以下是我的Spring启动Rest API应用程序。
Vendor.Java
package hello;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;
@Entity(name="vendor")
public class Vendor {
@Id
@Column(name="vendorId")
private int vendorId;
@Column(name="vendorName")
private String vendorName;
@Column(name="vendorPhone")
private int vendorPhone;
@Column(name="vendorBalance")
private int vendorBalance;
@Column(name="vendorChequeAmount")
private int vendorChequeAmount;
public int getVendorId() {
return vendorId;
}
public void setVendorId(int vendorId) {
this.vendorId = vendorId;
}
public String getVendorName() {
return vendorName;
}
public void setVendorName(String vendorName) {
this.vendorName = vendorName;
}
public int getVendorPhone() {
return vendorPhone;
}
public void setVendorPhone(int vendorPhone) {
this.vendorPhone = vendorPhone;
}
public int getVendorBalance() {
return vendorBalance;
}
public void setVendorBalance(int vendorBalance) {
this.vendorBalance = vendorBalance;
}
public int getVendorChequeAmount() {
return vendorChequeAmount;
}
public void setVendorChequeAmount(int vendorChequeAmount) {
this.vendorChequeAmount = vendorChequeAmount;
}
}
VendorRepository.Java
package hello;
import org.springframework.data.repository.CrudRepository;
import hello.Vendor;
// This will be AUTO IMPLEMENTED by Spring into a Bean called userRepository
// CRUD refers Create, Read, Update, Delete
public interface VendorRepository extends CrudRepository<Vendor, Integer> {
}
MainController.Java
package hello;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.GetMapping;
import org.springframework.web.bind.annotation.RequestParam;
import org.springframework.web.bind.annotation.ResponseBody;
import hello.Vendor;
import hello.VendorRepository;
@Controller // This means that this class is a Controller
@RequestMapping(path="/demo") // This means URL's start with /demo (after Application path)
public class MainController {
@Autowired // This means to get the bean called userRepository
// Which is auto-generated by Spring, we will use it to handle the data
private VendorRepository vendorRepository;
@GetMapping(path="/all")
public @ResponseBody Iterable<Vendor> getAllVendors() {
// This returns a JSON or XML with the users
return vendorRepository.findAll();
}
@GetMapping(path="/msg")
public @ResponseBody String getMsg(){
return "Hi";
}
}
Application.Java
package hello;
import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
@SpringBootApplication
public class Application {
public static void main(String[] args) {
SpringApplication.run(Application.class, args);
}
}
当我尝试访问@ http://localhost:8089/demo/all时,我收到以下错误
Whitelabel Error Page
This application has no explicit mapping for /error, so you are seeing this as a fallback.
Sat Feb 10 14:59:39 GST 2018
There was an unexpected error (type=Internal Server Error, status=500).
could not extract ResultSet; SQL [n/a]; nested exception is org.hibernate.exception.SQLGrammarException: could not extract ResultSet
请查看错误: -
Application.properties
server.port=8089
spring.jpa.hibernate.ddl-auto=none
spring.jpa.properties.hibernate.show_sql=true
spring.jpa.properties.hibernate.use_sql_comments=true
spring.jpa.properties.hibernate.format_sql=true
spring.jpa.properties.hibernate.type=trace
spring.jpa.hibernate.naming.strategy=org.hibernate.cfg.EJB3NamingStrategy
spring.datasource.url=jdbc:mysql://localhost:3306/web_customer_tracker
spring.datasource.username=springstudent
spring.datasource.password=springstudent
仍然,我收到错误:
这是什么错误?如何解决?
答案 0 :(得分:1)
我只是尝试使用小写的列名,但它确实有效!
@Id
@Column(name="vendorid")
private int vendorId;
@Column(name="vendorname")
private String vendorName;
@Column(name="vendorphone")
private int vendorPhone;
@Column(name="vendorbalance")
private int vendorBalance;
@Column(name="vendorchequeamount")
private int vendorChequeAmount;
答案 1 :(得分:0)
从错误消息中,似乎SQL查询中的列名生成错误。我建议在开发时在spring.jpa.show-sql=true
中设置application.properties
以在控制台中查看生成的查询。
错误很奇怪,因为当您将其名称设置为vendor_id
(werid数据库列命名策略,顺便说一句)时,它会抱怨列vendorId
。研究了一下,我遇到this question,根据它的第一个答案,它似乎是列名的某种错误。如上所述,请尝试在application.preferences
中添加此内容:
spring.jpa.hibernate.naming_strategy=org.hibernate.cfg.EJB3NamingStrategy
答案 2 :(得分:0)
如果您使用的是Hibernate v5,请在application.properties
中使用以下行spring.jpa.hibernate.naming.physical-strategy=org.hibernate.boot.model.naming.PhysicalNamingStrategyStandardImpl