i have switch problem in c3#

Question

1.00/5 (2 votes)

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in form1 it have comboBox1 and label1
in combo there items 1 , 2 ,3 , , etc when u select" 1 "then label1.text = "1" and so on

but this code not work

C#

private void comboBox1_SelectedIndexChanged(object sender, EventArgs e)
{

  switch (comboBox1.SelectedItem)
    {
      case 1:
            l1.Text = "11111111111";
          break ;
      case 2:
        l1.Text = "22222222222";
          break ;
      case 3:
          l1.Text = "33333333333";
          break ;


    }
}

Posted 15-Feb-12 1:02am

majid5487

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BobJanova · Accepted Answer · 2012-02-15T01:15:00

void comboBox1_SelectedIndexChanged(object sender, EventArgs e){
 l1.Text = comboBox1.SelectedItem.ToString();
}

Or you can use SelectedIndex, which will be a zero based index.

Is this a homework question? If so, you would be much better off actually learning the material instead of posting it here. You will find your course/job/employment prospects very hard if you are stuck at this basic level and don't have the initiative to learn it yourself.

Ibrahim Yusuf · Accepted Answer · 2012-02-15T01:06:00

Solution 1

You can't use switch keyword because ComboBox items is type object.

You can try

C#

private void comboBox1_SelectedIndexChanged(object sender, EventArgs e)
{

  switch (int.Parse(comboBox1.SelectedItem))
    {
      case 1:
            l1.Text = "11111111111";
          break ;
      case 2:
        l1.Text = "22222222222";
          break ;
      case 3:
          l1.Text = "33333333333";
          break ;


    }
}

Posted 15-Feb-12 1:06am

Ibrahim Yusuf

Comments

majid5487 15-Feb-12 7:31am

i changed to comboBox1.Text and its worked but why .SelectedItem will not compile :o
here it fixed code thanks you :D
switch (int.Parse(comboBox1.Text ))
{
case 1:
l1.Text = "11111111111";
break;
case 2:
l1.Text = "22222222222";
break;
case 3:
l1.Text = "33333333333";
break;

}

krumia 15-Feb-12 10:19am

Because, as Ibrahim has said, SelectedItem property returns something of object type, where string (note the lowercase 's') is treated as primitive type.