org.postgresql.util.PSQLException：错误：关系“员工”不存在位置：13答案

【问题标题】：org.postgresql.util.PSQLException: ERROR: relation "employee" does not exist Position: 13org.postgresql.util.PSQLException：错误：关系“员工”不存在位置：13
【发布时间】：2021-03-02 22:36:57
【问题描述】：

我正在尝试根据https://javainspires.blogspot.com/2020/07/spring-boot-2-spring-data-jdbc-jdbc.html 上的教程将数据库从 MySQL 更改为 PostgreSQL。我经常遇到标题中显示的错误。

Employee.java

package com. example.demo;

import javax.persistence.Entity;
import javax.persistence.Table;

public class Employee {

    private String username;
    private String email;
    private String password;

    public Employee() {
        super();
        // TODO Auto-generated constructor stub
    }

    public Employee(String username, String email, String password) {
        super();
        this.username = username;
        this.email = email;
        this.password = password;
    }

    public String getUsername() {
        return username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

}

员工控制器

package com. example.demo;

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.jdbc.core.JdbcTemplate;
import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.PostMapping;

@Controller
public class EmployeeController {

    @Autowired
    JdbcTemplate jdbcTemplate;
    
    @PostMapping(path = "addUser")
    public String addUser(Employee user) {
        
        String insert_query = "INSERT INTO employee (username,email,password) VALUES(?,?,?)"; 
        
        int rows = jdbcTemplate.update(insert_query,
                user.getUsername(), 
                user.getEmail(),
                user.getPassword());
        
        if(rows == 1) {
            return "success"; 
        } else {
            return "error"; 
        }
                
    
    }
    
}

HTML 文件： index.html

<!DOCTYPE html>
<html xmlns:th="http://www.thymeleaf.org">
<head>
<meta charset="ISO-8859-1">
<title>Java Inspires</title>
</head>
<body>

 <form th:action="@{/addUser}" th:object="${user}" method="post" >
 
  <p>User Name :<input type="text" name="username" /></p>
  <p>Email :<input type="email" name="email" /></p>
  <p>Password :<input type="password" name="password" /></p>
  <p><input type="submit" value="Sign Up" /></p>
 </form>
</body>
</html>

Application.properties

spring.datasource.url=jdbc:postgresql://localhost:5432/database
spring.datasource.username=postgres
spring.datasource.password=admin
spring.datasource.driver-class-name=org.postgresql.Driver

schema.sql

CREATE TABLE employee
(
    username varchar(100) NOT NULL, 
    email varchar(100) NOT NULL,
    password varchar(100) NOT NULL
);

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>2.4.3</version>
        <relativePath/> <!-- lookup parent from repository -->
    </parent>
    <groupId>com.example</groupId>
    <artifactId>JDBC</artifactId>
    <version>0.0.1-SNAPSHOT</version>
    <name>JDBC</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-jdbc</artifactId>
        </dependency>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-thymeleaf</artifactId>
        </dependency>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-web</artifactId>
        </dependency>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-tomcat</artifactId>
            <scope>provided</scope>
        </dependency>
        <dependency>
            <groupId>org.postgresql</groupId>
            <artifactId>postgresql</artifactId>
            <scope>runtime</scope>
        </dependency>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-test</artifactId>
            <scope>test</scope>
        </dependency>
    </dependencies>

    <build>
        <plugins>
            <plugin>
                <groupId>org.springframework.boot</groupId>
                <artifactId>spring-boot-maven-plugin</artifactId>
            </plugin>
        </plugins>
    </build>

</project>

【问题讨论】：

标签： java spring postgresql spring-boot

【解决方案1】：

我删除了 schema.sql 文件并在 PgAdmin4 中手动创建了它。上面的代码在那之后工作了。

【讨论】：

【解决方案2】：

您可以尝试将架构名称添加到您的查询中。

INSERT INTO employee (username,email,password) VALUES(?,?,?)

从这里到

INSERT INTO SCHEMA_NAME.employee (username,email,password) VALUES(?,?,?)

如果这不起作用，您可以用@Entity and @Table(name = "TABLE_NAME" 注释员工类吗？在您的情况下，表名可能是用户。

并在方法参数Employeepublic String addUser(@RequestBody Employee user)添加@RequestBody注解

【讨论】：

尝试将架构名称更改为我的查询：
之后发生了什么？
尝试将架构名称更改为我的查询：不起作用使用实体、表和 ID 和列注释用户类也不起作用将用户更改为员工以避免任何可能的事故--没有工作@raj240
您能否尝试将 currentSchema=SCHEMA_NAME 添加到您的 spring.datasource.url 中，例如 spring.datasource.url=jdbc:postgresql://localhost:5432/database?currentSchema=SCHEMA_NAME，同时将 Employee 类注释为实体，同时指定表名。
Employee 类被注释为：@Entity @Table(name = "employee") public class Employee { @Id @Column(name="username") private String username; @Column(name ="email") private String email; @Column(name = "password") private String password;