有时遇到一种情况,.ShowDialog()不显示,也不报错;如下:
<span style="font-size:14px;"> private void button1_Click(object sender, EventArgs e)
{
Thread thread = new Thread(show);
thread.Start();
}
void show()
{
Control.CheckForIllegalCrossThreadCalls = false;
//this.Invoke(new Action(() =>
//{
if (saveFileDialog1.ShowDialog() == DialogResult.OK)
{ }
//}));
}</span>原因分析:这属于线程间操作的一种异常。界面呈现和新创建的thread分开在两个线程中。在thread线程中
不能够进行界面呈现,即显示.ShowDialog();
解决方法:1:添加参数this。
.ShowDialog(IWin32Window owner); //owner:表示将拥有模式对话框的顶级窗口
<span style="font-size:14px;"> private void button1_Click(object sender, EventArgs e)
{
Thread thread = new Thread(show);
thread.Start();
}
void show()
{
Control.CheckForIllegalCrossThreadCalls = false;
//this.Invoke(new Action(() =>
//{
if (saveFileDialog1.ShowDialog(this) == DialogResult.OK)
{ }
//}));
}</span>2:使用Invoke
private void button1_Click(object sender, EventArgs e)
{
Thread thread = new Thread(show);
thread.Start();
}
void show()
{
// Control.CheckForIllegalCrossThreadCalls = false;
this.Invoke(new Action(() =>
{
if (saveFileDialog1.ShowDialog() == DialogResult.OK)
{ }
}));
}原文:http://blog.csdn.net/ilipan/article/details/46364863