I have one protocol that recieve id and send picture back.Now I have question how can I send just create .gif ?? Here please take a look to my code
if($id!="")
{
........
$ext = substr( $picturename, strrpos($picturename, '.')+1 );
switch($ext)
{
case 'jpg':
$ext = 'jpeg';
break;
case 'jpeg':
$ext = 'jpeg';
break;
case 'png':
$ext = 'png';
break;
case 'gif':
$ext = 'gif';
break;
}
header('Content-Description: File Transfer');
header('Content-Type: image/'.$ext);
$file = "./images/".$picturename;
readfile($file);
//this work
}
else
{
// Create a 1x1 image
$im = imagecreatetruecolor(55, 30);
$trans = imagecolorallocate($im, 0, 0, 0);
// Make the background transparent
imagecolortransparent($im, $trans);
header('HTTP/1.1 403 Forbidden');
//How can I send .gif back with header ??
}
You can set multiple headers and output an image in the same manner. Attached is the code to generate a 1x1 pixel.
....
header('HTTP/1.1 403 Forbidden');
header('Content-Type: image/gif');
// output a 1x1 pixel
echo base64_decode("R0lGODdhAQABAIAAAPxqbAAAACwAAAAAAQABAAACAkQBADs=");
Just output imagegif($im);. Setting the 403 header doesn't do anything special, so you can output the image and that's what will be received.
If you don't need to dynamically include anything in the image (like text for example) you could just have a default 1x1 image that you return instead of using imagecreate function...?
As far as the 403 header, though, I'm not sure if browsers will probably display an image if it thinks it was an error response from the server - so you might get broken images.
Related
Following code:
class FilesController extends Controller {
public function icon($fileId) {
$this->uses("FileAttachment");
$FA = new FileAttachment($fileId);
$fnth = $FA->path . 'th___' . $FA->filename;
$this->View->render = false;
if (file_exists($fnth)) {
$imageinfo = getimagesize($fnth);
switch ($imageinfo[2]) {
case IMAGETYPE_JPEG:
$image = imagecreatefromjpeg($fnth);
Header("Content-type: image/jpg");
imagejpeg($image);
break;
case IMAGETYPE_GIF:
$image = imagecreatefromgif($fnth);
Header("Content-type: image/gif");
imagegif($image);
break;
case IMAGETYPE_PNG:
$image = imagecreatefrompng($fnth);
//Header("Content-type: image/png");
imagepng($image);
break;
case IMAGETYPE_BMP:
$image = imagecreatefromwbmp($fnth);
Header("Content-type: image/bmp");
imagewbmp($image);
break;
}
}
imagedestroy($image);
}
}
produces broken images. The image is shown in a browser just as a broken one. The file exists, otherwise there would come nothing. My Windows shows the pictures correctly. What can be the cause?
PS. If I remove Header, the browser displays the image's binary content, so it actually looks OK...
Your code is rather pointless - you're forcing PHP (and gd) to do a bunch of useless work to load up the file, decompress it into memory, and then recompress it for output.
Why not simply have:
$info = getimagesize($path_to_file);
header('Content-type: ' . $info['mime']);
readfile($path_to_file);
?
As for your code, check the entire output of the function. If there's ANY php warnings/errors being produced, they'd get embedded in the output and general cause "corrupt" images.
imagecreatefromgif(); //returns an image resource
Have you tried to add at the bottom ?
readfile($image); //reads a file and writes it to the output buffer.
I had the below code that load image from DB. There are more than 600 rows of image has been inserted into the DB. I need the script that can perform these action:
Step 1) Load the image from DB
Step 2) process the image by putting the watermark
Step 3) Output the image to the browser.
I had the below code, that load and show the image. but I don't have any idea how to do the watermark.
