I need to save multiple pictures with Laravel. The pictures are stored separately on disk and the links to the places are stored in a JSON file in the database. So each users has an column in the database with:
{"images": ["/user/57/house-11-1.png", "/use/57/house-12-2.png"]}
So when the user uploads a file and click on save this code happens:
$path = $this->processImage($request, $user->id, $house->id);
/* uploads the image to the server and pass path of the image*/
if ($path) {
$jsonstring = $house->images;
dd($jsonstring);
$arr = json_decode($jsonstring);
$arr['images'] = [$path];
$json = json_encode($arr);
dd($json);
$house->images = $json;
$house->save();
}
The laravel model that is used is called userHouse and saves images as:
class userHouse extends Model implements Changeable
{
protected $casts = [
'images' => 'array',
];
...
}
processImage function:
private function processImage($request, $userID, $houseId)
{
$path = null;
$number = rand(1, 99);
if ($request->hasFile('image')) {
$image = Image::make($request->file('image'))
->resize(750, null, function ($constraint) {
$constraint->aspectRatio();
})
->encode('png');
$path = "/vendors/{$userID}/horse-{$houseId}-{$number}.png";
Storage::disk('fileadmin')->put($path, $image->encoded);
}
return $path;
}
the error I get is:
[2019-05-27 08:01:15] local.ERROR: Serialization of 'Illuminate\Http\UploadedFile' is not allowed {"userId":57,"email":"info#test.com","exception":"[object] (Exception(code: 0): Serialization of 'Illuminate\\Http\\UploadedFile' is not allowed at /Users/dsfsdf/ah-website/user/laravel/framework/src/Illuminate/Session/Store.php:128)
[stacktrace]
How can I add the url to the JSON string in the database? Uploading single images works.
What I could think of is that the processImage method cannot handle multiple uploaded files in the request.
Also, IMO, this method should actually receive the UploadedFile object instead of the entire request.
So, the code would be similar to:
$images = $house->images;
foreach ($request->images as $uploadedFile) {
$images[] = $this->processImage($uploadedFile, $user->id, $house->id);
}
$house->images = $images;
$house->save();
Also, if you defined the images attribute as array in the model, there is no need to decode/encode it as laravel already does it.
Related
I'm trying to return query data and storage Images from laravel controller. My code looks like following :
class ClaimController extends Controller{
.....
public function show(Claim $claim)
{
$front_image = Storage::disk('do_spaces')->get($claim->images->front_image); //image file
$back_image = Storage::disk('do_spaces')->get($claim->images->back_image); //image
// return $front_image [works]
$claim = Claim::all();
//this throws error
return response()->json([
'claim' => $claim,
'images' => [$front_image, $back_image]
]);
}
}
Now I far I understand return->response()->json([]) doesn't send image file. How I can return all data together to frontend app ?
you have the option to return the images from original server which they are stored in or you can return an encoded version of the image as string and in front end reconstruct it.
$path = 'myfolder/myimage.png';
$type = pathinfo($path, PATHINFO_EXTENSION);
$data = file_get_contents($path);
$base64 = 'data:image/' . $type . ';base64,' . base64_encode($data);
And then use it in front like this :
var image = new Image();
image.src = 'data:image/png;base64,iVBORw0K...';
document.body.appendChild(image);
This is the link for second part.
You can use Storage::url()
$st=\Illuminate\Support\Facades\Storage::disk('do_spaces')->url($claim->images->front_image);
dd(url($st));
Storage::url() return image path and then url() will return full url path
"http://127.0.0.1:8000/storage/imagefolder/filename"
Make sure to symbolic link to storage folder by running following command
php artisan storage:link
Also you can do the following without storage url
asset('storage/'.$claim->images->front_image)
I need to store images in a backend for logged in users. The stored images need to be protected and not visible from the outside (public). I choosed a "storage" folder for this.
