If I use php file as source to image, where:
$file = $_GET["file"];
$file_get = get_file_contents("from/".$file);
$fopen = fopen("to/".$file,"w+");
fwrite($fopen, $file_get);
fclose($fopen);
header("Location:to/".$file);
And if I use many images of that kind on one page, like:
<img src="image.php/?file=img.jpg>
<img src="image.php/?file=img2.jpg>
<img src="image.php/?file=img3.jpg>
...
I found that code in image.php doesn't run asynchronously. Images are downloaded one by one. How can I avoid it?
I see there some problems in your code. The first is that you have a big security whole when you use the $_GET input directly in your code to get an image.
The next one is why do you fetch the content from one file and write them to another file to redirect to them? That is not really fast if you write every time the file to another location.
If you get the content echt echo the content and set the correct header to show the image.
header('Content-type:image/png');
readfile($fullpath);
Its much easier and you have a less IO to show files. Otherwise you can use a script like PHPThumb which generated smaller versions and cache the files.
http://phpthumb.sourceforge.net/
Related
I am trying to display img_a.png or img_b.png based on the access level of the user (e.g. signed in or not). Of course the content (in this case img_a and img_b) should not be available to the general public.
I tried a few solutions, none of them helps me and that's why I look for help here. What I tried so far is:
Based on the user checks I tried adding the resources folder to "open_basedir" which is a lame option but was looking the easiest solution. All ended up by raising a warning that resource is not in the basedir folder through it's obviously there.
Attempted to put the resources in the public folder and restrict them via .htaccess. But in this case they got restricted not only for the unwanted audience but for everyone.
Final and closest attempt was to put back the images outside the webroot and write a class that validates the access and serves the image like:
class Image {
...
public function getImage() {
...
header('Content-Type: '.$this->type);
readfile($this->item);
which afterwards is displayed in the initial script through:
echo "<img src=".$image->getImage($file).">";
The problem above is that the headers were already sent so I could either stream the image or the html output of the .php page but not both. Do I have a way out there?
Create a script that checks whatever user attribute you want, determines what image to serve, and read/send that image. Use the script URL as the <img src=... attribute, ie
<img src='/scripts/user_image.php'>
Something like this will work for PNG images. Similar functions exist for GIF, JPG, etc.
<?php
// do stuff here to determine file name of image to send
if($_SESSION['userlevel']=="ADMIN"){
$imageFilename="admin_image.png";
}
// Create Image From Existing File
$image = imagecreatefrompng($imageFilename);
//Set the Content Type
header('Content-type: image/png');
// Send Image to Browser
imagepng($image);
// Clear Memory
imagedestroy($image);
exit;
?>
OK, per your comment, I think you are referencing things wrong.
My working script is exact as above, only the if-then is commented out. I'm just assigning the filename to the variable. I've named the script user_image.php.
A simple index.html file to reference the image -
<html>
<head><title>test</title></head>
<body>
Hello World!<br />
<img src="user_image.php">
</body>
</html>
And it Just Works. See it in action at https://itsjustcrap.com/img or grab source at https://itsjustcrap.com/img/src.zip
I'll leave it up for a few days, but other than a png image and a commented out if-then, the source is exact same as what is in this post.
Actually I think I found a solution to the headers confict I had.
Displaying the image as data like:
<img src="data:image/png;base64, <?=base64_encode(file_get_contents($file))?>" alt="img">
instead of feeding it inside the source attribute with readfile() php script like:
<img src="image.php?<?=$file?>" alt="my_image">
where the last one was giving me a headers conflict when inside html page.
Discovered the solution over here How to display Base64 images in HTML?
I am trying to read multiple image files from a folder (.htaccess protected) and display in a HTML page using php readfile().
The problem is I can see only the first image is read and the next is not shown in the browser. The code is as below
<?php
$image1 = 'files/com_download\256\50\www\res\icon\android\icon-36-ldpi.png';
$image2 = 'files/com_download\256\50\www\res\icon\android\icon-48-mdpi.png';
$imginfo = getimagesize($image1);
header("Content-type: ".$imginfo['mime']);
readfile($image1);
$imginfo = getimagesize($image2);
header("Content-type: ".$imginfo['mime']);
readfile($image2);
?>
I could see the first image 'icon-36-ldpi.png' successfully read and displayed in the browser and the second image is not read and not displayed in the browser.
Am I missing something? Any advice please.
Sorry if I am doing stupid but the requirement is to read multiple image files and render in the browser like a grid view. I cannot use img tag because of security reasons.
You can't dump both images out at once. Why not make two images in your html so the browser makes two calls to your script. Then use a GET param to pass the filename you want to display.
---Edit---
Important Security Note
There is an attack vector which you open up when doing soething like this. Someone could easily view your source html and change the parameter to get your image script to output any file they want. They could even use "../../" to go up directories and search for well known files that exist. e.g. "../../../wp_config.php". Now the attacker has your wordpress database credentials. The correct way to prevent against this is to always validate the input parameter properly. For example, only output if the file name ends with ".jpg"
I have a random image generator for my site. The problem is, it takes a really long time.. I was wondering if anybody could help to speed it up in any ways. The site is http://viralaftermath.com/, and this is the script:
header('Content-type: image/jpeg;');
$images = glob("images/" . '*.{jpg,jpeg,png,gif}', GLOB_BRACE);
echo file_get_contents($images[array_rand($images)]);
This is a pretty resource-intensive way to do this, as you are passing the image data through PHP and not specifying any caching headers, so the image has to be reloaded every single time you open the page.
