Android Question random label

NJDude

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• Jun 18, 2014

• #2

Yes it is possible, have you tried?

 

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Douglas Farias

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• Jun 18, 2014

• #3

with Label1.Left = caca&"%y" xD
CACA is a random rnd(0,99) result

and this dont work
Rnd(0,99)%y

 

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NJDude

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• Jun 18, 2014

• #4

You cannot do VarName & "%" but you are on the right path.

 

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Douglas Farias

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• Jun 18, 2014

• #5

if help someone

B4X:

Dim lado,alto As Int
lado = GetDeviceLayoutValues.Width
alto = GetDeviceLayoutValues.Height
r1 = Rnd(0,lado)
r2 = Rnd(0,alto)

    Label1.Left = r1
    Label1.top = r2

 

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stevel05

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• Jun 18, 2014

• #6

Just limit the width and height so the label doesn't disappear off the screen:

B4X:

r1 = Rnd(0,lado - Label1.Width)
r2 = Rnd(0,alto - Label1.Height)

 

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Douglas Farias

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• Jun 18, 2014

• #7

xD
when I went to post you posted first
thx man

i see this problem because i have my label out of screen

 

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NJDude

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• Jun 18, 2014

• #8

Or you could limit the area too, check the attached sample.

 

Attachments

• SillySample.zip

  9.6 KB · Views: 172

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NJDude

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• Jun 18, 2014

• #9

stevel05 said:

Just limit the width and height so the label doesn't disappear off the screen:

B4X:

r1 = Rnd(0,lado - Label1.Width)
r2 = Rnd(0,alto - Label1.Height)

Click to expand...

The label will disappear or look incomplete in case the number is zero.

 

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stevel05

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• Jun 18, 2014

• #10

The label will disappear or look incomplete in case the number is zero.

Click to expand...

0,0 is top left, you should be able to see all of the label. It may not be where you want it though. Limiting to borders is a good idea.

 

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