我在我的项目中使用控制器、用户、仓库和服务创建了一个登录用户。在控制器中,如果电子邮件ID和密码不为空,则登录,否则会抛出异常“用户不存在”。我得到了下面的错误,你能帮助我修复这个问题吗?
错误:
Error creating bean with name 'registrationRepository': FactoryBean threw exception on object creation; nested exception is java.lang.IllegalArgumentException: Failed to create query for method public abstract com.adventure.network.model.User com.adventure.network.repository.RegistrationRepository.findByEmailIdAndByPassword(java.lang.String,java.lang.String)! No property byPassword found for type User!
控制器:
@RestController
public class RegistrationController {
@Autowired
private RegistrationService service;
@PostMapping("/login")
public User loginUser(@RequestBody User user) throws Exception {
String tempEmailId = user.getEmailId();
String tempPassword = user.getPassword();
User userObject = null;
if(tempEmailId!=null && tempPassword!=null) {
userObject = service.fetchUserByEmailIdByPassword(tempEmailId, tempPassword);
}
if(userObject == null) {
throw new Exception("User is not exict");
}
return userObject;
}
服务项目:
@Service
public class RegistrationService {
@Autowired
private RegistrationRepository repo;
public User fetchUserByEmailIdByPassword(String email, String password) {
return repo.findByEmailIdAndByPassword(email, password);
}
}
储存库:
public interface RegistrationRepository extends JpaRepository<User, Integer> {
public User findByEmailIdAndByPassword(String emailId, String password);
}
pom.xml
<?xml version="1.0" encoding="UTF-8"?>
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 https://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<parent>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-parent</artifactId>
<version>2.3.3.RELEASE</version>
<relativePath/> <!-- lookup parent from repository -->
</parent>
<groupId>com.adventure</groupId>
<artifactId>network</artifactId>
<version>0.0.1-SNAPSHOT</version>
<name>network</name>
<description>Demo project for Spring Boot</description>
<properties>
<java.version>11</java.version>
</properties>
<dependencies>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-data-jpa</artifactId>
</dependency>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-web</artifactId>
</dependency>
<dependency>
<groupId>com.h2database</groupId>
<artifactId>h2</artifactId>
<scope>runtime</scope>
</dependency>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-test</artifactId>
<scope>test</scope>
<exclusions>
<exclusion>
<groupId>org.junit.vintage</groupId>
<artifactId>junit-vintage-engine</artifactId>
</exclusion>
</exclusions>
</dependency>
</dependencies>
<build>
<plugins>
<plugin>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-maven-plugin</artifactId>
</plugin>
</plugins>
</build>
</project>
2条答案
按热度按时间5vf7fwbs1#
根据您的错误消息
No property byPassword found for type User!
,它正在存储库中查找名为byPassword
的属性,您需要将其更改为andPassword
,如下所示。Spring将解析by后面提到的每个属性,您只需要在方法的开头使用by
您也可以将名称更改为更易读的名称
xggvc2p62#