$dbconn = #mysql_connect($mysql_server,$mysql_manager_id,$mysql_manager_pw) or exit("SERVER Unavailable");
#mysql_select_db($mysql_database,$dbconn) or exit("DB Unavailable");
$sql = "SELECT type,content FROM upload WHERE id=". $_GET["imgid"];
$result = #mysql_query($sql,$dbconn) or exit("QUERY FAILED!");
$contenttype = #mysql_result($result,0,"type");
$image = #mysql_result($result,0,"content");
header("Content-type: $contenttype");
echo $image;
mysql_close($dbconn);
?>
Please help...
You could ether learn how to manipulate images on your own from php.net or you just get a package like the one below:
http://pear.php.net/package/Image_Tools
(Tools collection of common image manipulations. Available extensions are Blend, Border, Marquee, Mask, Swap, Thumbnail and Watermark.)
How about calling your image the same way you do for the SELECT type, content?
Select the image with an imagepath from your database and then style it so it floats over your information. you could also hardcode the watermark image as it is always the same image you can have repeated. You won't see your information but if you put an opacity on the image, you can see through it:
#img.watermark {
float:left;
opacity:0.1;
z-index:1;
}
This is just one idea on how to do it, but should work quite nicely!
take a look at
imagecopymerge()
from php's graphics librabry, it should do what you're looking for.
http://www.php.net/manual/en/function.imagecopymerge.php
Finally I get the solution, here is the code:
<?php
$dbconn = #mysql_connect($mysql_server,$mysql_manager_id,$mysql_manager_pw) or exit("SERVER Unavailable");
#mysql_select_db($mysql_database,$dbconn) or exit("DB Unavailable");
$sql = "SELECT id, original_name, type, content FROM upload WHERE id=". $_GET["imgid"];
$result = #mysql_query($sql,$dbconn) or exit("QUERY FAILED!");
$fileID = #mysql_result($result,0,"id");
$contenttype = #mysql_result($result,0,"type");
$filename = #mysql_result($result,0,"original_name");
$image = #mysql_result($result,0,"content");
$fileXtension = pathinfo($filename, PATHINFO_EXTENSION);
$finalFileName = $fileID.".".$fileXtension;
// put the file on temporary folder
$filePutPath = "/your/temporary/folder/".$finalFileName;
// put the contents onto file system
file_put_contents($filePutPath, $image);
// get the watermark image
$stamp = imagecreatefrompng('../images/watermark.png');
switch($fileXtension)
{
case 'JPEG':
case 'JPG' :
case 'jpg' :
case 'jpeg':
$im = imagecreatefromjpeg($filePutPath);
break;
case 'gif' :
case 'GIF' :
$im = imagecreatefromgif($filePutPath);
break;
case 'png' :
case 'PNG' :
$im = imagecreatefromgif($filePutPath);
break;
default :
break;
}
list($width, $height) = getimagesize($filePutPath);
// set area for the watermark to be repeated
imagecreatetruecolor($width, $height);
// Set the tile (Combine the source image and the watermark image together)
imagesettile($im, $stamp);
// Make the watermark repeat the area
imagefilledrectangle($im, 0, 0, $width, $height, IMG_COLOR_TILED);
header("Content-type: $contenttype");
// free the memory
imagejpeg($im);
imagedestroy($im);
// delete the file on temporary folder
unlink($filePutPath);
mysql_close($dbconn);
?>
i know that
$localfile = $_FILES['media']['tmp_name'];
will get the image given that the POST method was used. I am trying to read an image which is in the same directory as my code. How do i read it and assign it to a variable like the one above?
The code you posted will not read the image data, but rather its filename. If you need to retrieve an image in the same directory, you can retrieve its contents with file_get_contents(), which can be used to directly output it to the browser:
$im = file_get_contents("./image.jpeg");
header("Content-type: image/jpeg");
echo $im;
Otherwise, you can use the GD library to read in the image data for further image processing:
$im = imagecreatefromjpeg("./image.jpeg");
if ($im) {
// do other stuff...
