Servlet 代码中的结果集处理错误

发布于 2024-12-25 03:26:30 字数 1918 浏览 0 评论 0原文

我在 servlet 中编写了一段代码用于登录检查，如果我没有注释掉 String s4 =，我不知道为什么会收到类似 java.sql.SQLException: No data found 的错误rs.getString(1) 和 out.println(s4) 行如果我注释掉这行，我没有收到任何错误。

为什么我会收到这样的错误？我找不到答案。

import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.*;
import javax.servlet.http.*;
import java.sql.*;

public class login extends HttpServlet {

    Connection conn;
    Statement stmt;
    ResultSet rs;
    String s = "";

    public void init() {
        try {
            Class.forName("sun.jdbc.odbc.JdbcOdbcDriver");
            conn = DriverManager.getConnection("Jdbc:Odbc:edsn");
            s = "Your information is connected ......";
        } catch (Exception e) {
            s = "Exception 1....." + e.getMessage();
        }
    }

    public void doPost(HttpServletRequest req, HttpServletResponse res)
            throws IOException, ServletException {
        res.setContentType("text/html;charset=UTF-8");
        PrintWriter out = res.getWriter();
        out.println(s);
        try {

            String ID = req.getParameter("T1");
            String query = "select * from user_db ";
            stmt = conn.createStatement();
            rs = stmt.executeQuery(query);
            out.println("user" + " " + "pass");
            while (rs.next()) {

                try {
                    if ((rs.getString(1)).equals(ID)) {
                        String s4 = rs.getString(1);
                        out.println(s4);

                        out.println("<html><body><h> login Pass.....:(</h></body></html>");

                    }

                } catch (Exception e) {
                    out.println(e);
                }
            }

        } catch (Exception e) {
            out.println("Unable To Show the info... . . ." + e.getMessage());
        }

    }
}

原文

I write a code in servlet for login checking I don't know why I get an error like java.sql.SQLException: No data found, if I had not commented out the String s4 = rs.getString(1) and out.println(s4) line if I commented out this lines I did not get any error.

Why do I get an error like this? I cannot find out the answer.

import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.*;
import javax.servlet.http.*;
import java.sql.*;

public class login extends HttpServlet {

    Connection conn;
    Statement stmt;
    ResultSet rs;
    String s = "";

    public void init() {
        try {
            Class.forName("sun.jdbc.odbc.JdbcOdbcDriver");
            conn = DriverManager.getConnection("Jdbc:Odbc:edsn");
            s = "Your information is connected ......";
        } catch (Exception e) {
            s = "Exception 1....." + e.getMessage();
        }
    }

    public void doPost(HttpServletRequest req, HttpServletResponse res)
            throws IOException, ServletException {
        res.setContentType("text/html;charset=UTF-8");
        PrintWriter out = res.getWriter();
        out.println(s);
        try {

            String ID = req.getParameter("T1");
            String query = "select * from user_db ";
            stmt = conn.createStatement();
            rs = stmt.executeQuery(query);
            out.println("user" + " " + "pass");
            while (rs.next()) {

                try {
                    if ((rs.getString(1)).equals(ID)) {
                        String s4 = rs.getString(1);
                        out.println(s4);

                        out.println("<html><body><h> login Pass.....:(</h></body></html>");

                    }

                } catch (Exception e) {
                    out.println(e);
                }
            }

        } catch (Exception e) {
            out.println("Unable To Show the info... . . ." + e.getMessage());
        }

    }
}

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瀟灑尐姊 2025-01-01 03:26:30

为什么要写这样的代码？翻遍整张桌子真是太浪费了……
仅IO...
为什么不改变这一点：

ResultSet rs = null;
PreparedStatement st = null;
try {...
String ID = req.getParameter("T1"); 
            String query = "select 1 from user_db where col_name = ?"; 
            st = conn.prepareStatement(query);
            st.setString(1, ID);  
            rs = st.executeQuery();
if (rs.next()) {
  out.println(ID); 
  out.println("<html><body><h> login Pass.....:(</h></body></html>"); 
}
..

} finally {
if (rs != null) try { rs.close();}catch (Exception e) {}
if (st != null) try { st.close();}catch (Exception e) {}
}

注意准备好的语句被缓存并且更适合频繁使用

你让数据库做它擅长的事情 - 搜索数据

select 1 而不是 select * 不会带回你并不真正需要的数据

jdbc 工作越多列越多以及您返回的一般数据，因此仅获取您所返回的数据
需要

并添加一个finally块以始终正确关闭数据库连接

Why write the code like this ? It's very wasteful going over the whole table...
The IO alone...
Why not change to this:

ResultSet rs = null;
PreparedStatement st = null;
try {...
String ID = req.getParameter("T1"); 
            String query = "select 1 from user_db where col_name = ?"; 
            st = conn.prepareStatement(query);
            st.setString(1, ID);  
            rs = st.executeQuery();
if (rs.next()) {
  out.println(ID); 
  out.println("<html><body><h> login Pass.....:(</h></body></html>"); 
}
..

} finally {
if (rs != null) try { rs.close();}catch (Exception e) {}
if (st != null) try { st.close();}catch (Exception e) {}
}

notice prepared statements are cached and better for frequent use

you let the db do what its good at - search the data

select 1 instead of select * does not bring back data you dont really need

jdbc works harder the more columns and data in general you return, so only get what you
need

and add a finally block to always close your db connections properly

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