Hola a todos,
estoy intentando generar una imagen desde una captura de pantalla hecha en flash, en un servidor con php 4 y windows me genera la imagen correctamente pero en un php5 y linux me genera una imagen en blanco, alguien sabe que puede ser? Este es el codigo de php que uso:
error_reporting(E_ALL);
/**
* Get the width and height of the destination image
* from the POST variables and convert them into
* integer values
*/
$w = (int)$_POST['width'];
$h = (int)$_POST['height'];
// create the image with desired width and height
$img = imagecreatetruecolor($w, $h);
// now fill the image with blank color
// do you remember i wont pass the 0xFFFFFF pixels
// from flash?
imagefill($img, 0, 0, 0xFFFFFF);
$rows = 0;
$cols = 0;
// now process every POST variable which
// contains a pixel color
for($rows = 0; $rows < $h; $rows++){
// convert the string into an array of n elements
$c_row = explode(",", $_POST['px' . $rows]);
for($cols = 0; $cols < $w; $cols++){
// get the single pixel color value
$value = $c_row[$cols];
// if value is not empty (empty values are the blank pixels)
if($value != ""){
// get the hexadecimal string (must be 6 chars length)
// so add the missing chars if needed
$hex = $value;
while(strlen($hex) < 6){
$hex = "0" . $hex;
}
// convert value from HEX to RGB
$r = hexdec(substr($hex, 0, 2));
$g = hexdec(substr($hex, 2, 2));
$b = hexdec(substr($hex, 4, 2));
// allocate the new color
// N.B. teorically if a color was already allocated
// we dont need to allocate another time
// but this is only an example
$test = imagecolorallocate($img, $r, $g, $b);
// and paste that color into the image
// at the correct position
imagesetpixel($img, $cols, $rows, $test);
}
}
}
// print out the correct header to the browser
header("Content-type:image/jpeg");
// display the image
imagejpeg($img, "mark.jpg",90);
