我有一个应用程序,它从两个表单(表单2到表单1)发送信息,并且我有一个进入表单2的菜单屏幕,但是当我尝试打开表单2时出现错误(CS7036)。基本上,我会说
“错误CS7036,没有给出与 要求“ difficulty.difficulty(Easy)的形式参数” incoming””
(困难是“形式2”,而容易是“形式1”
从菜单
private void btnStart_Click(object sender, EventArgs e)
{
this.Close();
difficulty diff = new difficulty();
diff.ShowDialog();
}
困难>
public partial class difficulty : Form
{
public difficulty(Easy incoming)
{
InitializeComponent();
}
private void btnEasy_Click(object sender, EventArgs e)
{
this.Close();
Easy easy = new Easy();
easy.ShowDialog();
}
}
我如何使它起作用,这样我才能进入菜单,然后又困难又容易,没有任何错误?任何帮助表示赞赏。
答案 0 :(得分:0)
在实例化类时,您没有传入添加到构造函数中的参数。
这是您代码的固定版本:
private void btnStart_Click(object sender, EventArgs e) {
//use whatever easy you have here
Easy easy = new Easy();
difficulty diff = new difficulty(easy);
diff.ShowDialog();
this.Hide();
}
您需要预先创建一个Easy
类才能传入。如果要在该类中使用该实例,可以执行以下操作:
public partial class difficulty : Form {
private Easy easy;
public difficulty(Easy incoming)
{
easy = incoming;
InitializeComponent();
}
private void btnEasy_Click(object sender, EventArgs e) {
....
}
}
然后在任何类函数中,引用easy
变量以获取数据。