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);
Related
This code works, but it's returning directory names and filenames. I haven't found a parameter that tells it to return only files or only directories.
Can glob.glob do this, or do I have to call os.something to test if I have a directory or file. In my case, my files all end with .csv, but I would like to know for more general knowledge as well.
In the loop, I'm reading each file, so currently bombing when it tries to open a directory name as a filename.
files = sorted(glob.glob(input_watch_directory + "/**", recursive=True))
for loop_full_filename in files:
print(loop_full_filename)
Results:
c:\Demo\WatchDir\
c:\Demo\WatchDir\2202
c:\Demo\WatchDir\2202\07
c:\Demo\WatchDir\2202\07\01
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_51.csv
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_52.csv
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_53.csv
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_54.csv
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_55.csv
c:\Demo\WatchDir\2202\07\05
c:\Demo\WatchDir\2202\07\05\polygonData_2022_07_05__12_00.csv
c:\Demo\WatchDir\2202\07\05\polygonData_2022_07_05__12_01.csv
Results needed:
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_51.csv
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_52.csv
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_53.csv
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_54.csv
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_55.csv
c:\Demo\WatchDir\2202\07\05\polygonData_2022_07_05__12_00.csv
c:\Demo\WatchDir\2202\07\05\polygonData_2022_07_05__12_01.csv
For this specific program, I can just check if the file name contains.csv, but I would like to know in general for future reference.
Line:
files = sorted(glob.glob(input_watch_directory + "/**", recursive=True))
replace with the line:
files = sorted(glob.glob(input_watch_directory + "/**/*.*", recursive=True))
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();
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)
I'm simply trying to overwrite the contents of a pre-generated (written with allocUnsafe(size)) 1GB file via a 4 byte buffer at an iterating offset, and before I open the file descriptor, fs.stat and the Windows file system show the correct size. As soon as I open the file descriptor, it appears both in fs.stat and in the file system the file is empty:
let stats = fs.statSync(dataPath)
let fileSizeInBytes = stats["size"]
let fileSizeInMegabytes = fileSizeInBytes / 1000000
console.log("fileSizeInMegabytes", fileSizeInMegabytes) // => fileSizeInMegabytes 1000
fd = fs.openSync(dataPath, 'w')
stats = fs.statSync(dataPath)
fileSizeInBytes = stats["size"]
fileSizeInMegabytes = fileSizeInBytes / 1000000
console.log("fileSizeInMegabytes", fileSizeInMegabytes) // => fileSizeInMegabytes 0
Why is opening the file descriptor emptying my file? Surely I'm missing something obvious, but I can't see it.
Opening the file using the w flag truncates the file, i.e. removes any contents.
You should use r+ to read and write to the file without wiping it clean.
For more info, check out the Node docs and the answers on this question.
In an application, I can get the path to a file which resides in a directory as a string:
"/path/to/the/file.txt"
In order to write another another file into that same directory, I want to change the string "/path/to/the/file.txt" and remove the part "file.txt" to finally only get
"/path/to/the/"
as a string
I could use
string = "/path/to/the/file.txt"
string.split('/')
and then glue all the term (except the last one) together with a loop
Is there an easy way to do it?
You can use os.path.basename for getting last part of path and delete it with using replace.
import os
path = "/path/to/the/file.txt"
delete = os.path.basename(os.path.normpath(path))
print(delete) # will return file.txt
#Remove file.txt in path
path = path.replace(delete,'')
print(path)
OUTPUT :
file.txt
/path/to/the/
Let say you have an array include txt files . you can get all path like
new_path = ['file2.txt','file3.txt','file4.txt']
for get_new_path in new_path:
print(path + get_new_path)
OUTPUT :
/path/to/the/file2.txt
/path/to/the/file3.txt
/path/to/the/file4.txt
Here is what I finally used
iter = len(string.split('/'))-1
directory_path_str = ""
for i in range(0,iter):
directory_path_str = directory_path_str + srtr.split('/')[i] + "/"