I am uploading an image and while saving to the database I am adding path to the location along with the filename.
eg : product/name/L4I84qXltYvCCDsAeQzi7fzZfKF3WTRa1LPOPGxj.jpeg
Why adding location where I can call them through asset()?
The table is also used by API application where it requires the url format like this.
My Controller Cdde
//! Product Image
if ($request->hasFile('product_image')) {
$product_image = $request->file('product_image');
$product_image_extension = $product_image->getClientOriginalExtension();
$product_image_filename = Str::slug($request->input('product_code')) . '.' . Str::random(3) . "." . $product_image_extension;
$product_image_location = public_path('assets/products/' . $product_image_filename);
Image::make($product_image)->resize(300, 300)->save($product_image_location);
}
Products::create([
//Other Fields
'image' => 'assets/products/' . $product_image_filename
]);
I am wondering how do I remove the image and add a new image
For comparison, I need the image file name. Since I am using the path, I am getting the path prefixed to the Image Name.
use Illuminate\Support\Facades\File
$image_path = "/images/filename.ext"; // Value is not URL but directory file path
if(File::exists($image_path)) {
File::delete($image_path);
}
use File;
#unlink(public_path() . 'uploads/documents/' . $users->document1);
Related
I need assistance to more understand the concept so I can become a better developer. I want to learn how to refactor the code and erase all duplications.
What's the best practices for image uploads? Renaming them correctly?
I have a block of code that handles two attachments:
if( $request->hasFile('LFImage') ) {
$destination = public_path('app/lostFound/lostItems' . $lostFound->LFImage);
if( File::exists($destination) )
{
File::delete($destination);
}
$file = $request->file('LFImage');
$extension = $file->getClientOriginalExtension();
$filename = $lostFound->LFNumber . '-' . $lostFound->lostItem . '.' . $extension;
$file->move('app/lostFound/lostItems', $filename);
$lostFound->LFImage = $filename;
}
if( $request->hasFile('handoverStatement') ) {
$destination = public_path('app/lostFound/handoverStatements' . $lostFound->handoverStatement);
if( File::exists($destination) )
{
File::delete($destination);
}
$file = $request->file('handoverStatement');
$extension = $file->getClientOriginalExtension();
$filename = $lostFound->lostItem . '-' . $lostFound->LFNumber . '.' . $extension;
$file->move('app/lostFound/handoverStatements', $filename);
$lostFound->handoverStatement = $filename;
}
They're exactly the same except with the upload directory.
How can I make it as a one code block across the entire application with changeable file name and location depending on the form?
Some file names require random strings, how can I "Edit" the random string to the file that was uploaded?
Best practice when uploading and storing files in Laravel is using Storage.
It has all needed methods to work with files, you can save the file like this:
use Illuminate\Support\Facades\Storage;
Storage::put('images/', $request->file('LFImage'));
In the documentation provided above, you can find other examples like renaming and moving files
In order to access these files from web as well, you can use the command php artisan storage:link, which creates a symbolic link to storage folder in your public folder. After you create the symbolic link, you can generate URL to the file like this:
asset('storage/test.txt')
To avoid duplications, you can create a function in your controller to create a file. You will then just call this function with different files to keep the file creation code in one place.
you can simply write this
if ($request->hasFile('logo')) {
deleteImageFromDirectory(setting('logo'), "Settings");
$data['logo'] = uploadImageToDirectory( $request->logo , "Settings");
}
and define uploadImageToDirectory function in your helper functions or create a trait
function uploadImageToDirectory($imageFile, $directory = '' ){
$imageName = $imageFile->getClientOriginalName(); // Set Image name
$imageFile->storeAs("/Images/$directory", $imageName, 'public');
return $imageName;
}
I have stored my image with following code
if (isset($image)) {
// make unique name for image
$currentDate = Carbon::now()->toDateString();
$imagename = $slug . '-' . $currentDate . '-' . uniqid() . '.' . $image->getClientOriginalExtension();
// check category dir is exists
if (!Storage::disk('public')->exists('category')) {
Storage::disk('public')->makeDirectory('category');
}
// resize image for category and upload
$category = Image::make($image)->resize(1600,479)->stream();
Storage::disk('public')->put('category/'.$imagename,$category);
// check category slider dir is exists
if (!Storage::disk('public')->exists('category/slider')) {
Storage::disk('public')->makeDirectory('category/slider');
}
// resize image for category slider and upload
$slider = Image::make($image)->resize(500,333)->stream();
Storage::disk('public')->put('category/slider/'.$imagename,$slider);
} else {
$imagename = "default.png";
}
try with this code
<img src="images/icons8-team-355979.jpg" alt="Profile Image">
Failed to load resource: the server responded with a status of 404 (Not Found)
this is the error
Try this
<img src="{{url('category/slider')}}/{{ $category->image }}" alt="{{$category->name}}">
The url function will go to '/public' folder. So, you may need to edit your '/config/filesystem.php' so that the uploaded photo is saved inside the '/public' directory.
For more information on this please check this document.
When you want to serve files from the public disk directly, you need to create a symlink to these files. This is described in the documentation here: https://laravel.com/docs/5.7/filesystem#the-public-disk.
