how to create multiple picturebox?

[lishannx]

ive a drag and drop program and i want to allow drag and drop mulitple times and not one time.

after being drag and drop, the created picturebox will also be created multiple times.

is it correct to use for loop?

pls guide.thank u.

rgds.

[Diamonddrake]

Could you please explain in detail what you are trying to accomplish?

[lishannx]

it is a drag and drop program.

1. user has to drag a picturebox to the panel.
2. if user drag the picturebox already, the location will be empty. thus, i created a picturebox to clone the picturebox that had already been dragged.
3. now, ive to allow the user to be able to continue dragging and dropping more den once.

how am i to do abt it?

rgds.

[Diamonddrake]

we have already covered this.
http://www.daniweb.com/forums/thread273621.html

This is the same question. I don't understand.

[lishannx]

it is not the same question.

now i have no problem drag and drop picturebox to panel.

i jus wan to make mulitple similar picturebox to drag and drop.

[Diamonddrake]

I have posted a Visual Studio project that should cover any confusion.
Good luck.

[lishannx]

thanks for yr help. it help me alot.

but now i ned to do smth like that for mousehover and mouseleave events and im having errors.

could u pls help? thank u.

[Diamonddrake]

We aren't here to write your software for you. If you have erroneous code. Post it and we can find out why its not working.

[cVz]

Your first problem is the fact that you write barely understandable questions... structure your questions correctly and learn proper English.

by the looks of things, you are either a web developer that is trying to code software or you are a web developer that used to use php and has no real clue as to what JavaScript Is.....

If you are using windows forms (SOFTWARE , NOT WEB) then you will find the answers in your events tab (next to properties).

if you are using ASP , C# and javascript, there are multiple ways of doing this, however i think the most time effective at this time would be to use attributes...

in your form load event type something like this:

pictureboxId.Attributes.Add("onMouseOver", "JS EVENT HERE");
// the above will act as a mouse over event. 
// Now lets do the mouse out event

pictureboxId.Attributes.Add("onMouseOut", "JS Event Here");

Play with it a bit .... and see where the road takes you

[lishannx]

Your first problem is the fact that you write barely understandable questions... structure your questions correctly and learn proper English.

by the looks of things, you are either a web developer that is trying to code software or you are a web developer that used to use php and has no real clue as to what JavaScript Is.....

If you are using windows forms (SOFTWARE , NOT WEB) then you will find the answers in your events tab (next to properties).

if you are using ASP , C# and javascript, there are multiple ways of doing this, however i think the most time effective at this time would be to use attributes...

in your form load event type something like this:

pictureboxId.Attributes.Add("onMouseOver", "JS EVENT HERE");
// the above will act as a mouse over event. 
// Now lets do the mouse out event

pictureboxId.Attributes.Add("onMouseOut", "JS Event Here");

Play with it a bit .... and see where the road takes you

wh0 are u and where are u from ? how would u made such a judgement? wad proper english?
u are the one having english problem. i am here to get help with my error not for u to judge on my english.

if u don understand my english den go back to PRIMARY ONE and learn proper english! YIKE. disgusting.

[lishannx]

///////the MOUSEHOVER AND MOUSELEAVE codes.
     void PB_MouseHover(object sender, EventArgs e)
        {
            mouseLocation = this.PointToClient(Cursor.Position);
            pictureBox3.Location = mouseLocation;
            pictureBox3.Show();
        }

        void PB_MouseLeave(object sender, EventArgs e)
        {
            pictureBox3.Hide();
        }

[Diamonddrake]

What's wrong with it? what is it suppose to that it isn't doing?

[lishannx]

It does not work.