java 提交以创建方法public abstract的查询

gt0wga4j  于 2022-12-02  发布在  Java
关注(0)|答案(2)|浏览(168)

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>
5vf7fwbs

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。

public interface RegistrationRepository extends JpaRepository<User, Integer> {

    public User findByEmailIdAndPassword(String emailId, String password);
}

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

public interface RegistrationRepository extends JpaRepository<User, Integer> {

    public User findUserByEmailIdAndPassword(String emailId, String password);
}
xggvc2p6

xggvc2p62#

Optional<Product> findByCodeAndName(String pro_code,String pro_name);

@Column(name = "pro_code",unique = true)
     @JsonProperty(value = "pro_code")
    private String pro_code;
     
     @Column(name = "pro_name",unique = true)
     @JsonProperty(value = "pro_name")
    private String pro_name;

Help me all brother i can't create this query

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