Spring Boot,JPA和MySQL返回空结果
问题描述:
好吧我是Java Spring的新手,我只是试图从MYSQL数据库访问一些数据。不幸的是,我似乎无法加载任何数据 - 我敢肯定这是一个愚蠢的错误。 (我已经按照多种来源,包括this one)Spring Boot,JPA和MySQL返回空结果
我使用: java的春天开机,MySQL和JPA,gradle这个
这里是我的build.gradle
:
buildscript {
repositories {
mavenCentral()
}
dependencies {
classpath("org.springframework.boot:spring-boot-gradle-plugin:1.3.2.RELEASE")
}
}
apply plugin: 'java'
apply plugin: 'eclipse'
apply plugin: 'idea'
apply plugin: 'spring-boot'
jar {
baseName = 'gs-serving-web-content'
version = '0.1.0'
}
repositories {
mavenCentral()
}
sourceCompatibility = 1.8
targetCompatibility = 1.8
dependencies {
compile("org.springframework.boot:spring-boot-starter-web")
compile("org.springframework.boot:spring-boot-starter-thymeleaf")
compile("org.springframework.boot:spring-boot-starter-data-jpa")
compile("com.h2database:h2") //I get embedded db error if I take this out
testCompile("junit:junit")
compile('mysql:mysql-connector-java:5.1.35')
}
task wrapper(type: Wrapper) {
gradleVersion = '2.3'
}
控制器:
@Controller
public class PersonController {
@Autowired
private PersonRepository personRepository;
@RequestMapping(value = "/hello", method = RequestMethod.GET)
public String greeting(...){
List<Person> allPeople = personRepository.findAll(); //this comes back with 0 values
...
}
PersonRepository:
public interface PersonRepository extends JpaRepository<Person, Long> {
List<Person> findAll(); //returns empty list
//Person findByAge(int Age); //throws an exception if I leave this on: "...nested exception is java.lang.IllegalArgumentException: Could not create query metamodel for method public abstract hello.model.Person hello.repository.PersonRepository.findByAge(int)!"
}
人:
@Entity
@Table(name="person")
public class Person {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private long PersonID;
private String Firstname;
private String Lastname;
private int Age;
//getters and setters
}
Application.yaml
spring:
datasource:
url: jdbc:mysql://localhost:3306/datatest
username: root
password: ***
driverClassName: com.mysql.jdbc.Driver
我的IntelliJ连接成功MySQL数据库(数据库标签下) ,我可以从标签罚款查询它,但我无法通过弹簧加载数据。考虑到我无法删除嵌入的h2数据库,除非有例外,我的设置中肯定有问题,我也不能添加findByAge
,findByFirstname
等组合(全都抛出异常)。
更新1:我申请HD1的Person类的变化,但我仍然没有得到任何数据传回:
@Entity
@Table(name="person")
public class Person {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private long personID;
private String firstName;
private String lastName;
private int age;
//getters and setters updated
PersonRepository:
List<Person> findAll();
Person findByAge(int age); //no more errors here, but returns null
Person findByLastName(String lastName); //no more errors here, but returns null
更新2:控制器方法:
@RequestMapping(value = "/hello", method = RequestMethod.GET)
public String greeting(@RequestParam(value = "name", required = false, defaultValue = "world") String name, Model model) {
List<Person> allPeople = personRepository.findAll();
...
System.out.print(allPeople.size()); //always 0, but I have many entries in db
//if I try to create and save new entry in the db:
Person person = new Person();
person.setFirstName("New");
person.setLastName("Person");
person.setAge(99);
personRepository.save(person); //does not show up in DB, why?
return "greeting";
}
答
显然,你不跟着你的链接教程,因为属性名称是驼峰那里和这里的首字母大写。 Spring需要camelCase的属性,因为entity scanner uses camelCase for accessor/mutator generation。
[根据注释]
在PersonRepository,尝试取消注释注释行。如果仍有问题,请在下次编辑中发布完整的堆栈跟踪和runlog。
答
你应该使用小写的属性
private long personID;
private String firstname;
private String lastname;
private int age;
我测试过,并得到同样的异常。但有其最高审计机关嵌套异常:
Unable to locate Attribute with the the given name [attributename]
感谢您的回答。你是对的,完全错过了骆驼案件。我已经更新了我的答案和类,它实际上允许我使用'findByAge'和'findByLastName'(而不是像以前那样获取异常),但两者都返回null。 – gudthing
仍然在等待运行日志 – hd1
确定[这里是](https://gist.github.com/gudthing/d29c377bd64cd639461f) – gudthing