Help to get get image size for images having space in their name

[vikaspa]

I am using following fucntion to get proportionate width for user specific height

 

 

 

However I noticed that in case image name is having space it gives an error

 

Example Function works ok for image name say "pict1.jpg"

 

 

 

However gives error for

 

pict 1.jpg (space in name beforer 1)

 

 

 

  list($width, $height, $type, $attr) = getimagesize(addslashes($imagename));

                    if ($width ==0)

                      {$imgstr="No Image"; }

    else 

    {

                    $h=$height*($reqwidth/$width) ;

                      if ($width >  $reqwidth)

      { $imgstr="<img src='".addslashes($imagename)."' width=".$reqwidth." height=".$h." border=0 >";}

      else

        { $imgstr="<img src='".addslashes($imagename)."' width=".$width." height=".$height." border=0 >";}

                }

return $imgstr;

}

 

 

 

[andy_b42]

i think the problem might be that spaces are no recognised url characters, try adding this line:

 

$imagename = str_replace(" ", "%20", $imagename);

 

before

 

list($width, $height, $type, $attr) = getimagesize(addslashes($imagename));

 

this will replace spaces with %20which is the url character for a space