images on the fly problem

[otrorichard]

Hi every one
I'm creating some images on the fly, after cut, resize, generate the image I add some little images in the final image.

my problem starts here

$temp='mImages/punto4.png'; 
$mark = imageCreateFromPNG($temp);

the code above works very good, but if i try to change the second line with a variable, for example :

$myVariable=4;
$temp='mImages/punto' . $myVariable . '.png';
$mark = imageCreateFromPNG($temp);

I can't obtain my image, what do you think is wrong?

thanks in advanced

richard

[gid]

I'd say your copy of php is haunted, that SHOULD work as far as I can tell... 🙁

Maybe $myVariable contains some spaces at the beginning or end? Try:

$myVariable = trim($myVariable);

maybe... I dunno, if that's not the case, try dumping each character of $myVariable using ord() and noting the ascii value, looking for any nulls or extra chacters that's in the variable and not visible to you for some reason. I dunno, though, I'm just guessing, it's hard to tell without staring at some code.

[otrorichard]

Originally posted by gid
I'd say your copy of php is haunted, that SHOULD work as far as I can tell... 🙁

Maybe $myVariable contains some spaces at the beginning or end? Try:

$myVariable = trim($myVariable);

maybe... I dunno, if that's not the case, try dumping each character of $myVariable using ord() and noting the ascii value, looking for any nulls or extra chacters that's in the variable and not visible to you for some reason. I dunno, though, I'm just guessing, it's hard to tell without staring at some code.

thanks, I put only this piece of code cause, without the "varible part" everything work very well.

I forgot tell you a detail, a pass this value like url example images.php?myVariable=2

richard

[marcel]

hi,
after
$temp='mImages/punto' . $myVariable . '.png';

try an echo:
echo $temp;
to see the value of $temp var.

Hope this helps.

PM

[gid]

Originally posted by marcel
hi,
after
$temp='mImages/punto' . $myVariable . '.png';

try an echo:
echo $temp;
to see the value of $temp var.

Hope this helps.


PM

That makes sense, I was kinda assuming he already tried something that simple, but I guess I shouldn't assume anything

[otrorichard]

thanks,

but this code code is part of a reference of an <img> tag

like <img src="myImage.php?myvariable=1" ....... >

I almost solve the problem but recomendation are welcome

richard

[marcel]

attention:
names variable are CAseSenSiTiVe

PM