Let's say I have a user enter the URL of an image.
After URL validation, etc. I want to get the image and store it in a PHP variable. Not the image path, but the actual image itself.
I am adding this image-holding variable in between other strings, so I cannot change use header() before echoing the image. Is it possible to use <img> HTML tags? I really don't want to copy the images to my own server...
How do I:
Store the image in a variable such that,
Echo the image from the variable without changing headers.
Edit:
I said above that I am putting this image inside another variable, e.g.:
$str = "blah blah" . $var_holding_img . "more text";
Is it possible to insert something in the string above that will be replaced with the images? Can parse the variable $str later to replace some random text like "abg30j-as" with the image...
I found an answer to my own question:
First, I created another PHP file, called img.php:
<?php
$url = $_GET['imgurl'];
/*
Conduct image verification, etc.
*/
$img_ext = get_ext($url); //Create function "get_ext()" that gets file extension
header('Content-type: image/' . $img_ext);
echo file_get_contents($url);
?>
Then, in the original PHP file, I used this PHP code:
<?php
$var_holding_img = '<img src="img.php?imgurl=http://example.com/image.png"/>';
$string = "This is an image:<br \>" . $var_holding_img . "<br \>displayed dynamically with PHP.";
echo $string;
?>
This way, the PHP file "img.php" can use the proper headers and the image can be inserted as HTML into any other PHP variable.
How do I:
Store the image in a variable such that,
Echo the image from the variable without changing headers.
You can do this in two ways.
In way one, you serialize the image in a string, and save the string in the session. Which is exactly the same as saving it server side, except that now the session GC should take care of clearing it for you. Then the IMG SRC you use will redirect to a script that takes the image and outputs it as image with proper MIME type.
In way two, if the image is small enough, you can encode it as BASE64 and output it into a specially crafted IMG tag:
http://www.sweeting.org/mark/blog/2005/07/12/base64-encoded-images-embedded-in-html
This saves you some time in connection, also. Of course the image must be reasonably small.
You can't save the actual image in a variable. Either you save the URL or copy the image (what you obvious don't want) to your server and save the path to the image
See answer 1, you can't echo the image itself, only link it
Edit: Okay obviously you can save images directly to a variable, but I don't recommend you to do this.
No, that isn't possible. If you want to serve something, it has to exist on the server.
Related
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'm having kind of controller http://samplesite.com/application/getimage written in php. It is a simple script that responds to GET with some parameters. According to these parameters I find an image in database (image is stored there as a BLOB) and I return it as follows:
header("Content-type: image/jpeg");
echo $image;
$image is a BLOB of image obtained from database.
When I open this URL http://samplesite.com/application/getimage/?id=200 my web browser displays an image.
Now I want to display this image inside a table in another php script. I want to perform it like
echo "<img src="http://mysite.com/application/getimage/?id=200" ...some params here... />";
but
this approach does not work. It displays no image or makes other problems.
What approach should I choose to make this work? I'm new to php normally work with Java EE. Thanks very much.
EDIT:
Of course there should be inside the echo \" and not only ". But this approach that I wrote here works fine. Problem was in syntax, I had one extra ).
Passing a PHP file AS an image (not recommended)
Your issue is that you haven't escaped speech marks, replace:
echo "<img src="http://mysite.com/application/getimage/?id=200" ...some params here... />";
With:
echo "<img src='http://mysite.com/application/getimage/?id=200' ...some params here... />";
or:
echo "<img src=\"http://mysite.com/application/getimage/?id=200\" ...some params here... />";
However, there is a far better way to do this:
Using image BLOBs properly
You're on the right track -- you can use an image BLOB instead of a file to put an image inline, but there's a little more work to it. First, you need to base64_encode the blob before you pass it. You should omit the header as you're passing text, not an image:
echo base64_encode($image);
Next, you need to set up the image tag like this, declaring that you are using data, the mime type and finally including the output of your getimage script as declared by RFC2557:
<img src="data:image/jpeg;base64,<?php include('http://mysite.com/application/getimage/?id=200'); ?>"/>
That should fix the issue.
