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M
majab93208718
Inspiring
March 29, 2020
解決済み

Show/hide - Basic functionality

  • March 29, 2020
  • 返信数 1.
  • 1415 ビュー

Hi, any idea why this code does not work as I think it should?

 

 

var _this = this;

_this.IN_mem.visible = true;
_this.IN_fig.visible = false;

stage.enableMouseOver(3);
_this.IN_btn.on('mouseover', function()
{
_this.IN_mem.visible = !_this.IN_mem.visible;
_this.IN_fig.visible = !_this.IN_fig.visible;
});

stage.enableMouseOver(3);
_this.IN_btn.on('mouseout', function()
{
_this.IN_mem.visible = !_this.IN_mem.visible;
_this.IN_fig.visible = !_this.IN_fig.visible;
});

 

 

What I would like is to have IN_mem visible and IN_fig invisible at the beginning.

 

Then when I mouseover the IN_btn I would like the reverse to happen: to have IN_mem invisible and IN_fig visible.

 

Then when I mouseout I would like to have the initial position: IN_mem visible and IN_fig invisible.

 

But what happens is:

 

When I mouseover:

  • IN_mem disappears for a second and at the same time IN_fig appears for a second. Then they both go to their previous stages (IN_mem becomes visible again; IN_fig becomes invisible again)
  • But I would like them to STAY (IN_mem invisible and IN_fig visible) in their new statue for as long as I mouseout.

When I mouseout nothing happens.

 

I do not understand that behaviour.

Isn't 

 

 

_this.XXX.visible = !_this.XXX.visible;

 

 

supposed to do the oposite of the current visibility status and STAY in that status. Not return to the previous status after one second?

 

I also don't understand why nothing happens when I mouseout. But that's a secondary question, I suppose. I believe I should first be able to understand why mouseover isn't working as I think it should.

 

Thanks for any hint that you might have.

このトピックへの返信は締め切られました。
解決に役立った回答 ClayUUID

You don't need to enable mouseover twice.

 

You only need to use the _this alias in your event handler functions. It doesn't hurt to use it everywhere, but it's unnecessarily verbose.

 

You shouldn't be using the ! operater at all, because that just confuses things. You don't want a toggle function, you want each event to set your objects to a specific visibility state. Just use true and false.

返信数 1

C
ClayUUID解決！
Legend
March 29, 2020

You don't need to enable mouseover twice.

 

You only need to use the _this alias in your event handler functions. It doesn't hurt to use it everywhere, but it's unnecessarily verbose.

 

You shouldn't be using the ! operater at all, because that just confuses things. You don't want a toggle function, you want each event to set your objects to a specific visibility state. Just use true and false.

M
majab93208718作成者
Inspiring
March 29, 2020

Hi, thanks so much for your reply.

I've changed my code to (unfortunatelly I don't understand what you mean by _this alias, but I guess it's not a priority at the momemnt):

 

var _this = this;

_this.IN_mem.visible = true;
_this.IN_fig.visible = false;

stage.enableMouseOver(3);

_this.IN_btn.on('mouseover', function()
{
_this.IN_mem.visible = false;
_this.IN_fig.visible = true;
});

_this.IN_btn.on('mouseout', function()
{
_this.IN_mem.visible = true;
_this.IN_fig.visible = false;
});

 

When I run this code, same thing happens as before:

 

When I mouseover:

  • IN_fig becomes visible for a second and disapperas again. At the same time IN_mem becomes invisible for a second and appears again.
  • Arent' they suppose to STAY in their new visibility state till I mouseout?

 

When I mouseout:

  • Nothing happens.

 

It's a strange behaviour, isn't it? Do you have any idea what might be going wrong?

 

C
ClayUUID
Legend
March 29, 2020

_this is an alias to this.

 

What you're descibing shouldn't be happening. Try putting in your event handler functions:

console.log("mouseover");

and

console.log("mouseout");

I'll trust you to know which one goes where.

 

Then open the browser console (F12) while your page is running and you'll be able to see when each event handler fires.

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