C# Listview selectionchanged激发两次,如何控制?
在显示表单后,此加载开始。我知道这是故意的。但我不喜欢再表演了C# Listview selectionchanged激发两次,如何控制?,c#,listview,selectedindexchanged,C#,Listview,Selectedindexchanged,在显示表单后,此加载开始。我知道这是故意的。但我不喜欢再表演了 private void HandleLSVVacunacion_SelectedIndexChanged(object sender, EventArgs e) { ListView SourceControl = (ListView)sender; if (SourceControl.SelectedItems.Count > 0) {
private void HandleLSVVacunacion_SelectedIndexChanged(object sender, EventArgs e)
{
ListView SourceControl = (ListView)sender;
if (SourceControl.SelectedItems.Count > 0)
{
int IdVacunacion = Convert.ToInt32(LSVVacunacion.SelectedItems[0].SubItems[0].Text.Trim());
Formularios.FrmVacunacion f = new Formularios.FrmVacunacion();
f.ID = IdVacunacion;
f.HC = hc;
f.ShowDialog();
CargarEsquemaVacunacion(); // reload The ListView with changes; and show again the form!
}
}
如果
CargarEsquemaVacunacion
中的代码导致事件处理程序重新输入,则可以删除事件处理程序,重新加载列表并重新应用事件处理程序
if (SourceControl.SelectedItems.Count > 0)
{
.....
f.ShowDialog();
try
{
LSVVacunacion.SelectedIndexChanged -= HandleLSVVacunacion_SelectedIndexChanged;
CargarEsquemaVacunacion();
}
finally
{
LSVVacunacion.SelectedIndexChanged += HandleLSVVacunacion_SelectedIndexChanged;
}
}
请注意,此代码位于try/finally块中,以确保在异常情况下,事件处理程序也会重新应用于事件SelectedIndexChanged调用HandlelsVacunacion_SelectedIndexChanged(发送方,e);