[Resolved] variable output from a function prob

viveleroi0

I have a function that calculates an image size scaled... and I have two variables that I try to use outside of the function.

Here is the function:

function Image_Tool($Use_Image)
{

$Src_Img_Size = getimagesize("http://www.jazbot.com$Use_Image");
$width=$Src_Img_Size[0];
$height=$Src_Img_Size[1];

if($height>="$width")
{
	$x=100*($width/$height);
	$Scale_Width="$x";
	$Scale_Height="100";
}

else
{
	$x=100*($height/$width);
	$Scale_Width="100";
	$Scale_Height="$x";
}

}

And here is my code:

Image_Tool($Image1);
$Image_SRC1="<a href=\"images.php?File=$Image1\"><img src=\"$Image1\" width=\"$Scale_Width\" height=\"$Scale_Height\" border=\"0\"></a>";

Now when I do this, everything works except $Scale_Width and $Scale_Height always print no value at all. Yet when I test it by using echo"$Scale_Height"; in the function itself, it prints just fine.

Why can I not get the two variables to keep their value outside of the function?

AstroTeg

First, you're making PHP do a bit more work then it needs to by doing this:

$Scale_Height="$x";

When really you should just do:

 $Scale_Height=$x;

(Minus the quotes around $x) Both lines of code do the same thing, but the 2nd version prevents PHP from trying to parse a string and then insert $x into the string and then return it to $Scale_Height (think of the quotes as a middle man in a transaction and in this case, you don't need a middleman).

Also, the reason why $Scale_Width does not work outside of your function is because of $Scale_Width has what's called "function scope".

When you define a variable, that variable typically lives within the scope of the definition. Since you defined it in the function, it will only live within that function. Oddly enough, this is a good thing.

I think what you'll want to do is have the function return the width. Its pretty simple: at the end of the function, just put in "return $width;" (minus the quotes of course). Then when you call the function, you have something like: $width = Image_Tool(...); And $width will get the returned value.

For a function like this though, you may want to return both values. In doing so, you will need to figure out how you want to return that data back (easiest is probably put the data in an array and return the array - fancy would be to use a class instead).

If you're not sure what's up on how the "return" works, check out php.net and check out the documentation. It will have more examples then what I can put here on what you can do with return.

viveleroi0

Wow - thanks a lot! That is some good information. I am trying to work it out ands look it up on the net... but I'm having trouble calling the array from the function.

I added the following to the above function:

$Image_Array=array("$Scale_Width", "$Scale_Height");
		return $Image_Array;

(before the end of the function, after the IFs.)

I added this to where I am calling function:

	$Scale_Width=$Image_Array[0];
	$Scale_Height=$Image_Array[1];

(Added at the top... right after I called the function).

Now I never get any errors, but basically I am still getting empty values returned.

I am sure I am doing the array/returning/calling wrong.

What should I change?

AstroTeg

You have to pass the value back from the function to a local variable (this is all outside of the function now).

So you might do something like:

$Image_Array = Image_Tool($Image1);

And then your second chunk of code should work.

And again, you have:

array("$Scale_Width", "$Scale_Height");

No quotes!

Just:

array($Scale_Width, $Scale_Height);

Less typing for you
Better PHP performance

viveleroi0

Thanks - I got it working. You've been a great help.