A much better approach would be to have glob() list the files within the HTML page that you're using to embed the image. Then randomize that list, and emit an <img> tag pointing to the actual file name that you determined randomly.
When you are linking to a static image instead of the PHP script, you also likely benefit from the web server's caching defaults for static resources. (You could use PHP to send caching headers as well, but in this scenario it really makes the most sense to randomly point to static images.)
$images = glob("images/" . '*.{jpg,jpeg,png,gif}', GLOB_BRACE);
# Randomize order
shuffle ($images);
# Create URL
$url = "images/".basename($images[0]);
echo "<img src='$url'>";
Profile your code and find the bottlenecks. I can only make guesses.
echo file_get_contents($file);
This will first read the complete file into memory and then send it to the output buffer. It would be way nicer if the file goes directly into output buffer. readfile() is your friend. It would be even better to avoid buffering completely. ob_end_flush() will help you there.
A next candidate is the image directory. If searching for one image takes a significant time, you have to optimize that. This can be achieved by an index (e.g. with a database).
How can I display an image and pass it as an input parameter in an executable in php without saving the image in a folder. The user gives the image path as input and I am using ajax to display the image when it is selected when I save it to a folder it works but how can I display it without saving it in a folder? My code now is
move_uploaded_file($_FILES["file"]["tmp_name"],"upload/".$_FILES["file"]["name"]);
//echo "Stored in "."upload/".$_FILES["file"]["name"];
echo "<img src='upload/".$_FILES["file"]["name"]."' class='preview'>";
I tried
<img src=$_FILES["file"]["tmp_name"]. class='preview'>
but it didnt work. As I will have thousands of input from thousands of user I dont want to save it. Is there any optimised and efficient method to do this?
I think, its not possible to show image without saving it. You could try to save the image in temp folder on the server side and clean this folder periodically to avoid much space consumption.
The src attribute of the <img> tag should be an URL accessible by the client.
You try to give a local path (ex: path/to/your/file.jpg) of a temporary file as URL, it will not working.
info: The uploaded image is save on the local disk on a temp directory, and could be deleted by PHP later.
If you want to show the image without moving it at a place reacheable by a URL, you can try to load its content as base64 content
$imagedata = file_get_contents($_FILES["file"]["tmp_name"]);
$base64 = base64_encode($imagedata);
and use in your HTML
<img src="data:image/png;base64, <?php echo $base64; ?>" />
I don't think you can show the image without saving it.
You need to save the file either to the filesystem or to memory if you want to later output
Your problem here is that $_FILES only exists in the script that the image was sent to. so when you initiate another http request for img source, php no longer has any clue what file your trying to read.
You need a way to tell which image to be read on http request.
One thing you can do is that you can save the file in a place accessible by the client and then just have php delete it after you output it. So once the image is outputted it will be deleted and no longer be existing in the file system.
Another approach would be to get the image from the memory by directly writing the contents to httpresponse.
You can do this way
$image = file_get_contents($_FILES["file"]["tmp_name"]);
$enocoded_data = base64_encode($image);
and when you show your image tag :
<img src="data:image/png;base64, <?php echo $enocoded_data ; ?>" />
Hope any of these helps you
I would like to generate a dynamic image from a script, and then have it load to the browser without being persistent on the server.
However, I cannot call this by setting the image's src="script.php", since that would require running the script that just generated the page and its data all over again, just to get the final data that will generate the graph.
Is there a way to do this that is similar to setting image's src="script.php", but which is called from within another script, and just sends the image without saving it? I need access to the data that is used in the generation of the markup, in order to create this dynamic image.
Or, if not, what is the easiest way to destroy the image once the page is loaded? a quick ajax call?
Is there any way to cache certain data for some limited time frame in order for it to be available to some other script?
Any ideas would be greatly appreciated, as I'm having a really hard time finding the right solution to this...
Thanks!
You can inline the image into a <img> tag if you need to.
Like
<?php
$final_image_data; // Your image data, generated by GD
$base64_data = base64_encode($final_image_data);
echo "<img src=\"data:image/png;base64,{$base64_data}\" ... />";
?>
That should work on all modern browsers, and IE8. Doesn't work well with some email clients tho (Outlook, for one).
Also, another solution I found is to store the image in a session variable which is then called from a php script in the image tag. This would allow a user specific image to be served, and then removed from memory by the script... This also avoids messy img src="" tags...
Hopefully that is helpful to someone.
Use a rewrite rule.
RewriteRule ^magicimage.jpg$ /myscript.php
Then simply echo your image data from gd, instead of writing it to disk -- which is as simple as not providing a filename to the appropriate image*() function
myscript.php
<?php
$im = imagecreatetruecolor($w, $h);
//...do gd stuff...
header('Content-type: image/jpeg');
//this outputs the content directly to the browser
//without creating a temporary file or anything
imagejpeg($im);
And finally, utilize the above
display.php
<img src="magicimage.jpg">