eval in a function...

[spaceDAISY]

I'm seeing some slightly odd behaviour, and i'm wondering if anyone feels like poking their nose in it with me...

Basic premise is this. I'm storing the string "$randicon" in a db field, and this string is retrieved and stored in the variable $items[1]->image. I'm trying to eval it, as $randicon is set as, well, a random image name earlier in the script. So far, this works fine for me:

$image = $items[1]->image; 
eval ("\$image = \"$image\";");
echo $image;

All of which is a little bit too much to place in-line 100 times or more in my site - so I decided to make it a function, for that extra bit of clarity:

function evalImage($image) {
        eval ("\$image = \"$image\";");
        print $image;
}

Then, of course, to call it as this:

evalImage($items[1]->image);

yet it doesn't work. The string "$randicon" gets passed to the function alright, but $image ends up being blank. I'm a little confused as to this, it seems like a fairly straight forward thing to do, yet it's not working. I've tried the ob_ series of functions aswell, all to no avail. Anyone seen this behaviour before or have some light to possibly shed on it?

[sarah]

Any variables or arrays that are used in $image, will need to included in the global variables for the function. That might be the issue...

[seby]

i think you have to define $items[1]->image
in the function, $image isn't assigned to something in the function... from what i can see.

[spaceDAISY]

I'll be honest here, i come from the world of perl, and i'm clearly misunderstanding how php functions work if this is the case...

When calling a function, you pass it a value, and that value should get taken as, well, $image, right? I mean, that's how it seems to be working. If i go "print $image;" in the function before calling the eval, it DOES print the string "$randicon", so I think i'm on the right path... it just seems to disappear into thin air once it gets eval()'d

[seby]

a function can't access outside variables if its not a global variable, globalizing a variable means it can be used throughout the script, and within functions..

try this.

function evalImage($image) { 
global $image;
        eval ("\$image = \"$image\";"); 
        print $image; 
}

[sarah]

Are you using objects and classes? What does $items[1]->image pull from?

[spaceDAISY]

nup, didn't work. it still prints nothing. if i go like this:

function evalImage($image) {  

        print $image;
        eval ("\$image = \"$image\";");  

        print "Hii!" . $image;  

}

then it prints the string "$randiconHii!". The weird thing is that the eval works fine if i don't pass the string to a function. Odd indeed!

[spaceDAISY]

Originally posted by sarah
Are you using objects and classes? What does $items[1]->image pull from?

it's built like this:

for($i = 0; $i < 3; $i++)
{
        $row = mysql_fetch_object($res);
        $items[$i] = new item;
        $items[$i]->title = fixtxt($row->title);
        $items[$i]->abstract = fixtxt($row->abstract);
        $items[$i]->id = $row->id;
        $items[$i]->image = $row->img_small;
}

so basically, it's the first 3 rows returned from a select. i'm using the [1] cause, well, that's just the first piece of code i tried using.

[sarah]

What happens when you add $items to the global variables? (Just curious...)

function evalImage($image) { 
global $items;
        eval ("\$image = \"$image\";"); 
        print $image; 
}

[spaceDAISY]

nada :/

[sarah]

http://us4.php.net/manual/en/function.eval.php has some info on using eval with classes...do a search for "->". Hope it helps...

[examancer]

OK... why that wasn't working didn't make sense to me either, but i think I got it...

ok, the problem isn't that $image isn't avialable globally... you passed a variable to a function, and the function received it and called it $image. No problem there. No need for global. But, then you want to use eval() to chage $image from the string '$randicon' to $image = $randicon, right?

Well, $randicon was never made available to the function, which is why it showed up blank.

you'd need to do something like this:

function evalImage($image) { 
  global eval($image);
  eval ("\$image = \"$image\";"); 
  print $image; 
}

That should make it so that if $image is the string '$randicon' or the string '$obscurevariablename' the variable will be made available to that function.

EDIT: Ok, that doesn't work either. I'm working on it though... It is an intesting problem.

[examancer]

Ok, this is how you do it... for sure... I actually tested this one:

$randicon = 'sucess!';
$image = '$randicon';
evalImage($image);

function evalImage($image) {
	eval("GLOBAL $image;");
	eval ("\$image = \"$image\";"); 
	print "Test output: " . $image;
}

make sense?

[spaceDAISY]

ah HAH! yes, that DOES work 🙂 i get it, i guess i'm just not familiar enough with scoping rules under php - as i said earlier, i'm a perl monkey by trade, and yeah... i guess i assumed $randicon was global, as perl would have globalised the variable when set in this kind of context... essentially, i just have another php include above all this code that has a line going:

$randicon = "icon" . rand(1, 23) . ".jpg";

it's not embedded in another subroutine or anything, really, it's just there - case in point, it was in scope if i called it from the other page... and the include order is certainly "right", ie. the $randicon variable is set before the function is defined (although i don't think that should matter...)

aaanyway, thank you all for your help, it was certainly an interesting problem for me, and i think now it's time to go find some good documentation on variable scope under php 🙂