标题:请问我这个代码为什么会出现这样的提示:“检测道无法访问的代码”。。秋高 ...
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zhongyu
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请问我这个代码为什么会出现这样的提示:“检测道无法访问的代码”。。秋高手指点
using System;
using System.Collections.Generic;
using
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;

namespace WindowsFormsApplication4
{
    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
        }
       // public static string Usrname = "钟宇";
        public static string   Password = "asdf";
        private void textBox1_TextChanged(object sender, EventArgs e)
        {
            
        }

        private void textBox2_TextChanged(object sender, EventArgs e)
        {

        }

        private void button1_Click(object sender, EventArgs e)
        {

            if ((textBox1.Text.Trim().Length == 9))
            {
                textBox2.Focus();
            }
        }
      
        public bool checkinput()
        {
             if (textBox1.Text.Trim().Length > 9 && textBox1.Text.Trim ().Length  <= 6 )
            {
                MessageBox.Show("用户名不满足要求","提示");
                return false;               
            }
            
                 if (textBox1.Text.Trim().Length < '0' || textBox1.Text.Trim().Length > '9')
                 {
                     MessageBox.Show("用户名格式错误。", "提示");
                     return false;     
                 }
            
             if (textBox1.Text == "")
             {
                 MessageBox.Show("用户名格式错误。", "提示");
                 return false;
             }
             if (textBox2.Text == "")
             {
                 MessageBox.Show("密码格式错误。", "提示");
                 return false;
             }
             //if ((textBox2.Text != asdf ))
             //{
             //    MessageBox.Show("密码错误。", "提示");
             //    return false;
             //}

            if((textBox1 .Text.Trim ().Length  == 9))
            {
              textBox2.Focus();
            }

            return true ;
            Form2 an = new Form2();
            an.ShowDialog();
         }
    //    if (! checkinput(int))
    //    {
    //        return ;
    //    }
    //    else
    //{
        
         

    }
}
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2013-04-22 15:17



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