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谁能告诉我为什么准备好的语句返回null？下面是查询数据库的代码：

package beans;
import java.sql.*;
public class ConnectToDB {
private Connection connect;

private final String url = "jdbc:mysql://localhost/";
private final String DBuser = "root";
private final String DBpass = "root";
private final String DBname = "reservation";

private final String Driver = "com.mysql.jdbc.Driver";
public ConnectToDB(){
    try{
        Class.forName(this.Driver);
        this.connect = DriverManager.getConnection(this.url+this.DBname, this.DBuser, this.DBpass);
    }catch(  ClassNotFoundException | SQLException e){ e.printStackTrace(); }
}
public Connection getConnection(){
    return this.connect;
}

private String get_user_info(String username,int index){
    /*
     * user_info[0] = user_id;
     * user_info[1] = username;
     * user_info[2] = password;
     * user_info[3] = firstname;
     * user_info[4] = middle_name;
     * user_info[5] = lastname;
     * user_info[6] = client_rights;
     * 
     */
    String user_info[] = new String[7];
    PreparedStatement pstmt = null;
    ResultSet rset = null;
    String query = null;
    try{
        query = "Select * from user where username = ?";
        pstmt = this.connect.prepareStatement(query);
        pstmt.setString(1, username);
        rset = pstmt.executeQuery();
        while(rset.next()){
            int user_id = rset.getInt("user_id");
            user_info[0] = String.valueOf(user_id);
            user_info[1] = rset.getString("username");
            user_info[2] = rset.getString("password");
            user_info[3] = rset.getString("firstname");
            user_info[4] = rset.getString("middle_name");
            user_info[5] = rset.getString("lastname");
            user_info[6] = rset.getString("user_rights");
        }
    }catch(SQLException e){ e.printStackTrace(); }
    finally{
        try{
            pstmt.close();
            rset.close();
        }catch(SQLException e){ e.printStackTrace(); }
    }
    return user_info[index];
}

public int getUserId(String username){
    String user_id_from_db = get_user_info(username, 0);
    int user_id = Integer.parseInt(user_id_from_db);
    return user_id;
}

public String getUsername(String username){ return get_user_info(username, 1); }
public String getPassword(String username){ return  get_user_info(username, 2); }
public String getFirstname(String username){ return get_user_info(username, 3); }
public String getMiddlename(String username){ return get_user_info(username, 4); }
public String getLastname(String username){ return  get_user_info(username, 5); }
public String getUserRights(String username){ return get_user_info(username, 6); }

public boolean userExists(String username){
    boolean queryStatus = false;
    if(username.equalsIgnoreCase(getUsername(username)))
        queryStatus = true;
    else
        queryStatus = false;
    return queryStatus;
}   
}

然后是调用查询的代码：

<jsp:useBean id="user" class="beans.ConnectToDB" scope="session" />
<jsp:useBean id="aes" class="beans.AES" scope="session" />

String getUsername = request.getParameter("username");
        String getPassword = request.getParameter("password");  

        final String passphrase = "#asdf@1234#";    
        byte[] password_byte = getPassword.getBytes();  
        byte[] passphrase_byte = passphrase.getBytes();
        byte[] encrypt_password = aes.encrypt(password_byte, passphrase_byte);      

        if((getUsername != null && !getUsername.isEmpty()) || (getPassword != null && !getPassword.isEmpty())){
            String username_from_db = user.getUsername(getUsername);
            String password_from_db = user.getPassword(getUsername);

            byte[] pass_db_byte = password_from_db.getBytes();
            byte[] encrypted_pass_db = aes.encrypt(pass_db_byte, passphrase_byte);

            if(getUsername.equalsIgnoreCase(username_from_db) && encrypt_password.equals(encrypted_pass_db)){
                response.sendRedirect("home_page.jsp");
            }
        }
        else{ response.sendRedirect("index.jsp"); }

当我调用getUsername(String username)方法时，它返回 null 这里是抛出的异常：

org.apache.jasper.JasperException: An exception occurred processing JSP page /authenticate_user.jsp at line 29
26:             byte[] encrypt_password = aes.encrypt(password_byte, passphrase_byte);      
27:             
28:             if((getUsername != null && !getUsername.isEmpty()) || (getPassword != null && !getPassword.isEmpty())){
29:                 String username_from_db = user.getUsername(getUsername);
30:                 String password_from_db = user.getPassword(getUsername);
31: 
32:                 byte[] pass_db_byte = password_from_db.getBytes();


Stacktrace:
org.apache.jasper.servlet.JspServletWrapper.handleJspException(JspServletWrapper.java:568)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:470)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:390)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:334)
javax.servlet.http.HttpServlet.service(HttpServlet.java:728)

root cause

java.lang.NullPointerException
beans.ConnectToDB.get_user_info(ConnectToDB.java:61)
beans.ConnectToDB.getUsername(ConnectToDB.java:72)
org.apache.jsp.authenticate_005fuser_jsp._jspService(authenticate_005fuser_jsp.java:110)
org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:70)
javax.servlet.http.HttpServlet.service(HttpServlet.java:728)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:432)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:390)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:334)
javax.servlet.http.HttpServlet.service(HttpServlet.java:728)
4

2 回答 2

1

你发的东西太多了。

你只需要这样：

beans.ConnectToDB.get_user_info(ConnectToDB.java:61)

在文本编辑器中打开 ConnectToDB.java，到第 61 行，并检查该行上的所有对象引用。其中之一是 null ，因为您没有正确初始化它，或者假设在获得它时它总是非 null 。找出哪个并正确初始化它。问题解决了。

我不认为这是一个好的解决方案。如果您有 JSP，那么您就有一个 servlet/JSP 引擎，它应该有一个 JNDI 数据库连接池和命名服务。您应该设置它并外部化您的数据库连接参数。它们不属于您的代码。池将比您更好地管理连接。

您也没有正确关闭资源。它们应该在 finally 块中关闭，以与创建相反的顺序，包装在单独的 try/catch 块中。我会编写一个可以调用的静态实用程序方法。

package persistence;

public class DatabaseUtils {
    private DatabaseUtils() {}

    // Similar for ResultSet and Connection
    public static void close(Statement st) {
        try {
            if (st != null) {
                st.close();
            }
        } catch (Exception e) {
            // Log the exception
        }
    }
}
于 2013-05-31T12:17:27.200 回答
0

您捕获的异常是由以前的异常引起的。块的第一行抛出异常try/catch：

    pstmt = this.connect.prepareStatement(query);
    pstmt.setString(1, username);
    rset = pstmt.executeQuery();

rset因此，在块开始执行之前没有设置任何值finally，并且当您尝试使用空值时会引发新的异常。更改您catch以捕获所有异常，您将找到根本原因。

当然，在使用finallyorcatch块时，请注意，如果抛出异常，则并非try块中的所有代码都已执行，因此可能未定义某些变量。

请注意，如果您使用的是 Java 7，则可以使用try with resources： http: //docs.oracle.com/javase/tutorial/essential/exceptions/tryResourceClose.html

此外，在 catch 和 finally 子句中抛出异常

于 2013-05-31T13:39:47.643 回答