I came up with this in my Controller:
public function update(Request $request, $id)
{
//Show the image
echo '<img src="'.$_POST['img_val'].'" />';
//Get the base-64 string from data
$filteredData=substr($_POST['img_val'], strpos($_POST['img_val'], ",")+1);
//Decode the string
$unencodedData=base64_decode($filteredData);
//Save the image
$storagepath = storage_path('app/images/users/' . Auth::user()->id);
$imgoutput = file_put_contents($storagepath.'/flyer.png', $unencodedData);
return view('backend.flyers.index')->withImgoutput($imgoutput)
//->withStoragepath($storagepath);
}
after hitting the save button, which triggers the update() I am able to see the image in my view, and it is also stored in my folder (current users=10) "storage/app/images/users/10/flyer.png"
my question is how can I access the image path?
I want to show the stored image with img src="">. I have no idea what to put inside "src= ..."
While dealing with user file uploads in web applications, the major aspect is about user's content's security.
One should use secure way to upload private files of a user in web applications.
As in your case, you want to access user's image outside public folder.
This can be done in a most secure way as given below.
First of all create a directory right in the root directory of Laravel (where the public folder is located), let the directory's name be uploads. Use this directory to upload private user files.
In the case of images create an another directory inside uploads as uploads/images/ inside uploads directory so that you can have a different storage locations for different type of files.
Remember to upload the image in images directory with a different name and without their extensions so that it looks like a extension-less file.
Keep the file name and its extension in the database which can be used later to retain image's location.
Now you need to create a separate route to show user's image.
Route::get('users/{id}/profile_photo', 'PhotosController#showProfilePhoto')->name('users.showProfilePhoto');
PhotosController.php
class PhotosController extends Controller {
private $image_cache_expires = "Sat, 01 Jan 2050 00:00:00 GMT";
public function showProfilePhoto($id) {
$user = User::find($id);
$path = base_path() . '/uploads/images/';
if($user && $user->photo) // Column where user's photo name is stored in DB
{
$photo_path = $path . $user->photo; // eg: "file_name"
$photo_mime_type = $user->photo_mime_type; // eg: "image/jpeg"
$response = response()->make(File::get($photo_path));
$response->header("Content-Type", $photo_mime_type);
$response->header("Expires", $this->image_cache_expires);
return $response;
}
abort("404");
}
}
The method above inside PhotosController - showProfilePhoto($user_id) will run as soon as you access the route named - users.showProfilePhoto.
Your HTML code will look like this.
<img src="<?php echo route('users.showProfilePhoto', array('id' => $user->id)); ?>" alt="Alter Text Here">
The above code will work like a charm and the image will be shown to the user without declaring/publishing the proper image path to public.
According to me this is the secure way to deal with file uploads in web applications.
You can do this like this:
Route::get('images/{filename}', function ($filename)
{
$path = storage_path() . '/' . $filename;
if(!File::exists($path)) abort(404);
$file = File::get($path);
$type = File::mimeType($path);
$response = Response::make($file, 200);
$response->header("Content-Type", $type);
return $response;
});
Reference:
Laravel 5 - How to access image uploaded in storage within View?
Or Alternatively you can use this library: https://github.com/thephpleague/glide
Just use composer to install it in your project
By default, this will render images from your storage, and allow you to do all sorts of things with it such as cropping, color correction etc.
Reference:
http://glide.thephpleague.com/
https://laracasts.com/discuss/channels/laravel/laravel-5-how-can-we-access-image-from-storage?page=1
Atimes you might have some images you do not wish to store in public directory for some various reasons.
Although storing your images has lots of advantages.
There are many ways you can achieve this, however I have this simple solution.
You should create a helper class like so if already don't have one
<?php namespace App\Services;
class Helper
{
public function imageToBase64($path)
{
$type = pathinfo($path, PATHINFO_EXTENSION);
$data = file_get_contents($path);
return 'data:image/' . $type . ';base64,' . base64_encode($data);
}
}
Then in your view (blade)
#inject('helper', 'App\Services\Helper')
<img width="200" height="250" src="{{$helper->imageToBase64(storage_path('app/images/users/' . Auth::user()->id)}}">
It will work 100% work. Open file filesystem in app/config/filesystem.php and write like that
'profile' => [
'driver' => 'profile',
'root' => '/home/folder/public_html/projectname/public/profiles',
],
Add this file at top
use Illuminate\Support\Facades\Storage;
My variable name is
$directoryName = 'profile';
$imageName = $request->image; // image is array of base64 encoded urls
$directory_path ='profiles';
Below function save your file in public/profiles folder.
