MySQL的例外 - com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException

Question

我得到的錯誤：

'Unknown column 'customerno' in 'field list' '.

但是，在我的客戶表中存在的列。那爲什麼我會得到這個異常？

代碼：

import java.sql.*; 

public class Classy { 

    static String myQuery = 
      "SELECT customerno, name" + 
      "FROM customers;"; 

    public static void main(String[]args) 
    { 
     String username = "cowboy"; 
     String password = "1234567"; 
     try 
     { 
      Connection con = DriverManager.getConnection("jdbc:mysql://localhost:3306/Business", username, password); 
      Statement stmt = con.createStatement(); 
      ResultSet rs = stmt.executeQuery(myQuery); 

      while(rs.next()) 
      { 
       System.out.print(rs.getString("customerno")); 
      } 

     } catch(SQLException ex){System.out.println(ex);} 

    } 

} 

Answer 1

與你的語法問題是你有name和之間沒有空格FROM

String myQuery = 
     "SELECT customerno, name" + // problem is here 
     "FROM customers;"; 

後name

String myQuery = 
     "SELECT customerno, name " + // add space here 
     "FROM customers"; 

Answer 2

import java.sql.*; 

public class Classy { 

static String myQuery = 
     "SELECT customerno, name" + 
     "FROM customers;"; 

public static void main(String[]args) 
{ 
    String username = "cowboy"; 
    String password = "1234567"; 
    try 
    { 
     Connection con = DriverManager.getConnection("jdbc:mysql://localhost:3306/Business", username, password); 
     Statement stmt = con.createStatement(); 
     ResultSet rs = stmt.executeQuery(myQuery); 

     while(rs.next()) 
     { 

//嘗試與數值存在於數據庫中更改此， e.g如果它是第2列做類似 rs.getString（1）;

  System.out.print(rs.getString("customerno")); 
     } 


    }catch(SQLException ex){System.out.println(ex);} 

} 

} 

Answer 3

看看你的查詢真的是什麼。這：

static String myQuery = 
     "SELECT customerno, name" + 
     "FROM customers;"; 

等同於：

static String myQuery = "SELECT customerno, nameFROM customers;"; 

現在可以看到有什麼不對？我很驚訝它抱怨customerno，而不是缺少FROM部分...

請注意，我懷疑你不想要;要麼。我會寫這一切一一線只是爲了可讀性，當你可以，也限制了可訪問性，並使其最終：

private static final String QUERY = "SELECT customerno, name FROM customers";