Well, I've uploaded an app to Heroku, and I've discovered that I can't upload files to it. Then I started to use Dropbox as storage option, and I've done a few tests, of send and retrieve link, and all worked fine.
Now, the problem is to use the uploadFile() method on DropboxAdapter. He accepts an resource as the file, and I did'nt work well. I've done a few tests, and still no way. Here is what I am doing, if anyone could me point a solution, or a direction to this problem, please. :)
Here is my actual code for the update user (Update the user image, and get the link to the file).
$input = $_FILES['picture'];
$inputName = $input['name'];
$image = imagecreatefromstring(file_get_contents($_FILES['picture']['tmp_name']));
Storage::disk('dropbox')->putStream('/avatars/' . $inputName, $image);
// $data = Storage::disk('dropbox')->getLink('/avatars/' . $inputName);
return dd($image);
In some tests, using fopen() into a file on the disk, and doing the same process, I've noticed this:
This is when I've used fopen() on a file stored on the public folder
http://i.imgur.com/07ZiZD5.png
And this, when i've die(var_dump()) the $image that I've tried to create. (Which is a suggestion from this two links: PHP temporary file upload not valid Image resource, Dropbox uploading within script.
http://i.imgur.com/pSv6l1k.png
Any Idea?
Try a simple fopen on the uploaded file:
$image = fopen($_FILES['picture']['tmp_name'], 'r');
https://www.php.net/manual/en/function.fopen.php
You don't need an image stream but just a filestream, which fopen provides.
Related
I am trying to download a file that I stored on S3 to my local Laravel installation to manipulate it. Would appreciate some help.
I have the config data set up correctly because I am able to upload it without any trouble. I am saving it in S3 with following pattern "user->id / media->id.mp3" --> note the fact that I am not just dumping files on S3, I am saving them in directories.
After successfully uploading the file to S3 I update the save path in my DB to show "user->id / media->id.mp3", not some long public url (is that wrong)?
When I later go back to try and download the file I am getting a FileNotFoundException at S3. I'm doing this.
$audio = Storage::disk('s3')->get($media->location);
The weird thing is that in the exception it shows the resource that it cannot fetch but when I place that same url in a browser it displays the file without any trouble at all. Why can't the file system get the file?
I have tried to do a "has" check before the "get" and the has check comes up false.
Do I need to save the full public URL in the database for this to work? I tried that and it didn't help. I feel like I am missing something very simple and it is making me crazy!!
Late answer but important for others,
$s3_file = Storage::disk('s3')->get(request()->file);
$s3 = Storage::disk('public');
$s3->put("./file_name.tif", $s3_file);
The response of $s3_file will be a stream, you can save that stream data to file using Laravel put file method, you will find this stream file in storage/public directory.
You can give your Content-Type as desired and Content-Disposition as 'attachment' because your files are coming from S3 and you have to download it as an attachment.
$event_data = $this->ticket->where('user_id', $user_id)->first();
$data = $event_data->pdf;
$get_ticket = 'tickets/'. $data;
$file_name = "YOUR_DESIRED_NAME.pdf";
$headers = [
'Content-Type' => 'application/pdf',
'Content-Disposition' => 'attachment; filename="'. $file_name .'"',
];
return \Response::make(Storage::disk('s3')->get($get_ticket), 200, $headers);
Say, you have AWS S3 as your default storage.
And you want to download my_file.txt from S3 to my_laravel_project\storage\app\my_file.txt
And you want to make it a one-liner
Storage::disk('local')->put('my_file.txt', Storage::get('my_file.txt'));
I am using plupload to upload file in my php based website, with large file uploading the file becomes a file named 'blob' without any suffix. I know this is a binary file that contains the raw data, question is how to retrieve the data and save it back as an image file, say .png/.jpg or etc? I tried:
$imageString = file_get_contents($blogPath);
$image = imagecreatefromstring($imageString);
But it gives me some 'Data is not in recognized format...' error, any thoughts? Thanks in advance.
Your call to imagecreatefromstring() should work just fine if your file_get_contents() is working. Use var_dump($imageString) to verify. Did you mean to name your variable $blobPath instead of $blogPath?
You don't need to load this image though. Just rename the file.
rename($blobPath, 'new/path/here.jpg');
http://php.net/manual/en/function.rename.php
I am storing the uploaded image files for late use, like attaching them to posts or products(my site is e-commerce CMS). I figured that my image file didn't get fully uploaded to the server, the image before upload is 6mb, but the blob file is just 192kb, so my best guess is that what get uploaded is just a chunk instead of the whole package, and yet that brought up another question: how should I take all the pieces and assemble them as one complete image file? As mentioned earlier, I am using plupload for js plugin and php as backend, the backend php code to handle uploading goes like this:
move_uploaded_file($_FILES["file"]["tmp_name"], $uploadFolder . $_FILES["file"]["name"]);
Instead of doing that you should do this to display image to the browser
<img src="data:image/jpeg;base64,'.base64_encode( $row['blob_image'] ).'"/>
I'm not sure what imagecreatefromsting does or how it encodes the image.
