(adsbygoogle = window.adsbygoogle || []).push({}); i am new to C# and i need the code to make a button open another form

He kinda just powned you Correy.

he might be able to code, but anyone can really..
he just thought he had something on me which he thought wrong =D

True. But don't underestimate this kid. He's admin at a site of which I'm staff, and he has some of the best coding tutorials I've ever seen. Especially the most noob friendly. I honestly think you should take a look, then rethink what you said.

i don't need to re-think nothing, i said this guy may be able to code.. he may be admin, he may post good tutorials.. but he can't tell me wrong about things which i know is true my absolute full percent.

All right. Have fun knowing it all.

I just wanted to point out that actually not anyone/everyone can write code, and even amongst those that can, there are even fewer that actually understand computers and understand the code that they are writing. I don't doubt that Correy is a capable programmer, however I initially replied to point out a flaw in his statement.

name.Show() and name.ShowDialog() will both open the form, and something that I've already proven, while they may act slightly different and be declared slightly differently, not only will both work fine, name.Show() is actually the proper and preferred format compared to name.ShowDialog(), which should only be used in the form being opened is truly a dialog interface as opposed to a whole new form.

right, so if i put in my coding.. form1.Show(); your saying it will show, am i correct?

---------- Post added at 10:21 PM ---------- Previous post was at 10:16 PM ----------



and what are you trying to implement?

Your method of using ShowDialog() works just fine. You stated earlier that Show() does not work, Show(), the method itself, does work, it does indeed show a form, however it may need to be used in a slightly different way than ShowDialog(). I further pointed out that ShowDialog() is meant for dialogs (hence the name) as opposed to running two forms at once. If your form is to be used as a dialog, then you should use ShowDialog(), otherwise, use Show(). This is because ShowDialog() will only allow the user to manipulate the second form that was just opened, and not the first form, whereas Show() will allow you to manipulate both forms at once. If the question was "How do I open a form as a dialog?" or "How do I open a dialog?" your response (ShowDialog()) would be correct, however, if the question is "How do I open another form?" (and that was indeed a correct paraphrasing of the question) then one should recommend Show() because its technically more correct, and will allow for more sophistication overall than ShowDialog().

That's all I'm trying to say, and that's all I've been trying to say. ShowDialog() and Show() are slightly different, but both work, and both can open a form. The semantics by which they are used is really irrelevant to the overall solution.


And don't hate on bananaman, I'm sure he's not trying to "implement" anything.


edit:

BAdmaNgLiTcHa was correct in his very first reply:


"name" is the name given to the object of the form.

After looking back at this I realise I had made a mistake what I replied with was in-fact wrong.
I implemented for name to create a new form which was not the question asked which was to show a already made form.
The correct script is:

    Form name = new name();

name.show();

God sometimes I can make the silliest of mistakes.