从TableLayoutPanel中访问组合框的SelectedIndex

时间:2012-09-04 08:59:36

标签: c# .net unit-testing moq tablelayoutpanel

我有以下设置:

ComboBox cb1 = new ComboBox();
ComboBox cb2 = new ComboBox();
ComboBox cb3 = new ComboBox();
ComboBox cb4 = new ComboBox();
ComboBox cb5 = new ComboBox();

TableLayoutPanel tlpComboBox = new TableLayoutPanel();
tlpComboBox.ColumnCount = 5;
tlpComboBox.ColumnStyles.Add(new System.Windows.Forms.ColumnStyle(System.Windows.Forms.SizeType.Percent, 20F));
tlpComboBox.ColumnStyles.Add(new System.Windows.Forms.ColumnStyle(System.Windows.Forms.SizeType.Percent, 20F));
tlpComboBox.ColumnStyles.Add(new System.Windows.Forms.ColumnStyle(System.Windows.Forms.SizeType.Percent, 20F));
tlpComboBox.ColumnStyles.Add(new System.Windows.Forms.ColumnStyle(System.Windows.Forms.SizeType.Percent, 20F));
tlpComboBox.ColumnStyles.Add(new System.Windows.Forms.ColumnStyle(System.Windows.Forms.SizeType.Percent, 20F));

tlpComboBox.Controls.Add(cb1, 4, 0);
tlpComboBox.Controls.Add(cb2, 3, 0);
tlpComboBox.Controls.Add(cb3, 2, 0);
tlpComboBox.Controls.Add(cb4, 1, 0);
tlpComboBox.Controls.Add(cb5, 0, 0);

然后我尝试检查tlpComboBox中的组合框的顺序。要使用文本框执行此操作,我执行了以下操作:

for (i = 0; i < 5; i++)
{
    Assert.AreNotEqual(tlpText.Controls[i].Text.ToString(), i.ToString());
}

不幸的是,在Controls [i]之后无法访问SelectedIndex。在使用组合框时,我有点困惑为什么。

for (i = 0; i < 5; i++)
{
    Assert.AreNotEqual(tlpComboBox.Controls[i]. "SelectedIndex isnt accessible", i.ToString());
}

有什么想法吗?

编辑:

http://i.imgur.com/7lhzE.jpg

1 个答案:

答案 0 :(得分:4)

您需要将Controls[i]投射到ComboBox

//sample meta code written in notepad
for (i = 0; i < 5; i++)
{
    Assert.AreNotEqual(
        ((ComboBox )(tlpComboBox.Controls[i])).SelectedIndex , 
        i.ToString());
}

这是因为Controls[i]类型(我猜)Control没有SelectedIndex属性。