我正在尝试为选件质量做一个案例，但是我收到“ ==”的操作员错误，此问题的解决方案是什么？
这是背后的代码
private void myComboBoxThatICreatedInXaml_SelectionChanged(object sender, SelectionChangedEventArgs e)
{
if (myComboBoxThatICreatedInXaml.SelectedValue.ToString == Low)
{
QualityChoices.Add(YouTubeQuality.QualityHigh);
case
(myComboBoxThatICreatedInXaml.SelectedValue.ToString == Medium)
{
QualityChoices.Add(YouTubeQuality.QualityMedium);
}
case
(myComboBoxThatICreatedInXaml.SelectedValue.ToString == High)
{
QualityChoices.Add(YouTubeQuality.QualityHigh);
}
}
这是我的xaml代码。
<ComboBox x:Name="myComboBoxThatICreatedInXaml" SelectionChanged="myComboBoxThatICreatedInXaml_SelectionChanged" >
<ComboBoxItem Tag="LW">Low</ComboBoxItem>
<ComboBoxItem Tag="MD">Medium</ComboBoxItem>
<ComboBoxItem Tag="HG">High</ComboBoxItem>
</ComboBox>
您在这里遇到了一些C＃语法问题。您需要()该ToString方法，并在字符串文字周围加上引号：
if (myComboBoxThatICreatedInXaml.SelectedValue.ToString() == "Low")
如果要使用switch语句，则如下所示：
switch(myComboBoxThatICreatedInXaml.SelectedValue.ToString())
{
case "Low":
QualityChoices.Add(YouTubeQuality.QualityHigh);
break;
case "Medium":
QualityChoices.Add(YouTubeQuality.QualityMedium);
break;
case "High":
QualityChoices.Add(YouTubeQuality.QualityHigh);
break;
default:
break;
}
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