Read file data without saving it in Flask

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I am writing my first flask application. I am dealing with file uploads, and basically what I want is to read the data/content of the uploaded file without saving it and then print it on the resulting page. Yes, I am assuming that the user uploads a text file always.

Here is the simple upload function i am using:

@app.route("/upload/", methods=["GET", "POST"])
def upload():
    if request.method == "POST":
        file = request.files["file"]
        if file:
            filename = secure_filename(file.filename)
  ["UPLOAD_FOLDER"], filename))
            a = "file uploaded"

    return render_template("upload.html", data = a)

Right now, I am saving the file, but what I need is that "a" variable to contain the content/data of the file .. any ideas?

Read file data without saving it in Flask _files: Questions

How do I list all files of a directory?

5 answers

How can I list all files of a directory in Python and add them to a list?


Answer #1

os.listdir() will get you everything that"s in a directory - files and directories.

If you want just files, you could either filter this down using os.path:

from os import listdir
from os.path import isfile, join
onlyfiles = [f for f in listdir(mypath) if isfile(join(mypath, f))]

or you could use os.walk() which will yield two lists for each directory it visits - splitting into files and dirs for you. If you only want the top directory you can break the first time it yields

from os import walk

f = []
for (dirpath, dirnames, filenames) in walk(mypath):

or, shorter:

from os import walk

filenames = next(walk(mypath), (None, None, []))[2]  # [] if no file


Answer #2

I prefer using the glob module, as it does pattern matching and expansion.

import glob

It does pattern matching intuitively

import glob
# All files ending with .txt
# All files ending with .txt with depth of 2 folder

It will return a list with the queried files:

["/home/adam/file1.txt", "/home/adam/file2.txt", .... ]


Answer #3

os.listdir() - list in the current directory

With listdir in os module you get the files and the folders in the current dir

 import os
 arr = os.listdir()
 >>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]

Looking in a directory

arr = os.listdir("c:\files")

glob from glob

with glob you can specify a type of file to list like this

import glob

txtfiles = []
for file in glob.glob("*.txt"):

glob in a list comprehension

mylist = [f for f in glob.glob("*.txt")]

get the full path of only files in the current directory

import os
from os import listdir
from os.path import isfile, join

cwd = os.getcwd()
onlyfiles = [os.path.join(cwd, f) for f in os.listdir(cwd) if 
os.path.isfile(os.path.join(cwd, f))]

["G:\getfilesname\", "G:\getfilesname\example.txt"]

Getting the full path name with os.path.abspath

You get the full path in return

 import os
 files_path = [os.path.abspath(x) for x in os.listdir()]
 ["F:\documentiapplications.txt", "F:\documenticollections.txt"]

Walk: going through sub directories

os.walk returns the root, the directories list and the files list, that is why I unpacked them in r, d, f in the for loop; it, then, looks for other files and directories in the subfolders of the root and so on until there are no subfolders.

import os

# Getting the current work directory (cwd)
thisdir = os.getcwd()

# r=root, d=directories, f = files
for r, d, f in os.walk(thisdir):
    for file in f:
        if file.endswith(".docx"):
            print(os.path.join(r, file))

os.listdir(): get files in the current directory (Python 2)

In Python 2, if you want the list of the files in the current directory, you have to give the argument as "." or os.getcwd() in the os.listdir method.

 import os
 arr = os.listdir(".")
 >>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]

To go up in the directory tree

# Method 1
x = os.listdir("..")

# Method 2
x= os.listdir("/")

Get files: os.listdir() in a particular directory (Python 2 and 3)

 import os
 arr = os.listdir("F:\python")
 >>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]

Get files of a particular subdirectory with os.listdir()

import os

x = os.listdir("./content")

os.walk(".") - current directory

 import os
 arr = next(os.walk("."))[2]
 >>> ["5bs_Turismo1.pdf", "5bs_Turismo1.pptx", "esperienza.txt"]

next(os.walk(".")) and os.path.join("dir", "file")

 import os
 arr = []
 for d,r,f in next(os.walk("F:\_python")):
     for file in f:

 for f in arr:

>>> F:\_python\
>>> F:\_python\programmi.txt

next(os.walk("F:\") - get the full path - list comprehension

 [os.path.join(r,file) for r,d,f in next(os.walk("F:\_python")) for file in f]
 >>> ["F:\_python\", "F:\_python\programmi.txt"]

os.walk - get full path - all files in sub dirs**

x = [os.path.join(r,file) for r,d,f in os.walk("F:\_python") for file in f]

