Trying to save an external URL image, to my s3.
But when I use code below, it returns error.
$contents = file_get_contents($url);
$filename = Str::slug($findproduct->title)."-".uniqid().".".$extension;
Storage::disk('s3')->putFileAs('product', $contents, $filename, ['visibility' => 'public']);
Here's error:
fopen() expects parameter 1 to be a valid path, string given
fopen() is the first function called by putFileAs API Docs.
It attempts to open the File supplied as parameter 2 to putFileAs(), in your case this is $contents. (FWIW, newer versions of Laravel also accept a string as a file path here)
It expects this to be either a File or UploadedFile object.
file_get_contents() returns a string instead.
Try using this instead:
$contents = new File($url);
Related
I am trying to lock a file with the flock function but I am not able to do it. I work with Laravel 8 and Storage Class.
The code is as follows:
$disk = Storage::disk('communication');
$file_name = 'received.json';
$file_exists = $disk->exists($file_name);
if($file_exists){
flock($disk->get($file_name), LOCK_EX);
...
}
The problem I'm having is that when I invoke the get() function on the file path, it returns the contents of the file (a string), which causes the following error:
flock() expects parameter 1 to be resource, string given
I need to know how to get file as a resource and not the content of the file.
Could someone help me and tell me how to do it?
Thank you very much in advance.
You can use Storage::readStream() method
if($file_exists){
$stream=Storage::disk('communication')->readStream($file_name);
flock($stream, LOCK_EX);
}
As per php doc
flock(resource $stream, int $operation, int &$would_block = null): bool
First param needed stream.flock() allows you to perform a simple
reader/writer model which can be used on virtually every platform
(including most Unix derivatives and even Windows).
Ref:https://www.php.net/manual/en/function.flock.php
I am trying to generate an Excel file with CodeIgniter and the PhpSpreadsheet library from a view. The reports that I need to make are not a list but much more complex and I can generate them more quickly with a view and sending parameters. This is my code:
$data = $this->model->bringdata();
$view = $this->load->view("data_view", $ data);
$spreadsheet = new Spreadsheet();
$writer = new \PhpOffice\PhpSpreadsheet\Writer\Html($spreadsheet);
$writer-> save($view);
And the following error is shown
Severity: Warning
Message: fopen (): Filename can not be empty
Filename: Writer / Html.php
Line Number: 160
In the PhpSpreadsheet documentation it shows that an html page must be explicitly sent:
$writer->save("05featuredemo.htm");
Is there any way to do what I need?
Thanks for the help
It's a simple solution, you need to pass true to the third argument of load->view in order for it to return a string rather than sending it to the browser.
$view = $this->load->view("data_view", $ data, true);
Added after comment
I made the mistake of not looking into how the Html writer works.
The HTML writer writes a spreadsheet as an html file. It does not create a spreadsheet. As I understand it (and I might be mistaken) you can use \PhpOffice\PhpSpreadsheet\Reader\Html to read an html file into a spreadsheet.
In your case I think you will have to save the string returned from $this->load->view() to a temporary file. CodeIgniter's file_helper might be useful here.
You then read the temporary file to create a spreadsheet. After that you can use $writer to save the spreadsheet somewhere as any of the file types the PhpSpreadsheet supports.
$view = $this->load->view("data_view", $data, true);
$this->load->helper("file");
$fileName = "temp_file_name.html";
$path = "full/path/to/some_writable_folder/";
$path_file = $path . $fileName;
if (write_file($path_file, $view))
{
//create spreadsheet the temp html
$reader = new \PhpOffice\PhpSpreadsheet\Reader\Html();
$spreadsheet = $reader->load($path_file);
//write out to html file
$writer = new \PhpOffice\PhpSpreadsheet\Writer\Html($spreadsheet);
$writer->save($path."05featuredemo.htm");
//delete the temporary file
unlink($path_file);
}
else
{
//handle the failure to write the temp file
}
I haven't tested this and I'm not sure you really want to save to an html file, but I think the above is close to the answer.