When you have done this, you can link to the assets using the asset() function.
echo asset('category/slider/icons8-team-355979.jpg');
I write a edit function to update news's info, delete previous image from web root and insert new image:
code is below:
if(unlink($data['News']['image_url']['tmp_name'], WWW_ROOT . 'media/' . $data['News']['image_url']['name'])) //delete image from root and database
{
echo 'image deleted.....'; //success message
}
I can't delete old image and insert new image,how can i correct my function ?
Here your data can not find existing data. use this code
$data1 = $this->News->findById($newsid);
$this->request->data = $data1;
$directory = WWW_ROOT . 'media';
if(unlink($directory.DIRECTORY_SEPARATOR.$data1['News']['image_url']))
{
echo 'image deleted.....';
}
Pass filepath as first argument of unlink():
unlink(WWW_ROOT . 'media/' . $data['News']['image_url']['name'] . '/' . $data['News']['image_url']['tmp_name']);
Also make sure that you have proper permissions to perform this operation in directory containing image.
I can't get the picture to display/show when viewing, although the files are already stored in the database (table 'menu') http://i.imgur.com/wo1w90H.png. Also when I upload the images all at once, their file name would change automatically. I don't know how and why this happens. I use array to upload multiple images.
if (isset($_POST["Submit"])) {
--some code here--
if (isset($_POST["id_list"])) {
// if id list available
foreach($_POST["id_list"] AS $id) {
--some code here--
/* Handle file upload */
if ($_FILES['upload']['error'][$id] == 'UPLOAD_ERR_OK') {
$path = "images/newmenu/";
$path_parts = pathinfo($_FILES["upload"]["name"][$id]);
$extension = $path_parts['extension'];
$picture = md5(uniqid()) . "." . $extension;
if (move_uploaded_file($_FILES['upload']['tmp_name'][$id], $path . "/" . $picture)) {
$update = " UPDATE menu
SET MenuPicture='$picture'
WHERE MenuID=$id";
$mysqli->query($update) or die(mysqli_error($mysqli));
}
}
}
}
}
}
Below is the form and yes it does include enctype="multipart/form-data"
<input type="file" multiple name="upload[' . $id . ']" value="' . $record["MenuPicture"] . '">
Filename changes because you are generating it this way
$picture = md5(uniqid()) . "." . $extension;
uniqid() is based on current time and hashing it will cause the filename to change everytime
When I upload the images all at once, their file name would change automatically
It was due to this:
$picture = md5(uniqid()) . "." . $extension;
// And later
move_uploaded_file($_FILES['upload']['tmp_name'][$id], $path . "/" . $picture)
Basically, you are moving your uploaded file to a new filename for your image file, which is generated using uniqid() and hashed with md5(), with the file extension appended at the end.
I can't get the picture to display/show when viewing
How are you trying to display the picture? Is it from web browser, or you go straight to the directory and open from there? What error(s) did you get, if any?
Actually, have you tried to go to the directory and see whether the file is created inside the images/newmenu/ directory?
Also, for the target upload directory, you might want to append it with $_SERVER['DOCUMENT_ROOT'] so that the target directory is not dependent on where your script is located, but it's always based on the root.
By the way, you might know already, but there is an entry in PHP manual page on uploading multiple files
I am uploading a file using Codeigniter's File Uploading Library and trying to insert the URL into the database. Codeigniter only supplies the server_path when using $this->upload->data(), which isn't usable for displaying the image to users.
I could normally just do something like base_url('uploads) . '/' . $data['file_name'] but I am storing the images in a folder for each post.
For example, I am getting C:/xampp/htdocs/site/uploads/32/image.jpg as the full_path, how can I convert this to http://mysite.com/uploads/32/image.jpg
The only thing that comes to mind is using a regular expression, but I feel like there has to be PHP function or Codeigniter function to help with this?
How can I convert the server path into the correct URL?
You can use $_SERVER['HTTP_HOST']; to get the url. Using the ID for your post, you can construct your path like so:
$url = "http://" . $_SERVER['HTTP_HOST'] . "/" . $id . "/" . basename($data['file_name']);
the basename() function returns everything after the last / in your path.
you only need to save the filename to the database and just use the post id and filename:
$url = "http://mysite.com/uploads/" . $postID . "/" . $fileName;
to get the file name use: How to get file name from full path with PHP?
<?php
$path = "C:/xampp/htdocs/site/uploads/32/image.jpg";
$file = basename($path); // $file is set to "image.jpg"
$file = basename($path, ".jpg"); // $file is set to "image"
?>
In CodeIgniter you should use $config['base_url']
You can use the ENVIRONMENT to setup different base_url, for example:
switch(ENVIRONMENT)
{
case 'development':
$config['base_url'] = 'http://localhost/';
break;
case 'testing':
$config['base_url'] = 'http://testing.server.com/';
break;
default:
$config['base_url'] = 'http://liveserver.com/';
}
See config.php
Now simply replace your local path with the base_url, str_replace should do the job (documentation here)
$newpath = str_replace("C:/xampp/htdocs/site/", $config['base_url'], $localpath);
Also, if you're interested in getting parts of the path, you could use explode (documented here) with / to create an array with every "section" of your path
$pathElements = explode('/', $localpath);
In this case, $pathElements[0] is C:, $pathElements[1] is xampp, etc...