>>> ["F:\_python\", "F:\_python\progr.txt", "F:\_python\"]

os.listdir() - get only txt files

 arr_txt = [x for x in os.listdir() if x.endswith(".txt")]
 >>> ["work.txt", "3ebooks.txt"]

Using glob to get the full path of the files

If I should need the absolute path of the files:

from path import path
from glob import glob
x = [path(f).abspath() for f in glob("F:\*.txt")]
for f in x:

>>> F:acquistionline.txt
>>> F:acquisti_2018.txt
>>> F:ootstrap_jquery_ecc.txt

Using os.path.isfile to avoid directories in the list

import os.path
listOfFiles = [f for f in os.listdir() if os.path.isfile(f)]

>>> ["a simple", "data.txt", ""]

Using pathlib from Python 3.4

import pathlib

flist = []
for p in pathlib.Path(".").iterdir():
    if p.is_file():

 >>> error.PNG
 >>> exemaker.bat
 >>> guiprova.mp3
 >>> thumb.PNG

With list comprehension:

flist = [p for p in pathlib.Path(".").iterdir() if p.is_file()]

Alternatively, use pathlib.Path() instead of pathlib.Path(".")

Use glob method in pathlib.Path()

import pathlib

py = pathlib.Path().glob("*.py")
for file in py:


Get all and only files with os.walk

import os
x = [i[2] for i in os.walk(".")]
for t in x:
    for f in t:

>>> ["", "data.txt", "data1.txt", "data2.txt", "data_180617", "", "", "", "", "", "", "data.txt", "data1.txt", "data_180617"]

Get only files with next and walk in a directory

 import os
 x = next(os.walk("F://python"))[2]
 >>> ["calculator.bat",""]

Get only directories with next and walk in a directory

 import os
 next(os.walk("F://python"))[1] # for the current dir use (".")
 >>> ["python3","others"]

Get all the subdir names with walk

for r,d,f in os.walk("F:\_python"):
    for dirs in d:

>>> .vscode
>>> pyexcel
>>> subtitles
>>> _metaprogramming
>>> .ipynb_checkpoints

os.scandir() from Python 3.5 and greater

import os
x = [ for f in os.scandir() if f.is_file()]

>>> ["calculator.bat",""]

# Another example with scandir (a little variation from
# This one is more efficient than os.listdir.
# In this case, it shows the files only in the current directory
# where the script is executed.

import os
with os.scandir() as i:
    for entry in i:
        if entry.is_file():

>>> error.PNG
>>> exemaker.bat
>>> guiprova.mp3
>>> thumb.PNG


Ex. 1: How many files are there in the subdirectories?

In this example, we look for the number of files that are included in all the directory and its subdirectories.

import os

def count(dir, counter=0):
    "returns number of files in dir and subdirs"
    for pack in os.walk(dir):
        for f in pack[2]:
            counter += 1
    return dir + " : " + str(counter) + "files"


>>> "F:\python" : 12057 files"

Ex.2: How to copy all files from a directory to another?

A script to make order in your computer finding all files of a type (default: pptx) and copying them in a new folder.

import os
import shutil
from path import path

destination = "F:\file_copied"
# os.makedirs(destination)

def copyfile(dir, filetype="pptx", counter=0):
    "Searches for pptx (or other - pptx is the default) files and copies them"
    for pack in os.walk(dir):
        for f in pack[2]:
            if f.endswith(filetype):
                fullpath = pack[0] + "\" + f
                shutil.copy(fullpath, destination)
                counter += 1
    if counter > 0:
        print("-" * 30)
        print("	==> Found in: `" + dir + "` : " + str(counter) + " files

for dir in os.listdir():
    "searches for folders that starts with `_`"
    if dir[0] == "_":
        # copyfile(dir, filetype="pdf")
        copyfile(dir, filetype="txt")

>>> _compiti18Compito Contabilità 1conti.txt
>>> _compiti18Compito Contabilità 1modula4.txt
>>> _compiti18Compito Contabilità 1moduloa4.txt
>>> ------------------------
>>> ==> Found in: `_compiti18` : 3 files

Ex. 3: How to get all the files in a txt file

In case you want to create a txt file with all the file names:

import os
mylist = ""
with open("filelist.txt", "w", encoding="utf-8") as file:
    for eachfile in os.listdir():
        mylist += eachfile + "

Example: txt with all the files of an hard drive

We are going to save a txt file with all the files in your directory.
We will use the function walk()

import os

# see all the methods of os
# print(*dir(os), sep=", ")
listafile = []
percorso = []
with open("lista_file.txt", "w", encoding="utf-8") as testo:
    for root, dirs, files in os.walk("D:\"):
        for file in files:
            percorso.append(root + "\" + file)
            testo.write(file + "
print("N. of files", len(listafile))
with open("lista_file_ordinata.txt", "w", encoding="utf-8") as testo_ordinato:
    for file in listafile:
        testo_ordinato.write(file + "

with open("percorso.txt", "w", encoding="utf-8") as file_percorso:
    for file in percorso:
        file_percorso.write(file + "


