我有一个Windows Forms应用程序,用于测试排序算法,它由一个表单组成。测试完成后,该表单会显示一个消息框,说明测试已完成,但是第一次测试后它会显示一次,第二次之后它会显示两次,第3次之后的3次,依此类推。 后台工作人员从“ RunWorkerCompleted”方法调用显示MessageBox的方法。
private void RunAlgorithmTests()
//this is called from an onclick method
{
backgroundWorker.DoWork += (s, e) =>
{
test.RunTests();
};
backgroundWorker.RunWorkerCompleted += (s, e) =>
{
ShowCompleteMessage();
};
backgroundWorker.RunWorkerAsync();
}
private void ShowCompleteMessage()
//show this message on successful test complete
{
string title = "Test završen!";
string message = "Test završen nakon" + swatch.Elapsed.Hours.ToString() + ":" + swatch.Elapsed.Minutes.ToString() + ":" + swatch.Elapsed.Seconds.ToString();
this.Text = "Aplikacija za testiranje algoritama sortiranja";
if (test.complete)
{
MessageBox.Show(message, title);
}
}
MessageBox只能在每次测试结束时出现一次,但每次测试运行都会增加它出现的次数。
答案 0 :(得分:0)
根据建议,仅一次收录这些事件。这是通过Form的Load()事件完成的:
private void Form1_Load(object sender, EventArgs e)
{
backgroundWorker.DoWork += (s, e) =>
{
test.RunTests();
};
backgroundWorker.RunWorkerCompleted += (s, e) =>
{
ShowCompleteMessage();
};
}
private void button1_Click(object sender, EventArgs e)
{
RunAlgorithmTests();
}
//this is called from an onclick method
private void RunAlgorithmTests()
{
backgroundWorker.RunWorkerAsync();
}
答案 1 :(得分:0)
问题在于,每次运行后台工作程序时,都会为这两个事件添加其他新的处理程序。第一次,您每个人都有一个处理程序。下次,您再添加一个,所以您有两个。两者都被称为。下次,您添加另一个,这三个都将被调用。
您要做的是确保仅发生一次。
我喜欢您将事件处理程序初始化放入其自己的方法中的想法,因此我们会坚持下去。
// Your Form class is probably called something else.
public Form1()
{
InitializeComponent();
// Since backgroundWorker was created in the form designer, it will have been
// initialized in InitializeComponent(). Therefore, this has to happen after
// InitializeComponent() is called.
InitializeBackgroundWorkerHandlers();
}
private void InitializeBackgroundWorkerHandlers()
{
backgroundWorker.DoWork += (s, e) =>
{
test.RunTests();
};
backgroundWorker.RunWorkerCompleted += (s, e) =>
{
ShowCompleteMessage();
};
}
// this is called from an onclick method
private void RunAlgorithmTests()
{
backgroundWorker.RunWorkerAsync();
}