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
Related
Here my problem, I do a query to MySQL (PDO) for give me the last 5 URLs of a table nammed avatar who contains the ID and the URL :
$response = $dbh->query("SELECT url FROM avatar ORDER BY id_URL DESC LIMIT 0,5 ");
And I did :
while ($donnees = $response->fetch())
{
$urlImage = $donnees['url']; //'url' contains the URL
$result = file_get_contents($urlImage);
header('Content-Type: image/png');
echo $result;
?>
But the header just return a small empty white square. However, the "$result = file_get_contents($urlImage);" takes properly the URL because when I do :
$urlImage = $donnees['url']; //'url' contains the URL
$result = file_get_contents($urlImage);
echo $result;
?>
It just shows the "encodage of the image" (a ton of special characters) but not displays the image.
I also try with "imagick" but it says to me that the class doesn't exist and I don't think that the imagecreatefrompng can be use with URL.
Thanks !
Can you try this and see if it works?
$image = file_get_contents($donnees['url']);
$finfo = new finfo(FILEINFO_MIME_TYPE);
header('content-type: ' . $finfo->buffer($image));
echo $image;
This is suppose to handle one Image. One php script can return one image. If you want to combine the images and render a big long image then probably you should look at http://image.intervention.io/
EDIT
What I understood after trying out the above code is that if you put file_get_contents before header then the raw characters are shown. However you put it after header then everything seems to be working
$image="http://www.hillspet.com/HillsPetUS/v1/portal/en/us/cat-care/images/HP_PCC_md_0130_cat53.jpg";
$filename = basename($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 $image;
Working fiddle
If you're trying to use a dynamic image source where your url is the image source, and it isn't working, then the problem might be that there's a space or extra character somewhere on the page, which will make the browser treat it like a document instead of an image in some cases.
Your problem is that the browser isn't understanding that it's supposed to be an image.
You could always do:
<img src="<?=$urlImage?>">
The php file that handles the images I display only allows one image format, either .jpg, .png, .bmp, etc but not all. The imageName stores the file name of the image stored in the database including its format. this is my code, so far it doesn't work yet and I'm not sure if that's allowed. Can you help me fix it please?
$con = mysqli_connect("localhost","root","","tickets");
$ticket = 109;
$result = mysqli_query($con,"SELECT image, imageName FROM tix WHERE tktNum=$ticket");
while($row = mysqli_fetch_array($result))
{
$image = $row['image'];
$imageName = $row['imageName'];
$format = substr( $imageName, -3 ); //gets the last 3 chars of the file name, ex: "photo1.png" gets the ".png" part
header('content-type: image/' . $format);
}
I just used this, just indicating it's an image but without specifying image type.
header("Content-type: image");
It seems to be working just fine regardless of file type. I tested with IE, Firefox, Safari and the Android browser.
The solution is to read in the file and decide which kind of image it is and basednd on it send out the appropriate header.
$filename = basename($file);
$file_extension = strtolower(substr(strrchr($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;
case "svg": $ctype="image/svg+xml"; break;
default:
}
header('Content-type: ' . $ctype);
Following code:
class FilesController extends Controller {
public function icon($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.
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();