有一个空白的文本框崩溃的程序,当我按下登录按钮
本文关键字:登录 按钮 程序 文本 空白 崩溃 有一个 | 更新日期: 2023-09-27 18:01:38
初级程序员。我有一个登录按钮,当我输入正确的凭据时,它完全正常工作,但是当我点击登录按钮时,有一个空的文本框,程序崩溃了,给了我"一个未处理的异常类型"系统。格式异常"发生在mscorlib.dll"。我尝试在else if中使用!= null,但这也不起作用。所以我的问题是我如何能够得到一个空的文本框显示"请输入一个有效的用户名和/或密码",而不是崩溃的程序?谢谢!
附加信息:输入字符串格式不正确。
private void btnLogin_Click(object sender, EventArgs e)
{
Entities2 db = new Entities2();
foreach (var usert in db.Teachers)
{
if (usert.TID == Convert.ToInt32(txtLogin.Text) && usert.Password == txtPassword.Text)
{
Teach teacher = new Teach();
teacher.ShowDialog();
}
else if (usert.TID != Convert.ToInt32(txtLogin.Text) && usert.Password != txtPassword.Text)
{
MessageBox.Show("Please Enter a Valid Username and/or Password");
}
}
}
private void btnLogin_Click(object sender, EventArgs e)
{
try
{
if(txtLogin.Text!="" && txtPassword.Text!="")
{
Entities2 db = new Entities2();
foreach (var usert in db.Teachers)
{
if (usert.TID == Convert.ToInt32(txtLogin.Text) && usert.Password == txtPassword.Text)
{
Teach teacher = new Teach();
teacher.ShowDialog();
}
else if (usert.TID != Convert.ToInt32(txtLogin.Text) && usert.Password != txtPassword.Text)
{
MessageBox.Show("Please Enter a Valid Username and/or Password");
}
}
else
{
if(txtLogin.Text=="")
{
MessageBox.Show("Please Enter a Username");
}
else if(txtPassword.Text=="")
{
MessageBox.Show("Please Enter a Password");
}
}
}
Catch(Exception ex)
{
MessageBox.Show("Please Enter a Valid Username and/or Password");
}
}
此外,这不是正确的方式来检查和匹配UserName密码。您可以使用以下方法
if(txtLogin.Text!="")
{
Entities2 db = new Entities2();
Teacher Tobj=db.Teachers.where(x=>x.TID==Convert.ToInt32(txtLogin.Text) && x.Password==txtPassword.Text).SingleOrDefault();
if (Tobj!=null)
{
Teach teacher = new Teach();
teacher.ShowDialog();
}
else
{
MessageBox.Show("Please Enter a Valid Username and/or Password");
}
}
检查文本框是否为"见下文:
private void btnLogin_Click(object sender, EventArgs e)
{
if (txtLogin.Text != "")
{
Entities2 db = new Entities2();
foreach (var usert in db.Teachers)
{
if (usert.TID == Convert.ToInt32(txtLogin.Text) && usert.Password == txtPassword.Text)
{
Teach teacher = new Teach();
teacher.ShowDialog();
}
else if (usert.TID != Convert.ToInt32(txtLogin.Text) && usert.Password != txtPassword.Text)
{
MessageBox.Show("Please Enter a Valid Username and/or Password");
}
}
}
}
private void btnLogin_Click(object sender, EventArgs e)
{
try
{
<body of your btnLogin_Click()>
}
catch(FormatException ex)
{
MessageBox.Show("Please Enter a Valid Username and/or Password");
}
catch(Exception ex2)
{
MessageBox.Show("Error: " + ex2.Message);
}
}
可以进行try解析。如果解析失败,它将返回false,你可以做一些错误处理(谢谢Phil)。
int userId = 0;
if(int.TryParse(txtLogin.Text, out userId)){
// err handling
}
尝试先解析输入的用户ID,如下所示:
private void btnLogin_Click(object sender, EventArgs e)
{
int userId;
if (int.TryParse(txtLogin.Text, out userId))
{
Entities2 db = new Entities2();
foreach (var usert in db.Teachers)
{
if (usert.TID == userId && usert.Password == txtPassword.Text)
{
Teach teacher = new Teach();
teacher.ShowDialog();
}
else
{
MessageBox.Show("Please Enter a Valid Username and/or Password");
}
}
}
else
{
MessageBox.Show("Please Enter a Valid User ID");
}
}