I want code that loads an image to a PHP server and then send it to browser.
For example I want sample.php to send an image to browser once it is requested.
in other words, I want to create a PHP file that acts like a proxy for an image.
why are you doing this?
why don't deliver the image directly?
if you are trying to display a random image you may as well just redirect to the image using
header("Location: address-of-image");
for delivering the file to your clients from your server and not from its original location you can just do. however your php.ini settings need to allow external file opens
readfile("http://www.example.com/image.jpg")
correct headers are not required if you are going to display the image in an img tag,
altough i would recommend it. you should check the filetype of the image or in most cases just set an octet-stream header so the browser doesnt assume an incorrect type like text or something and tries to display binary data.
to do so just do
header("Content-type: application/octet-stream")
one more thing to consider may be setting correct headers for caching...
You need to use
$image = fopen("image.png");
Modify the headers(not sure exacly if it's correct)
headers("Content-type: image/png");
And then send the image
echo fread($image, file_size("image.png"));
Related
I am redirecting to an image with a Location header from PHP, and in firefox when you view the image and right click to save it prompts to save with the name of the PHP redirect script, not the name of the image. This behaviour is not present in the other browsers.
Here is the code of the file:
<?php
header("Location: foo.jpg");
Is there anyway to get firefox to use the correct name when a user opens the save dialog?
jewlhuq's suggestion to bypass php altogether works.
<?php print("<script>window.location='image.jpg';</script>"); ?>
Using php to read the file's contents and dump those to the browser with the proper headers including the following
header('Content-Disposition: inline; filename="desired-filename.jpg"');
also works.
Which is better depends on your application. I used the first for the problem listed above, in another application I needed to serve an image with a different file name than the one it is actually saved with, for that I needed the latter.
That question might look silly but i would appreciate if i get a good answer.
I know what http header is and we can change it using header function in php.
Suppose i have a php file an_image.php and the code of it is as below :
<?php header('Content-type:image/jpeg'); ?>
<img src="image/flower.jpg">
Why am i getting a broken icon? By changing the header content type am i not changing the output as image?
As i think img tag is still an html output so as i'm trying to set an html content into an image content so i get the broken icon.
So what is the use of content-type:image/jpeg and where can it be used?
For example flower.jpg picture is in my image folder. If i create flower.php
and open the flower.jpg using a text editor and copy the code of it and paste it on flower.php and set the header content-type:image/jpeg and try to open it on browser it doesn't work saying syntax error.
Looking for a good explanation .
The correct content type for what you're outputting is text/html. You'd use the image/jpeg content type only if you were outputting the actual image file's contents itself.
<?php
header('Content-Type: image/jpeg');
readfile('image/flower.jpg');
Common uses include having a PHP script output a protected file after verifying a user's permissions allow it to be accessed, tracking pixels (save some data then serve a 1x1 image, for example), and serving dynamically generated images.
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 want to have a PNG picture, but when accessing it, it runs a PHP script, the PHP script should decide what picture to send (using some if statements and whatever). Then the PHP script should read the image file from somewhere on my web server and output it.
Here is the issue, if I get a .png file, and put PHP code in it, it won't work, however, if I use the .php extension, it works, and I can even embed the image into other websites, and the PHP can decide what image to send, but if I want to view that image directly (copy it's URL into my address bar) it doesn't work, it gives me the images plain contents (random jibberish).
Anyone know what to do?
Also This is my first question on Stack Overflow - please tell me if I am doing something wrong.
You need to send Content-Type headers.
For png:
header('Content-Type: image/png');
For others change png to jpg or gif or bmp or whatever.
Please note that header() function must be used before anything is written to output.
First, make sure you have your image image.png somewhere accessible to php.
Then create a php script image.php:
<?php
header('Content-Type: image/png');
readfile('image.png');
The script now acts like it was a PNG image.
It sounds like you know how to send the image, your issue is that you want the URL to look like it's a PNG image.
There are a couple of things you can do. First, if your web server supports URL rewriting (like Apache's mod_rewrite module), you can use a rewrite rule so that the user access the script as something like http://example.com/generated_image.png but your server will translate/rewrite this URL to point directly to your PHP script, so something like /var/www/image_generator.php.