// Output the result
header("Content-type: image/jpeg");
imagejpeg($im);
}
Finally, if you don't know the filename of the image you need (though if it's in the same location as your code, you should), you can use a glob() to find all the jpegs, for example:
$jpegs = glob("./*.jpg");
foreach ($jpegs as $jpg) {
// print the filename
echo $jpg;
}
If you want to read an image and then render it as an image
$image="path-to-your-image"; //this can also be a url
$filename = basename($image);
$file_extension = strtolower(substr(strrchr($filename,"."),1));
switch( $file_extension ) {
case "gif": $ctype="image/gif"; break;
case "png": $ctype="image/png"; break;
case "jpeg":
case "jpg": $ctype="image/jpeg"; break;
default:
}
header('Content-type: ' . $ctype);
$image = file_get_contents($image);
echo $image;
If your path is a url, and it is using https:// protocol then you might want to change the protocol to http
Working fiddle
My problem is when you use for exemple :
<img src="/img.jpg" />
the src has to be an image and the image has to be accessible to the person. I want to be able to control the access to those image (if the user is logged for example). IF the person has not access to the image, he can't access it. My script that controls the access to an image is a php file.
I know .htaccess can limit access to ressources, but I need to valid in the php file. Is there a way to do this or to load the image with javascript (using ajax request) and changing the source of the image to the location of the image in the temp folder?
Your image source doesn't necessarily have to point to the real image, neither does it have to be an image:
<img src="images.php?f=img.jpg" />
Then you can write a PHP script which does the required validations and return the image afterwards through the script
$image = basename($_GET['f']);
if (user_has_access()) {
// You have to determine / send the appropriate MIME-type manually
header('content-type: image/jpg');
readfile($image);
} else {
header('HTTP/1.1 403 Forbidden');
}
$image = basename($_GET['image']);
if (validation()) {
header('Content-type: image/jpeg');
readfile($image);
} else {
header('HTTP/1.1 403 Forbidden');
}
basename is mandatory, of a hacker will have every password stored on your server.
correct content type
sane memory consumption
<img src="/image.php?id=myImage.jpg">
And myImage.jpg:
<?php
$imageName = $_GET['id'];
$ctype="image/jpg";
$extension = substring($imageName, strstr($imageName, '.'));
switch($extension) {
case "gif": $ctype="image/gif"; break;
case "png": $ctype="image/png"; break;
case "jpeg":
case "jpg": $ctype="image/jpg"; break;
}
if (someValidation_here) {
header("Content-Type: $ctype");
$handle = fopen($imageName, "rb");
echo fread($handle, filesize($imageName));
fclose($handle);
}
?>
Serve your images with PHP
You can limit access using with PHP, if the PHP is outputting the image contents.
By using the src tag http://www.example.com, you are not using PHP, your web server is serving up the image on its own.
To output an image with PHP, make sure you set your header variable appropriately.
For example:
<?php
if($user->isAuthenticated())
{
$image = imagecreatefromjpeg ($server_image_path);
header('Content-Type: image/jpeg');
imagejpeg($image, NULL, 75);
imagedestroy($image);
}
else
{
header('HTTP/1.1 403 Forbidden');
}
?>
You can do this with PHP ... see imagejpeg for an example ... you then have the img as follows :
<img src="/myfile.php" />
or use your PHP file to grab an image and output it using :
header('Content-Type: image/jpg');
echo file_get_contents("yourimage.jpg");
You can ceep user_login_status in your $_SESSION, and on a view part check it
<?if ($_SESSION['user_status'] == 'login'){?>
<img src="/img.jpg" />
<?}else{?>
// some stuff instead of <img/>
<?}?>
I have a question about the images displaying using a function getImage_w($image,$dst_w), which takes the image URL ($image) and the destination width for this image ($size). Then it re-draws the image changing its height according to the destination width.
