I have created a custom product configurator that saves a canvas element as a base64 encoded image. When I echo the image in the browser as the image src it works fine.
So something like this works:
$base64Image = $_POST['dataUrl'];
echo '<img src="'.$base64Image.'" />';
My problem is that codeigniter wont add this base64 image src to the session, probably because its too big. I have tried some methods that people have used to write an image to the server and they all throw errors. Does anyone know how I can write this base64 string to an image on the server like 'myimage.png' in the images/custom folder?
Any help is appreciated.
You can also use
$decoded=base64_decode($base64Image);
file_put_contents('newImage.JPG',$decoded);
Reference Link: http://j-query.blogspot.in/2011/02/save-base64-encoded-canvas-image-to-png.html
Use tempnam() to get a unique file name in a directory writable by the script. You can map a session variable "thisUsersTempFile" to that file name.
Or you can store the association somewhere else if it is not temporary. If you need to clean up the tmp files, you would probably need to do that since I don't think you can hook the session destruction. You could poll for existing sessions and delete the tmpfiles associated with the sessions that were destroyed. You could use a cron job for that.
Related
I'm using GAE version 1.9.0 and I want to delete an image from the data storage and upload another image to its location. This is how I'm doing it right now.
unlink("gs://my_storage/images/test.jpg");
move_uploaded_file($_FILES['image']['tmp_name'],'gs://my_storage/images/test.jpg');
And then I want to get the Image serving URL of the latest uploaded image, and I do it like this.
$image_link = CloudStorageTools::getImageServingUrl("gs://my_storage/images/test.jpg");
The issue is, when the name of the deleted image("test.jpg") and the uploaded image("test.jpg") is the same, the old file is served when I call for the newly uploaded file(I think it is cached.)
Is there anyway I can permanently delete this file without caching it?
You should probably delete the original serving URL before creating another with the same name.
There's a deleteImageServingUrl() method in CloudStorageTools that you can use to do this.
Here it is how to do in php laravel.
$object = $post_media->media_cloud;
$objectname = substr($object,48,100);
$bucket = Storage::disk('gcs')->delete($objectname);
Here in $object i get google cloud image url from db
Then we take only object name from that url, by substr.
Since you have given in your config Storage class as Storage::disk('gcs')
so this will call the function delete by taking the objectname.
Hope it helps anyone.
Note : For multiple images either pass an array of objects, or repeat it foreach loop.
How can I display an image and pass it as an input parameter in an executable in php without saving the image in a folder. The user gives the image path as input and I am using ajax to display the image when it is selected when I save it to a folder it works but how can I display it without saving it in a folder? My code now is
move_uploaded_file($_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 $enocoded_data ; ?>" />
Hope any of these helps you
So im making a website with an image upload functionality and im storing the image name to the database. I took a screenshot of my mac and wanted to upload this photo "Screen shot 2011-02-18 at 6.52.20 PM.png". Well, thats not a nice name to store in mysql! How do people ususally rename photos in such a way that each photo uploaded has a unique name? Also, how would i make sure i keep the file extension in the end when renaming the photo.
I would drop the extension, otherwise Apache (or equivalent) will run a1e99398da6cf1faa3f9a196382f1fadc7bb32fb7.php if requested (which may contain malicious PHP). I would also upload it to above the docroot.
If you need to to make the image accessible above the docroot, you can store a safe copy that is ran through image functions or serve it from some PHP with header('Content-Type: image/jpeg') for example and readfile() (not include because I can embed PHP in a GIF file).
Also, pathinfo($path, PATHINFO_EXTENSION) is the best way to get an extension.
Ensure you have stored a reference to this file with original filename and other meta data in a database.
function getUniqueName($originalFilename) {
return sha1(microtime() . $_SERVER['REMOTE_ADDR'] . $originalFilename);
}
The only way this can generate a duplicate is if one user with the same IP uploads the same filename more than once within a microsecond.
Alternatively, you could just use the basename($_FILES['upload']['tmp_name']) that PHP assigns when you upload an image. I would say it should be unique.
Hash the image name. Could be md5, sha1 or even a unix timestamp.
Here is an (untested) example with a random number (10 to 99)
<?php
function generate_unique_name($file_name)
{
$splitted = split(".", $file_name);
return time() . rand(10,99) . "." . $splitted[count($splitted)-1];
}
?>
You could use an image table like:
id: int
filename: varchar
hash: varchar
format: enum('jpeg', 'png')
The hash can be something like sha1_file($uploaded_file) and used to make sure duplicate images aren't uploaded. (So you could have multiple entries in the image table with the same hash, if you wanted.) The id is useful so you can have integer foreign key links back to the image table.
Next store the images in either:
/image/$id.$format
or
/image/$hash.$format
The second format via the hash would make sure you don't duplicate image data. If you are dealing with lots of images, you may want to do something like:
/image/a/b/c/abcdef12345.jpg
where you use multiple layers of folders to store the images. Many file systems get slowed down with too many files in a single directory.
Now you can link to those files directly, or set up a URL like:
/image/$id/$filename
For example:
/image/12347/foo.jpg
The foo.jpg comes from whatever the user uploaded. It is actually ignored because you look up via the id. However, it makes the image have a nice name if the person chooses to download it. (You may optionally validate that the image filename matches after you look up the id.)
The above path can be translated to image.php via Apache's MultiView or ModRewrite. Then you can readfile() or use X-SendFile (better performance, but not always available) to send the file to the user.
Note that if you don't have X-SendFile and don't want to process things through PHP, you could use a RewriteRule to convert /image/$hash/foo.jpg into /image/a/b/c/$hash.jpg.
I am trying to save PhpThumb output. As what I could find on-line was not sufficient or too complex, I would like to ask if any one knows how to it?
$thumb_src="\"phpThumb/phpThumb.php?src=../apartmentsPhotos/".$num['ref']."/1.JPG&h=119&q=100\"";
echo" '<'img src=".$thumb_src />";
So what I want to do is to save the img src into an Image.
(So far I was creating the thumbnails on the fly but it seems that google and my web server don´t like it too much. Saving the thumbnails will ensure that in no time I will have all my thumbnails in real files and then I will use this function just for new content.)
From phpThumb's FAQ
The best way is to call phpThumb as an object and call RenderToFile() to save the
thumbnail to whatever filename you want. See /demo/phpThumb.demo.object.php for an example. The other way is to use the 'file' parameter (see /docs/phpthumb.readme.txt) but this parameter is deprecated and does not work in phpThumb v1.7.5 and newer.
Once you have generated the URL with this line you posted:
$thumb_src="\"phpThumb/phpThumb.php?src=../apartmentsPhotos/".$num['ref']."/1.JPG&h=119&q=100\"";
Pass it as a $_GET variable to another page, call it serveThumb.php:
if (!isset($_GET['img']))
exit;
header('Content-type: application/pdf');
echo file_get_contents($_GET['img']);
You might have to add your own validation to serveThumb.php. Now you can save the result of serveThumb.php as a JPG.
Alternatively, save the contents of the image as a JPG file.
if (!isset($_GET['img']))
exit;
$img = file_get_contents($_GET['img']);
file_put_contents("myImage.jpg", $img);
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">