e doesn't exist in the current context

Question

1.00/5 (2 votes)

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I used two textbox like name txtSerial and txtID where I want to type only
number 0 to 9. This is ok. But I want to use same code in keypress in a other private function. But there e handler doesn't exist. Now how can I solve this problem? I need to use e handler in others private function. Please update me
a solution.

C#

private void txtSerial_KeyPress(object sender, KeyPressEventArgs e)
{
      number_handler();
}
private void txtID_KeyPress(object sender, KeyPressEventArgs e)
{
      number_handler();
}

private void number_handler()
{
    char var = e.KeyChar;

    if (var == 8)
        e.Handled = false;

    else if (var < '0' || var > '9')
         e.Handled = true;
}

Posted 9-Nov-12 5:52am

Member 8454009

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4 solutions

Solution 2

In both KeyPress handlers, "e" is a parameter, which you do not pass to the number_handler method.
Hence, it is not available, since the compiler does not check what calls what and make variables available in a chain.

If you need "e" in number_handler, then pass it through as a parameter, or better, hand it e.KeyChar and have it return true or false depending on whether it handled it or not. That way, it does not need to know it is called from an event handler, and can be a more generic method.

Posted 9-Nov-12 6:03am

OriginalGriff

Comments

Sergey Alexandrovich Kryukov 9-Nov-12 12:10pm

There are further problems, please see my answer.
--SA

Solution 4

Wow! solution 3 is greatly helped me. But there e is mistake in number_handler().

C#

private void txtSerial_KeyPress(object sender, KeyPressEventArgs e)
{
      number_handler(e);
}
private void txtID_KeyPress(object sender, KeyPressEventArgs e)
{
      number_handler(e);
}

private void number_handler(KeyPressEventArgs e)
{
    char var = e.KeyChar;

    if (var == 8)
        e.Handled = false;

    else if (var < '0' || var > '9')
         e.Handled = true;
}

Posted 9-Nov-12 6:35am

Member 8454009

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Sergey Alexandrovich Kryukov · Accepted Answer · 2012-11-09T06:01:00

Solution 1

Wow! And where do you think e is defined in number_handler? Add a parameter KeyPressEventArgs e to this method, for example.

Wait a minute. Do you really understand what are you trying to write? Don't use var — this is a keyword. Your code won't compile because 8 is not a char, you can use '\U+0008' to make it a character, as comparison with KeyPressEventArgs.KeyChar requires. No need to assign Handled to false — this is its initial value; only assign to true when required.

—SA

Posted 9-Nov-12 6:01am

Sergey Alexandrovich Kryukov

Updated 9-Nov-12 6:18am

v4

Comments

fjdiewornncalwe 9-Nov-12 12:16pm

+5.

Sergey Alexandrovich Kryukov 9-Nov-12 12:17pm

Thank you, Marcus.
--SA

Maciej Los 9-Nov-12 12:46pm

Not complete answer... but good enough for 5+! ;)

Sergey Alexandrovich Kryukov 9-Nov-12 13:04pm

Thank you very much, Maciej.
What would you consider a complete? If you say that this is writing the complete code, I can tell you this: as OP's code has bugs, we cannot say for sure what exactly it was supposed to do.
--SA

Maciej Los 9-Nov-12 14:02pm

Sergey, "Not complete answer.." - It was a joke, because of the second part of sentence "but good enough".
Cheers!
;)

Sergey Alexandrovich Kryukov 9-Nov-12 14:20pm

Whatever... thank you again.
--SA

arunrv · Accepted Answer · 2012-11-09T06:06:00

Solution 3

Create function like this

C#

private void txtSerial_KeyPress(object sender, KeyPressEventArgs e)
{
      number_handler(e);
}
private void txtID_KeyPress(object sender, KeyPressEventArgs e)
{
      number_handler(e);
}

private void number_handler(KeyPressEventArgs e)
{
    char var = e.KeyChar;

    if (var == 8)
        e.Handled = false;

    else if (var < '0' || var > '9')
         e.Handled = true;
}

It will work.

Posted 9-Nov-12 6:06am

arunrv

Updated 9-Nov-12 6:10am

v2

Comments

Thomas Daniels 9-Nov-12 12:10pm

But in the txtSerail_KeyPress and txtID_KeyPress methods, don't call number_handler();
but number_handler(e);
I added this to your answer.

Sergey Alexandrovich Kryukov 9-Nov-12 12:11pm

No, it won't. I explained why (or why else) it won't in my answer. You mechanically copied OP's code and fixed just one thing. You need to check up the solution before posting.
--SA