Defining' decimal rounded to two digits after dot' typing to texBox problem

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I will start with the code

C#

private void textBox2_KeyPress(object sender, KeyPressEventArgs e)
{
    if (e.KeyChar >= '0' && e.KeyChar <= '9' || e.KeyChar == '.' && !textBox2.Text.Contains(".") || e.KeyChar == '\b')
    {
        if (textBox2.Text.Contains("."))
        {
            string[] txt = textBox2.Text.Split('.');

            if (txt[1].Length > 1)
            {
                if (e.KeyChar >= '0' && e.KeyChar <= '9')
                {
                    e.Handled = true;
                }
            }
        }
        else
        {
            e.Handled = false;
        }
    }
    else
    {
        e.Handled = true;
    }
}

Explain: Basicaly what it does: when someone types to textBox2 it lets insert only numeric symbols, backspace and one dot (.), it also defines that if dot is inserted there can only be inserted two numbers after it.

The problem: When I enter digit example. 5.22 ir aby other with dot and two number after it, then i selec with mouse digit 5 and delete it, I can't put any number insted it, because there is .22 which by stated rules does not allow any new insertions.

How should I change this code to allow insert numbers to left side of dot in such situation? Any suggestions?

Posted 6-May-13 9:23am

Tautvydas Deržinskas

Updated 6-May-13 9:26am

v2

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Comments

Sergey Alexandrovich Kryukov 6-May-13 15:41pm

Listen to a good friendly advice: never do it!
—SA

2 solutions

Solution 1

You almost never actually need rounding. Instead, you should use proper string formatting when you present numbers on scree. Please see:
http://msdn.microsoft.com/en-us/library/kfsatb94.aspx[^],
http://msdn.microsoft.com/en-us/library/dwhawy9k.aspx[^],
http://msdn.microsoft.com/en-us/library/0c899ak8.aspx[^].

By the way, it's also much safer then explicit rounding. What if you accidentally use the rounded value in calculations. It could lead to incorrect calculation result; and such bug would be hard to find.

—SA

Posted 6-May-13 9:41am

Sergey Alexandrovich Kryukov

Comments

Tautvydas Deržinskas 6-May-13 16:08pm

The thing is that I am not presenting number to screen, I am checking how they are typed to textBox.

Sergey Alexandrovich Kryukov 6-May-13 16:21pm

I got it. Don't do it. Nobody care how many digits it was, you should care only about the value. You need to use double.TryParse, it will validate for double and return double value in case of success. Then you should validate the value itself. Another option: use masked text box; but it may be not flexible enough. For some domains, this is the very it.
—SA

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T Pat · Accepted Answer · 2013-05-06T19:18:00

C#

private void textBox1_KeyPress(object sender, KeyPressEventArgs e)
{
string[] txt = null;
           if (e.KeyChar >= '0' && e.KeyChar <= '9' || e.KeyChar == '.' && !textBox1.Text.Contains(".") || e.KeyChar == '\b')
           {
               if (textBox1.Text.Contains("."))
               {
                   txt = textBox1.Text.Split('.');


                   if (txt[1].Length > 1)
                   {
                       if (e.KeyChar >= '0' && e.KeyChar <= '9')
                       {
                           e.Handled = true;
                       }
                   }
                   if (txt[0].Length != leftlen)
                   {
                       if (e.KeyChar >= '0' && e.KeyChar <= '9')
                       {
                           e.Handled = false;
                       }
                       else
                       {
                           e.Handled = true;
                       }
                   }
               }
               else
               {
                   e.Handled = false;
               }
           }
           else
           {
               e.Handled = true;
           }
           if (txt != null)
           {
               leftlen = txt[0].Length;
           }
}

Most of the things you want is done in this

Just declare leftlen globally and initialize it on form load to -1
Hope it will solve your problem :)