function UploadImagesByBase64($directoryName, $imageName,$directory_path)
{
$data = array();
$image = $imageName;
foreach ($image as $image_64) {
if($image_64 !=null){
$extension = explode('/', explode(':', substr($image_64, 0, strpos($image_64, ';')))[1])[1]; // .jpg .png .pdf
$replace = substr($image_64, 0, strpos($image_64, ',')+1);
// find substring fro replace here eg: data:image/png;base64,
$image = str_replace($replace, '', $image_64);
$image = str_replace(' ', '+', $image);
$imageName = Str::random(10).time().'.'.$extension;
Storage::disk($directoryName)->put($imageName, base64_decode($image));
$data[] = $directory_path.'/'.$imageName;
}
}
$imageName = implode(',', $data);
return $imageName;
}
I'm facing following problem and can't seem to figure this one out.
I wrote an API endpoint accepting a POST with binary data (header: content-type:image/jpeg).
I know i can read out the raw string with file_get_content('php://input') or Laravel's $request->getContent().
PHP also has a function createimagefromstring($string) which also seems to read the string in correctly.
What i'd like to do is create an UploadedFile from this raw data , so that I can handle it with already written functions.
Is this possible?
Thank you in advance
I think I found it... Still curious if there are improvements that can be made..
$imgRaw = imagecreatefromstring( $request->getContent() );
if ($imgRaw !== false) {
imagejpeg($imgRaw, storage_path().'/tmp/tmp.jpg',100);
imagedestroy($imgRaw);
$file = new UploadedFile( storage_path().'/tmp/tmp.jpg', 'tmp.jpg', 'image/jpeg',null,null,true);
// DO STUFF WITH THE UploadedFile
}
You can try to use base64 encoding. Symfony have some nice stuff for this.
Your code will be smthng like this:
$base64Content = $request->request->get('base64Content'); // this is your post data
$yourFile = new UploadedBase64EncodedFile(new Base64EncodedFile($base64Content)); // this is an instance of UploadedFile
Hope it helps!
As per Laravel 8
Just follow the constructor:
* #param string $path The full temporary path to the file
* #param string $originalName The original file name of the uploaded file
* #param string|null $mimeType The type of the file as provided by PHP; null defaults to application/octet-stream
* #param int|null $error The error constant of the upload (one of PHP's UPLOAD_ERR_XXX constants); null defaults to UPLOAD_ERR_OK
* #param bool $test Whether the test mode is active
$file = new UploadedFile(
$pathIncludingFilename,
$fileName,
'image/jpeg',
null,
false
);
There is no need to manually create it, Symfony parses received $_FILES array for you. Http Request object has a FileBag property called $files with a get() method which returns an UploadedFile instance.
/** #var UploadedFile $file */
$file = $request->files->get('user-pictures-upload')[0];
$cmd = new UploadPictureCmd($file, $this->getUser()->getId());
Here is the example of generating images files using fzaninotto/faker in Symfony 4 Fixtures
class FileFixtures extends Fixture
{
private $faker;
private $parameterBag;
public function __construct(ParameterBagInterface $parameterBag)
{
$this->faker = Factory::create();
$this->parameterBag = $parameterBag;
}
public function load(ObjectManager $manager)
{
$tempFixturesPath = $this->parameterBag->get('kernel.project_dir') . '/tmp';
if (!file_exists($tempFixturesPath)) {
mkdir($tempFixturesPath);
}
$fileName = $this->faker->image($tempFixturesPath, 640, 480, 'cats', false, true);
$file = new UploadedFile($tempFixturesPath . '/' . $fileName, $fileName, 'image/jpeg', null, null, true);
//do something with $file
}
}
If it counts for anything, this is how I did it in Laravel 5.4. In my case, I wanted to be able to easily resize an image and be able to do something like this.
request()->file('image')->resize(250, 250)->store('path/to/storage');
This is what I did to the UploadedFile class.
Illuminate\Http\UploadedFile.php ~this file ships with the Laravel framework
public function resize($w, $h) {
$image = Intervention::make($this)->fit($w, $h)->save($this->getPathname());
$filesize = filesize($this->getPathname());
return new static(
$this->getPathname(),
$this->getClientOriginalName(),
$this->getClientMimeType(),
$filesize,
null,
false
);
}
Using Intervention, I resized the image that is stored in the /tmp/ folder when files are uploaded and then I saved it in the same place. Now all I do after that is create an instance of UploadedFile so that I can keep using Laravel's methods on request()->file('image'). Hope this helps.
