从代码创建JAVA数据库。陈述不会执行并创建表？

Question

我不知道为什么我的程序不是在创建表格，但是我也需要关于如何在创建代码时使用代码填充表格的一些想法？我还需要再向这个数据库添加两个表。

这是我得到的错误：

java.sql.SQLSyntaxErrorException: Table/View 'PIZZASIZE' does not exist. 
Caused by: ERROR 42X05: Table/View 'PIZZASIZE' does not exist. 
Caused by: java.lang.RuntimeException: Exception in Application start method 
Caused by: javafx.fxml.LoadException: file:/C:/Users/Allie/Documents/NetBeansProjects/Pizzeria_AllieBeckman/dist/run1674141987/Pizzeria_AllieBeckman.jar!/pizzeria_alliebeckman/FXMLDocument.fxml 

这是一个本应创建表的代码：

 // connect to the derby URL using the given username and password 
     connect = DriverManager.getConnection("jdbc:derby://localhost:1527/pizzeria;create=true", connectProps); 
     // current url for pre created database "jdbc:derby://localhost:1527/pizza" 
     // if connection is successful print that it succeeded. 
     System.out.println("database created"); 

     stmt = connect.createStatement(); 

     String sqlCreate = "CREATE TABLE PIZZASIZE " 
      + "(id int NOT NULL, " 
      + "size char(20) NOT NULL, " 
      + "PRIMARY KEY (id))"; 

     stmt.execute(sqlCreate); 

Answer 1

根据IDE你使用，你可以手动创建在控制台中的表格，而不需要经过编写代码的麻烦。以下是一些如何从表格中获取信息的示例。

Connection conn = CreatingDerbyDJB.dbConnection(); 

      try{ 
       String query = "INSERT INTO Items (Name,Color,ItemName,SchoolName, Description) VALUES(?,?,?,?, ?)"; 
       PreparedStatement pstmt = conn.prepareStatement(query); 


       pstmt.execute(); 
       conn.close(); 
       }catch(Exception e) 
      { 
        e.printStackTrace(); 
      }  } 

以下是Connection类的外观： package main;

import java.sql.Connection; 
import java.sql.DriverManager; 

import javax.swing.JOptionPane; 

public class CreatingDerbyDJB 
{ 
    public static final String DRIVER = "org.apache.derby.jdbc.EmbeddedDriver"; 
    public static final String JDBC_URL = "jdbc:derby:LostAndFoundDB"; 
    public static Connection dbConnection() 
    { 
     try 
     { 
      Class.forName(DRIVER).newInstance(); 
      Connection c = DriverManager.getConnection(JDBC_URL); 
      return c; 
     }catch(Exception e) 
     { 
      JOptionPane.showMessageDialog(null, e); 
      return null; 
     } 
    } 
} 

赞同这个答案如果这对你有帮助，我会很高兴解释的东西，如果它没有意义。 :)