How to get the original file name extension even if i change the extension name of files intensionally. for example one file is for xls format i changed its extension to .pdf so is there any option we can get the original extension of file upload ?
let avatar = req.files.avatar;
avatar.mv('./uploads/' + avatar.name);
One approach you can do is by getting the MIME type of the file. If you're using Linux then you could use the exec or execSync function to run the file command:
const execSync = require('child_process').execSync;
const mimeType = execSync('file --mime-type -b "' + '<file-path>' + '"').toString();
console.log(mimeType.trim());
If you input an excel file on the command above, it will read as text/html Linux reads like a text/html. For better MIME type recognition, you could install xdg-mime on your system and replace the command on the function above.
const mimeType = execSync('xdg-mime query filetype "' + '<file-path>' + '"').toString();
Related
I upload a file through the form, check it, and only after checking it I want to add it to my database.
form = BookForm(request.POST, request.FILES)
file = form.files
path = file.get('book_file').temporary_file_path()
in path - '/tmp/tmpbp4klqtw.upload.pdf'
But as soon as I want to transfer this file from the temporary storage to some other folder, I get the following error:
path = os.replace(path, settings.MEDIA_ROOT)
IsADirectoryError: [Errno 21] Is a directory: '/tmp/tmpbp4klqtw.upload.pdf' -> '/home/oem/bla/bla'
Can't understand why this file is not in reality? What can I do about it? Is it possible to set some special path for the "temporary file"?
UPD:
You should use path = os.replace(path, settings.MEDIA_ROOT + '/name-of-file.pdf') – Willem Van Onsem
os.replace(…) [python-doc] expects a filename as target if you specify a file as source, so you can move this to:
os.replace(path, f'{settings.MEDIA_ROOT}/name-of-file.pdf')
you can also make use of shutil.move(…) [python-doc] to specify the directory, this function will also return the filepath of the target file:
from shutil import move
target_file = move(path, settings.MEDIA_ROOT)
My Node.js program wants to read the contents of the file "test.txt" on a Windows machine. It checks with fs.existsSync() that the file exists and reads its content. But now I want the program instead to give an error or warning if the name of the file on disk is actually "TEST.txt" or any other name which differs in case from the name my program is looking for, e.g. "test.txt".
Is there a straightforward way to figure out that even though existsSync() tells me a file exists, the file on disk has a name which differs in case from the file-name I am using to look for it?
You can use fs.readdir to get a list of all files in directory and then compare the filename to see if matches as is.
var fs = require('fs');
var path = __dirname;
var filename = 'test.txt';
var files = fs.readdirSync(path);
var exists = files.includes(filename);
// true if file on disk is "test.txt",
// false if file on disk is "TEST.txt"
console.log(exists);
I have a problem with a corrupted excel file. So far I have used 7zip to open it as an archive and extract most of the data. But some important sheets cannot be extracted.
Using the l command of 7zip I get the following output :
7z.exe l -slt "C:\Users\corrupted1.xlsm" xl/worksheets/sheet3.xml
Output:
Listing archive: C:\Users\corrupted1.xlsm
--
Path = C:\Users\corrupted1.xlsm
Type = zip
Physical Size = 11931916
----------
Path = xl\worksheets\sheet3.xml
Folder = -
Size = 57217
Packed Size = 12375
Modified = 1980-01-01 00:00:00
Created =
Accessed =
Attributes = .....
Encrypted = -
Comment =
CRC = 553C3C52
Method = Deflate
Host OS = FAT
Version = 20
However when trying to extract it (or test it for that matter) I get :
7z.exe t -slt "C:\Users\corrupted1.xlsm" xl/worksheets/sheet3.xml
Output:
Processing archive: C:\Users\corrupted1.xlsm
Testing xl\worksheets\sheet3.xml Unsupported Method
Sub items Errors: 1
The method listed above says Deflate, which is the same for all the worksheets.
Is there anything I can do? What kind of corruption is this? Is it the CRC? Can I ignore it somehow or something?
Please help!
Edit:
The following is the error when trying to extract or edit the xml file through 7zip:
Edit 2:
Tried with WinZip as well, getting :
Extracting to "C:\Users\axpavl\AppData\Local\Temp\wzf0b9\"
Use Path: yes Overlay Files: yes
Extracting xl\worksheets\sheet2.xml
Unable to find the local header for xl\worksheets\sheet2.xml.
Severe Error: Cannot find a local header.
This might help:
https://superuser.com/questions/145479/excel-edit-the-xml-inside-an-xlsx-file
and this on too: http://www.techrepublic.com/blog/tr-dojo/recover-data-from-a-damaged-office-file-with-the-help-of-7-zip/
I am using rails 4 application, where i have a xls file in my applications public folder ,and am having a link in view page after clicking that link that xls file should download to system.no need to write anything in that file its just a default template.
I use:
view:
<%= link_to "Dowload Template", admin_job_templates_path, :file_name => 'Job_Upload_Template.xlsx' %>
controller :
def index
require 'open-uri'
file_path = "#{Rails.root}/public/Job_Upload_Template.xlsx"
send_file file_path, :filename => "Job_Upload_Template.xlsx", :disposition => 'attachment'
end
also tried send_data method
def index
require 'open-uri'
url = "#{Rails.root}/public/Job_Upload_Template.xlsx"
data = open(url).read
send_data data, :filename =>'Job Template'
end
Am getting the below error.
No such file or directory # rb_sysopen -
/home/amp/workspace/LA_Tracker/public/Job_Upload_Template.xlsx
am not able to dowload the file, Plese help.
It's telling you that file doesn't exist. Do you have the capitalization correct? Double check. Also, what happens if you type:
ls /home/amp/workspace/LA_Tracker/public/Job_Upload_Template.xlsx
into the command line? Can your OS find the file?
I'm trying to add functionality to a large piece of code and am having a strange problem with File Separators. When reading a file in the following code works on my PC, but fails when on a Linux server. When on PC I pass this and it works:
fileName = "C:\\Test\\Test.txt";
But when on a server I pass this and get "File Not Found" because the BufferedReader/FileReader statement below swaps "/" for "\":
fileName = "/opt/Test/Test.txt";
System.out.println("fileName: "+fileName);
reader = new BufferedReader(new FileReader(new File(fileName)));
Produces this output when run on the LINUX server:
fileName: /opt/Test/Test.txt
File Not Found: java.io.FileNotFoundException: \opt\Test\Test.txt (The system cannot find the path specified)
When I create a simple Test.java file to try and replicate it behaves as expected, so something in the larger code source is causing the BufferedReader/FileReader line to behave as if it's on a PC, not a Linux box. Any ideas what that could be?
I don't see where you used File.separator. Try this instead of hard coding the path separators.
fileName = File.separator + "opt" + File.separator + "Test" + File.separator + "Test.txt";