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Miliardsto

Change var name to string

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Miliardsto

I got that func

Func makeHelpImgGUI($title,$width,$height,$img)

$img = GUICtrlCreatePic("",20,40,$width,$height)
_ResourceSetImageToCtrl($img, "HERE")


EndFunc

and I call this func like that

makeHelpImgGUI("Image",1190, 800,$SETTINGS_JPG)

 

so what is the problem in the parameter where is - "HERE" I need value of img but passed as string

so $img = $SETTINGS_JPG and how make it "SETTINGS_JPG"

 

I tried something like that but not work

Func makeHelpImgGUI($title,$width,$height,$img)


$name_str = String($img)
$name_str = StringTrimLeft ($name_str, 1 )

$img = GUICtrlCreatePic("",20,40,$width,$height)
_ResourceSetImageToCtrl($img, $name_str)



EndFunc

 

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IAMK

Try Malik/MHz's comment.

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Miliardsto
9 minutes ago, Zedna said:

Maybe try 


_ResourceSetImageToCtrl($img, Eval($name_str))

instead of


_ResourceSetImageToCtrl($img, $name_str)

not work

  •  
 

do not know what to do with this . xd

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JLogan3o13

Read it, perhaps? And "not work" is about as little information as you could possibly provide.

How about you help us help you by A: putting forth some actual effort and B: if something doesn't work, explaining what exactly you're seeing?


√-1 2^3 ∑ π, and it was delicious!

How to get your question answered on this forum!

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spudw2k

Can't you just pass a string to begin with and not have to worry bout converting a variable name into a string?

makeHelpImgGUI("Image",1190, 800,"SETTINGS_JPG")

Func makeHelpImgGUI($title,$width,$height,$img_name)
    $img = GUICtrlCreatePic("",20,40,$width,$height)
    _ResourceSetImageToCtrl($img, $img_name)
EndFunc

Also, you don't use the $title variable in the function you shared,  You should remove it if it is not necessary.

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