That's the function (in libs/image.php file):
function getImage_w($image,$w){
/*** get The extension of the image****/
$ext= strrchr($image, ".");
/***check if the extesion is a jpeg***/
if (strtolower($ext)=='.jpg' || strtolower($ext)=='.jpeg'){
/***send the headers****/
header('Content-type: image/jpeg');
/**get the source image***/
$src_im_jpg=imagecreatefromjpeg($image);
/****get the source image size***/
$size=getimagesize($image);
$src_w=$size[0];
$src_h=$size[1];
/****calculate the distination height based on the destination width****/
$dst_w=$w;
$dst_h=round(($dst_w/$src_w)*$src_h);
$dst_im=imagecreatetruecolor($dst_w,$dst_h);
/**** create a jpeg image with the new dimensions***/
imagecopyresampled($dst_im,$src_im_jpg,0,0,0,0,$dst_w,$dst_h,$src_w,$src_h);
imagejpeg($dst_im);
}
In a file imagetest.php I have this code portion:
<?php
require 'libs/image.php';
echo '<h1>HELLO WORLD : some html</h1>
<img src="'.displayImg_w('http://www.sanctius.net/wp-content/uploads/2010/05/Avatar-20.jpg',200).'">
';
In the past, I used to write the URL with $_GET paramers defining the image. But now , I want to use the function directly in my code.
The problem is that the image is displaying correctly, but the Hello World HTML code is not translated by the browser (I know that the header are already sent by the first code) But I want to know how to display the image correctly without affecting the html code. and also without using get parameters that change the URL of the image to this undesired form :
libs/image.php?image=http://www.example.com/image&width=200
After my earlier, totally wrong answer, I hope to make up for it with this. Try this code:
<?php
function getImage_w($image,$w){
// Get the extension of the file
$file = explode('.',basename($image));
$ext = array_pop($file);
// These operations are the same regardless of file-type
$size = getimagesize($image);
$src_w = $size[0];
$src_h = $size[1];
$dst_w = $w;
$dst_h = round(($dst_w/$src_w)*$src_h);
$dst_im = imagecreatetruecolor($dst_w,$dst_h);
// These operations are file-type specific
switch (strtolower($ext)) {
case 'jpg': case 'jpeg':
$ctype = 'image/jpeg';;
$src_im = imagecreatefromjpeg($image);
$outfunc = 'imagejpeg';
break;
case 'png':
$ctype = 'image/png';;
$src_im = imagecreatefrompng($image);
$outfunc = 'imagepng';
break;
case 'gif':
$ctype = 'image/gif';;
$src_im = imagecreatefromgif($image);
$outfunc = 'imagegif';
break;
}
// Do the resample
imagecopyresampled($dst_im,$src_im,0,0,0,0,$dst_w,$dst_h,$src_w,$src_h);
// Get the image data into a base64_encoded string
ob_start();
$outfunc($dst_im);
$imgdata = base64_encode(ob_get_contents()); // Don't use ob_get_clean() in case we're ever running on some ancient PHP build
ob_end_clean();
// Return the data so it can be used inline in HTML
return "data:$ctype;base64,$imgdata";
}
echo '<h1>HELLO WORLD : some html</h1>
<img src="'.getImage_w('http://www.sanctius.net/wp-content/uploads/2010/05/Avatar-20.jpg',200).'" />
';
?>
This basically is not possible. The webbrowser requests the HTML page and expects HTML. Or it requests an image and expects an image. You cannot mix both in one request, just because only one Content-Type can be valid.
However, you can embed the image in HTML using data URI:
<img src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAUAAAAFCAYAAACNbyblAAAAHElEQVQI12P4/8/w38GIAXDIBKE0DHxgljNBAAO9TXL0Y4OHwAAAABJRU5ErkJggg==" alt="Red dot">
Be aware that the base64 encoding is quite ineffective, so make sure you definitly compress your output, if the browser supports it, using for example gzip.
So for you it likely looks like the following:
echo '<img src="data:image/jpeg;base64,' . base64_encode(displayImg_w('http://www.sanctius.net/wp-content/uploads/2010/05/Avatar-20.jpg',200)) . '">';
And make sure displayimg_w() does not output a header anymore.
Furthermore, displayimg_w() needs to be adjusted at the end to return the image data as string rather than direct output:
ob_start();
imagejpeg($dst_im);
return ob_get_flush();