I looked at the documentation for that function; you're missing:
$data = 'iVBORw0KGgoAAAANSUhEUgAAABwAAAASCAMAAAB/2U7WAAAABl'
. 'BMVEUAAAD///+l2Z/dAAAASUlEQVR4XqWQUQoAIAxC2/0vXZDr'
. 'EX4IJTRkb7lobNUStXsB0jIXIAMSsQnWlsV+wULF4Avk9fLq2r'
. '8a5HSE35Q3eO2XP1A1wQkZSgETvDtKdQAAAABJRU5ErkJggg==';
$data = base64_decode($data); <--- this operation
I am doing products import php script for prestashop 1.3.1 and I have one problem. I have URL of picture, but i dont know hoe to use it and make different images (thumbnails it is called I think).
If I have picture http://www.nordix.cz/img/p/824-2268.jpg what I must write in PHP to make thubnails?
Thank you so much for tips!
To process an image (create thumbs) first you have to copy it to a local directory. You can't do any processing on an image which is on another server or url. So here is how i did it in one of my PS project.
1) First check if the image exists or not. You can do it by using fopen in read mode, if it returns true, then the file exists. It is a good practice to do it because it will avoid unnecessary calls to the remote server.
$imageUrl = "http://www.nordix.cz/img/p/824-2268.jpg";
#fopen($imageUrl, "r");
2) Now you have the image as the fopen returned true, you need to copy the image to the PS temp directory as below
$tmpName = tempnam(_PS_TMP_IMG_DIR_, 'PS');
copy($imageUrl , $tmpName);
after the copy function downloads the image to PS temp directory, then you can process that image as you want. Remember that you have to make all processing on $tempName, as it is the file now. $tempName is like $_FILES['imageFieldName']['tmp_name'].
Thank you
Prestashop provides a set of function to process images. I've never worked on 1.3 but in 1.4 they are located in /images.inc.php (they made a class in 1.5). Take a look at this file and you will find all the function you will need, especially imageResize()
I have moved my images to Rackspace Cloud Files and am using their PHP API. I am trying to do the following:
Get an image from my "originals" container
Resize it, sharpen it, etc.
Save the resized image to the "thumbs" container
My problem is with #2. I was hoping to resize without having to copy the original to my server first (since the images are large and I'd like to resize dynamically), but can't figure out how. This is what I have so far (not much):
$container = $conn->get_container("originals");
$obj = $container->get_object("example.jpg");
$img = $obj->read();
Part of the problem is I don't fully understand what is being returned by the read() function. I know $img contains the object's "data" (which I was able to print out as gibberish), but it is neither a file nor a url nor an image resource, so I don't know how to deal with it. Is it possible to convert $img into an image resource somehow? I tried imagecreatefromjpeg($img) but that didn't work.
Thanks!
First, you cannot resize an image without loading it into memory. Unless the remote server offers some "resize my image for me, here are the parameters" API, you have to load the image in your script to manipulate it. So you'll have to copy the file from the CloudFiles container to your server, manipulate it, then send it back into storage.
The data you receive from $obj->read() is the image data. That is the file. It doesn't have a file name and it's not saved on the hard disk, but it is the entire file. To load this into gd to manipulate it, you can use imagecreatefromstring. That's analogous to using, for example, imagecreatefrompng, only that imagecreatefrompng wants to read a file from the file system by itself, while imagecreatefromstring just accepts the data that you have already loaded into memory.
You can try to dump the content of the $img variable into a writable file as per the below:
<?php
$filename = 'modifiedImage.jpg';
/*
* 'w+' Open for reading and writing; place the file pointer at the beginning of the file and truncate
* the file to zero length. If the file does not exist, attempt to create it.
*/
$handle = fopen($filename, 'w+');
// Write $img to the opened\created file.
if (fwrite($handle, $img) === FALSE) {
echo "Cannot write to file ($filename)";
exit;
}
echo "Success, wrote to file ($filename)";
fclose($handle);
?>
More details:
http://www.php.net/manual/en/function.fopen.php
http://www.php.net/manual/en/function.fwrite.php
Edit:
You might also want to double check the type of data returned by the read() function, because if the data is not a jpg image, if it's for example a png, the extension of the file needs to be changed accordingly.
I'm writing a web app that at one point allows a user to upload a photo to a flickr account (mine). I want to do this without saving the intermediate image on the server my web app is on.
What I've got so far is a page which implements phpFlickr and accepts a POST from a simple html form. I use $_FILES['file']['tmp_name'] as the path for phpFlickr to use. Here's the code:
<?php
require_once("phpFlickr.php");
$f = new phpFlickr("apikey", "secret", true);
$_SESSION['phpFlickr_auth_redirect'] = "post_upload.php";
$myPerms = $f->auth("write");
$token = $f->auth_checkToken();
$phid = $f->sync_upload($_FILES['file']['tmp_name']);
echo "Uploading Photo..." . $phid;
?>
I'm guessing that the tmp file is being lost because of the redirect that happens when $f->auth("write") is called, but I don't know. Is there a way to preserve it? Is there any way to do this without saving the file to the server?
Answer: There is No way to directly upload a file to Flickr without saving it as an intermediate file.
I've moved on to using move_uploaded_file() followed by a flickr API call, and its working perfectly.
I've also managed to get it to play nice with the excellent Jquery Uploadify, which lets me send multiple files to it in one go.