Another option would be to actually name your script "generated_image.png" but force your webserver to treat it like a PHP script. For instance, in Apache you could try something like:
<Location /generated_image.png>
ForceType application/x-httpd-php
</Location>
As a final note, if you're not actually worried about the URL, but worried about the file name that is used if the user decides to save it to disk, you can simply use the Content-Disposition HTTP header in your response. In PHP it would look something like this:
<?php
header("Content-Disposition: inline; filename="generated_image.png");
?>
With that, it doesn't matter what the URL is, if the user saves the image through their web browser, the web browser should offer "generated_image.png" as the default filename.
Simplest version I know...
<?php
header('Content-Type: image/png');
if(whatever)
{
$image=your_image_select_function();
}
// as suggested by sh1ftst0rm with correction of unmatched quotes.
header('Content-Disposition: inline; filename="'.$your_name_variable.'"');
readfile($image);
?>
Then, you treat it like an image file. That is, if this is "pngmaker.php" then, in your HTML document, you do
<img src="pngmaker.php">
You can even do
<img src="pngmaker.php/?id=123&user=me">
I am trying to display an image from a MySQL blob field. I have tried a few different things and none of them seem to work.
I have tried:
header("Content-type: $type"); img src = $blobData;
header("Content-type: $type"); echo($blobData);
<?php
header("Content-type: $type");
echo $blobData;
?>
This code looks perfectly OK. However, I heard a similar complain from another person and I was able to troubleshoot it by assuring that:
The php script does not output any extra character before or after sending the binary image data.
The php script is saved as a pure ASCII text file, not as a Unicode/UTF-8 encoded file. The Unicode/UTF-8 encoded PHP files might include a signature as the first bytes. These bytes will be invisible in your text editor but server will send these few extra bytes to the browser before the JPEG/GIF/PNG data. The browser will therefore find the wrong signature in the beginning of data. To workaround, create a blank text file in notepad, paste in the php code and save the file in ANSI encoding.
Another option you might consider (assuming you are on Apache):
Create an .htaccess file with a mod_rewrite for all image extensions (png, jpg, gif).
Have it redirect to a php script that looks up the image requested in the DB. If it is there, it echos out the header and BLOG. If it isn't there, it returns a standard 404.
This way you can have:
<img src="adorablepuppy.jpg" />
Which then gets redirected ala:
RewriteEngine on
RewriteRule \.(gif|jpg|png)$ imagelookup.php
This script does a query for the image, which (obviously) assumes that the requested image has a unique key that matches the filename in the URL:
$url = $_SERVER['REQUEST_URI'];
$url_parts = explode("/", $url);
$image_name = array_pop($url_parts);
Now you have just the image filename. Do the query (which I shall leave up to you, along with any validation methods and checks for real files at the address, etc.).
If it comes up with results:
header('Content-type: image/jpeg');
header('Content-Disposition: inline; filename="adorablepuppy.jpg"');
print($image_blog);
otherwise:
header("HTTP/1.0 404 Not Found");
FYI: I have no idea if this would be bad in terms of performance. But it would allow you to do what I think you want, which is output the image as though it were a flat image file on the server using a simple image element. I'm inclined to agree that BLOBs are not the best way to go, but this does avoid any cross-browser issues.
I believe that the issue that you are encountering is an issue with encoding. This resource claims that you can use the print function.
Just get the image from the database. And print it using the correct headers.
$image = mysql_fetch_array(...)
header("Content-type: image/jpeg"); // change it to the right extension
print $image['data'];
For performance reasons... this is not advisable. There are several reasons to put images in databases but the most common are:
a) keeping them indexed (duh!)
You can do this by storing the images flat on the server and just indexing the image filename.
b) keeping the image hidden/protected
Flickr and alike still store the images flat on the server and use a different approach. They generate a URL thats hard to find.
This link points to a protected image on my account. You can still access it once you know the correct URL. Try it!
farm2.static - a farm optimized for delivering static content
1399 - perhaps the server
862145282 - my username
bf83f25865_b - the image
In order to find all my secret images any user can hard hit Flickr with the above address and change the last part. But it would take ages and the user would probably be blocked for hammering the server with thousands of 404s.
That said there is little reason to store images on BLOBs.
Edit:Just a link pointing to someone that explained much better than I did why BLOB is not the way to go when storing images.