Second addition
If you want to save .xlsx files you need an Xlsx writer. Use the following instead of what's shown above.
//write out to excel file
$writer = new \PhpOffice\PhpSpreadsheet\Writer\Xlsx($spreadsheet);
$writer->save($ path. "doc.xlsx");
I confess that I have never used, or considered using the html reader so I'm not sure you'll get what you want. I honestly think that creating an html table so you can import it into excel is probably doing it the hard way.
I am trying to get the image as output, but it seems like set_output() function is not working properly with content type jpeg
My Code is given below...
$image = file_get_contents('assets/images/ThinkstockPhotos-145054512_small.jpg');
$this->output->set_content_type('jpeg')->set_output($image);
When I replace the image with a plain text file, in that case, it shows me the correct output
$file = file_get_contents('assets/images/test.txt');
$this->output->set_content_type('text')->set_output($file);
I have change content type from set_content_type('jpeg') to set_content_type('jpg') and set_content_type('gif') but stil it does not work and not show me on output.
What output I am getting now is shown in screenshot given below.
I was able to replicate the issue on my localhost setup only when I provided file_get_contents either a (1) bad path or (2) a non-existent image. I think you need to provide a full path to the image as with any directory/file related operation.
Try using FCPATH that is where you index.php lies and I assume your assets/ folder as well.
public function index() {
$file = FCPATH . 'assets/images/ThinkstockPhotos-145054512_small.jpg';
if (!is_file($file)) {
exit('File not found!');
}
$image = file_get_contents($file);
$this->output->set_content_type('jpeg')->set_output($image);
}
Note: if you get File not found! then assure that the file exists in /htdocs/assets/... where /htdocs/index.php is your main CI file
It is working good for full path like this
$file=$request->file('file');
$file->move('C:\xampp\htdocs\modo\images',$file->getClientOriginalName());
But i cant understand why it doesnt for root folder path :
$file->move('\modo\images',$file->getClientOriginalName());
You need to use base_path() method. This method returns the fully qualified path to the project root:
So in your case try the below code:
$file = $request->file('file');
$file->move(base_path('\modo\images'), $file->getClientOriginalName());
and if you want to return the public directory then use:
$path = public_path();
For more info read Laravel helper functions
You need to do it this way:
$file->move(base_path('\modo\images'),$file->getClientOriginalName());
I'm having a problem regarding passing parameter from url. The function I use sometimes give the correct result and sometimes not.
here is my controller functions:
public function link_gen(){
$text = "i have lost my password please help me";
$encrypted_text = $this->encrypt->encode($text);
$encrypted_url = urlencode($encrypted_text);
echo $encrypted_url. br();
echo br(). $this->retrive(urldecode($encrypted_url));
echo anchor('encryption/ret_back?username='.$encrypted_url, 'click me');
// echo anchor('encryption/ret_back/'.$encrypted_url, 'click me');
}
public function ret_back(){
// echo br(). $this->retrive(urldecode($str));
$user = $this->input->get('username');
echo br(). $this->retrive(urldecode($user));
echo $user. br();
echo "hellooooo". br();
}
For the sake of testing I'm encrypting a text then encode it in URL using urlencode(), then I pass this string to another function using URL, and then decode in the string in URL using urldecode() and then decode the text. But when I run this, sometimes the decoded text appears and sometimes it does not.
Can anybody tell me the reason and solution for this?
I want to use this in a 'forgot password' module.
It depends on how you create your links and how do you retrieve the data from url.
Anyway these are simple examples to get url data:
get from $_GET[] (http://site.com/?q=hello+world)
in this case you get the param q with echo $this->input->get('q',true);
get data from uri segments (http://site.com/id/230)
in this case you get id value via echo $this->uri->segment(2);
Then, if data in url exists it is not possible that the system didn't get that, so be sure data is in your url when opening url.