name;
        this.city = city;
        this.status = status;
    public int getId() {
        return id;
    public void setId(int id) {
        this.id = id;
    public String getName() {
        return name;
    public void setName(String name) {
        this.name = name;
    public String getCity() {
        return city;
    public void setCity(String city) {
        this.city = city;
    public String getStatus() {
        return status;
    public void setStatus(String status) {
        this.status = status;
This is my main application main class
package com.example.jpa;
import java.util.List;
import java.util.Optional;
import org.apache.catalina.core.ApplicationContext;
import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
import org.springframework.context.ConfigurableApplicationContext;
import org.springframework.data.jpa.repository.config.EnableJpaRepositories;
import com.example.jpa.entity.User;
import com.example.jpa.repository.UserRepo;
@SpringBootApplication
public class SpringApplicaton1Application {
    public static void main(String[] args) {
        ConfigurableApplicationContext context = SpringApplication.run(SpringApplicaton1Application.class, args);
        UserRepo repo = context.getBean(UserRepo.class);
        User user = new User();
        user.setId(1);
        user.setCity("Delhi");
        user.setName("Ramesh");
        user.setStatus("Java Developer");
        List<User> users= repo.getAllUser();
        users.forEach(n->{
            System.out.println(n);
This is my Application.property file
spring.datasource.name=test
spring.datasource.url=jdbc:mysql://${MYSQL_HOST:localhost}:3306/practice   
spring.datasource.username=root
spring.datasource.password=root
spring.datasource.driver-class-name=com.mysql.jdbc.Driver
spring.jpa.properties.hibernate.dialect=org.hibernate.dialect.MySQL55Dialect
spring.jpa.hibernate.use-new-id-generator-mappings= false
spring.jpa.hibernate.ddl-auto=update
                Your error message does not match your code; you did not define a query method findNameLike in this code. Also, (1) nativeQuery is an absolute last resort; your query can trivially be written in JPQL, and (2) findAll is already implicitly declared.
– chrylis -cautiouslyoptimistic-
                Jul 30, 2021 at 20:34
                As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.
– Community
                Oct 20, 2021 at 8:01
i was solving this problem by add these annotation in model class
@Table //is a corresponding table that matches that entity in the database
@Entity // for specifies class is an entity and is mapped to a database table. 
@Data // for getter and seter
public class UserModel {
        
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