2013-05-21 77 views
0

Screenshot showing error「找不到符號」,通過方法傳遞變量的

爲什麼我引用的是正在通過構造函數傳遞一個變量時,我得到一個錯誤?

下面是完整的Customer.java類:

import java.sql.ResultSet; 
import javax.servlet.http.HttpServletRequest; 

public class Customer { 

    // SQL Variables 
    private final String URL = "jdbc:mysql://localhost/books"; 
    private final String USER = "root"; 
    private final String PASSWORD = "[email protected]"; 

    // Class Variables 
    int CustomerID; 
    String customerEmail; 
    String customerPassword; 
    String customerFirstName; 
    String customerLastName; 
    String customerAddress; 
    String customerCity; 
    String customerProvince; 
    String customerPostalCode; 
    String customerPhoneNumber; 
    String customerCreditCardNumber; 
    String customerExpiryDate; 
    String customerCreditCardType; 

    // Class Constructors 
    public Customer(int customerID) // Constructor for an Existing Customer 
    { 
     ResultSet customerData = MySQLQuery("select * from Customers where CustomerID=" . customerID); 
    } 
    public Customer(HttpServletRequest post) // Constructor for a new customer 
    { 

    } 
    // Constructor Methods 
    public ResultSet MySQLQuery(String query) 
    { 
     return queryResultSet; 
    } 
    // Getters and Setters 
} 

這是在NetBeans 7.3在默認包內Customer.java類創建一個Java Web項目。

+2

Java字符串連接不同於PHP,您必須使用'+'連接字符串,而不是'.' – BackSlash

回答

5

當您的意思是+時,您似乎正在使用.

偶然一個老的PHP程序員? :-)

+0

該死的打我1秒 –

+0

是的,Java不是PHP –