Sending an image in my php form, it gets saved onto my server and at the
action.php page it gets displayed. Now when I try to:
echo '<div id="image"><img src="'.$target_path.'" width="280" height="280"></div>';
it works just fine... but if I add unlink($target_path); at the end of my php code it
will not even display the image even though it gets deleted AFTER displaying the image...
So the question is, how can I display the image and deleting it at the same time so my server does not gets stuffed with user pictures?
Try another thing: output the image base-64 encoded.
$contents = file_get_contents($file);
$base64 = base64_encode($contents);
echo '<div id="image"><img src="data:image/jpg;base64,'.$base64.'" width="280" height="280"></div>';
(instead of move_uploaded_file() etc, use as $file variable the $_FILES[...]['tmp_name'])
You can achieve this by creating a little script that gets the image-filename and will delete it after it has been retrieved:
<?php
$file = image_file_from_parameter($_GET['image']);
headers_for_file($file);
readfile($file);
unlink($file);
In your HTML output you then link to that script:
<img src="path/to/image.php?image=893sudfD983D" />
If you set nice caching headers, the user won't notice that your server did serve the file only once.
When you echo the url of an image with img src you're just sending the browser the url of an image, not the actual image data. The image needs to remain on the server if you want it to be viewable by this approach.
You could use bwoebi's solution to pass the actual image data instead of a link, but a better solution is just to keep the images on the server and periodically delete old files.
How can I display an image and pass it as an input parameter in an executable in php without saving the image in a folder. The user gives the image path as input and I am using ajax to display the image when it is selected when I save it to a folder it works but how can I display it without saving it in a folder? My code now is
move_uploaded_file($_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 $enocoded_data ; ?>" />
Hope any of these helps you
I need to send a GET request to my page pic.php, and I want to get a real picture in return.
For now I implemented this idea like this:
<?php
if ((isset($_GET['pic']))&&(isset($_GET['key']))){
$pic = $_GET['pic'];
$pic=stripslashes($pic);
header('Location: /content/'.$pic);
}
?>
But it's not really what I want - it redirects to image directly. What I want is to keep the same URL, but get a needed file depending on what values were submitted.
What is the best way to do that?
thx.
This example code snippet should do what you ask. I've also included code to only strip slashes if magic quotes is enabled on the server. This will make your code more portable, and compatible with future versions of PHP. I also added use of getimagesize() to detect the MIME type so that you output the proper headers for the image, and do not have to assume it is of a specific type.
<?php
if(isset($_GET['pic']))
{
//Only strip slashes if magic quotes is enabled.
$pic = (get_magic_quotes_gpc()) ? stripslashes($_GET['pic']) : $_GET['pic'];
//Change this to the correct path for your file on the server.
$pic = '/your/path/to/real/image/location/'.$pic;
//This will get info about the image, including the mime type.
//The function is called getimagesize(), which is misleading
//because it does much more than that.
$size = getimagesize($pic);
//Now that you know the mime type, include it in the header.
header('Content-type: '.$size['mime']);
//Read the image and send it directly to the output.
readfile($pic);
}
?>
I can see you doing this in two ways:
1) Return the URL to the image, and print out an image tag:
print '<img src=' . $img_url . ' />';
2) Alternatively, you could just pull the data for the image, and display it. For instance, set the header appropriately, and then just print the image data.
header("content-type: image/png");
print $img_data;
This assumes that you have the image data stored in a string $img_data. This method will also prevent you from displaying other things on the page. You can only display the image.
You can load the image, send the image headers, and display the image as such:
header('Content-Type: image/jpeg');
readfile('/path/to/content/pic.jpg');
Obviously the headers would depend on the filetype, but that's easy to make dynamic.
Not sure if I understand what you're after, but guessing that you want to load the picture in an img tag?
If I'm right you just do:
<img src=http://www.domain.com/pic.php?"<?php echo image here ?>" />
Basically you just make the source of the image the webpage you get directed to where the image is.