I need to send an image to server via an ajax request and it gets through just fine
and in my controller I can just use $_FILES["image"] to do stuff to it.
But I need to validate the image before I save it.
And in the Yii this can be achieved by doing something like this
$file = CUploadedFile::getInstance($model,'image');
if($model->validated(array('image'))){
$model->image->saveAs(Yii::getPathOfAlias('webroot') . '/upload/user_thumb/' . $model->username.'.'.$model->photo->extensionName);
}
But the problem is I don't have a $model, all I have is $_FILES["image"], now what should I put instead of the $model???
is there any other way where I can validate and save files without creating a model and just by Using $_FILES["image"]?
thanks for this awesome community... :)
Exists many ways how you can do upload. I want offer to you one of them.
1.You need to create model for your images.
class Image extends CActiveRecord {
//method where need to specify validation rules
public function rules()
{
return [
['filename', 'length', 'max' => 40],
//other rules
];
}
//this function allow to upload file
public function doUpload($insName)
{
$file = CUploadedFile::getInstanceByName($insName);
if ($file) {
$file->saveAs(Yii::getPathOfAlias('webroot').'/upload/user_thumb/'.$this->filename.$file->getExtensionName());
} else {
$this->addError('Please, select at least one file'); // for example
}
}
}
2.Now, need to create controller, where you will do all actions.
class ImageController extends CController {
public function actionUpload()
{
$model = new Image();
if (Yii::app()->request->getPost('upload')) {
$model->filename = 'set filename';
$insName = 'image'; //if you try to upload from $_FILES['image']
if ($model->validate() && $model->doUpload($insName)) {
//upload is successful
} else {
//do something with errors
$errors = $model->getErrors();
}
}
}
}
Creating a model might be overkill in some instances.
The $_FILE supervariable is part of the HTTP mechanism.
You can handle the copy by using the native PHP function move_uploaded_file()
$fileName = "/uploads/".myimage.jpg";
unlink($fileName);
move_uploaded_file($_FILES['Filedata']['tmp_name'], $fileName);
However, you lose the niceties of using a library that provides additional functionality and checks (eg file type and file size limitations).
I have setup a repository to create a new resident.
<?php namespace Crescent\Repos;
interface ResidentRepository {
public function create($input);
}
Then in my controller I have, which uses the intervention image class to resize an image and it uploads correctly to the directory, but how can I save the name of the file to the DB using this repo?
public function store()
{
if (Input::hasFile('photo')){
$res = new Residents;
$file = Input::file('photo');
$name = $file->getClientOriginalName();
$input = Input::all();
$image = Image::make(Input::file('photo')->getRealPath())->resize(200, 200);
$image->save(public_path() . '/uploads/residents/' . $input['photo']->getClientOriginalName());
$res->photo = $name; // This doesn't work
}
$this->resident->create(Input::all());
}
Everything else works all the data, but the image isn't storing the name just showing some temp dir/name like /tmp/phpIX7KcY
I see that you have done $res = new Residents; and $res->photo = $name; but you haven't done $res->save(); which would have saved the name to the table corresponding to Residents. Also since you haven't added anything else to $res, only the photo would be saved.
Replace the code in your controller with the following:
public function store()
{
$input = Input::all();
if (Input::hasFile('photo')){
$file = Input::file('photo');
$name = $file->getClientOriginalName();
$image = Image::make(Input::file('photo')->getRealPath())->resize(200, 200);
$image->save(public_path() . '/uploads/residents/' . $input['photo']->getClientOriginalName());
$input['photo'] = $name;
}
$this->resident->create($input);
}
If in your code $this->resident->create(Input::all()); saves all data properly except the photo, it is because by passing Input::all() you're saving everything exactly as it was received from the client and the filename received from the resizing operation isn't present in Input::all(). By assigning Input::all() to the variable $input and doing $input['photo'] = $name;, the location of the file on the server is stored instead of the location on the client. So, by doing $this->resident->create($input);, the location on the server is stored along with other data received from the client.