All the file of C: in one text file

This is a shorter version of the previous code. Change the folder where to start finding the files if you need to start from another position. This code generate a 50 mb on text file on my computer with something less then 500.000 lines with files with the complete path.

import os

with open("file.txt", "w", encoding="utf-8") as filewrite:
    for r, d, f in os.walk("C:\"):
        for file in f:
            filewrite.write(f"{r + file}

How to write a file with all paths in a folder of a type

With this function you can create a txt file that will have the name of a type of file that you look for (ex. pngfile.txt) with all the full path of all the files of that type. It can be useful sometimes, I think.

import os

def searchfiles(extension=".ttf", folder="H:\"):
    "Create a txt file with all the file of a type"
    with open(extension[1:] + "file.txt", "w", encoding="utf-8") as filewrite:
        for r, d, f in os.walk(folder):
            for file in f:
                if file.endswith(extension):
                    filewrite.write(f"{r + file}

# looking for png file (fonts) in the hard disk H:
searchfiles(".png", "H:\")

>>> H:4bs_18Dolphins5.png
>>> H:4bs_18Dolphins6.png
>>> H:4bs_18Dolphins7.png
>>> H:5_18marketing htmlassetsimageslogo2.png
>>> H:7z001.png
>>> H:7z002.png

(New) Find all files and open them with tkinter GUI

I just wanted to add in this 2019 a little app to search for all files in a dir and be able to open them by doubleclicking on the name of the file in the list. enter image description here

import tkinter as tk
import os

def searchfiles(extension=".txt", folder="H:\"):
    "insert all files in the listbox"
    for r, d, f in os.walk(folder):
        for file in f:
            if file.endswith(extension):
                lb.insert(0, r + "\" + file)

def open_file():

root = tk.Tk()
bt = tk.Button(root, text="Search", command=lambda:searchfiles(".png", "H:\"))
lb = tk.Listbox(root)
lb.pack(fill="both", expand=1)
lb.bind("<Double-Button>", lambda x: open_file())

Read file data without saving it in Flask join: Questions

Why is it string.join(list) instead of list.join(string)?

5 answers

Evan Fosmark By Evan Fosmark

This has always confused me. It seems like this would be nicer:

my_list = ["Hello", "world"]
# Produce: "Hello-world"

Than this:

my_list = ["Hello", "world"]
# Produce: "Hello-world"

Is there a specific reason it is like this?


Answer #1

It"s because any iterable can be joined (e.g, list, tuple, dict, set), but its contents and the "joiner" must be strings.

For example:

"_".join(["welcome", "to", "stack", "overflow"])
"_".join(("welcome", "to", "stack", "overflow"))

Using something other than strings will raise the following error:

TypeError: sequence item 0: expected str instance, int found


Answer #2

This was discussed in the String methods... finally thread in the Python-Dev achive, and was accepted by Guido. This thread began in Jun 1999, and str.join was included in Python 1.6 which was released in Sep 2000 (and supported Unicode). Python 2.0 (supported str methods including join) was released in Oct 2000.

  • There were four options proposed in this thread:
    • str.join(seq)
    • seq.join(str)
    • seq.reduce(str)
    • join as a built-in function
  • Guido wanted to support not only lists and tuples, but all sequences/iterables.
  • seq.reduce(str) is difficult for newcomers.
  • seq.join(str) introduces unexpected dependency from sequences to str/unicode.
  • join() as a built-in function would support only specific data types. So using a built-in namespace is not good. If join() supports many datatypes, creating an optimized implementation would be difficult, if implemented using the __add__ method then it would ve O(n¬≤).
  • The separator string (sep) should not be omitted. Explicit is better than implicit.

Here are some additional thoughts (my own, and my friend"s):

  • Unicode support was coming, but it was not final. At that time UTF-8 was the most likely about to replace UCS2/4. To calculate total buffer length of UTF-8 strings it needs to know character coding rule.
  • At that time, Python had already decided on a common sequence interface rule where a user could create a sequence-like (iterable) class. But Python didn"t support extending built-in types until 2.2. At that time it was difficult to provide basic iterable class (which is mentioned in another comment).

Guido"s decision is recorded in a historical mail, deciding on str.join(seq):

Funny, but it does seem right! Barry, go for it...
Guido van Rossum


Answer #3

Because the join() method is in the string class, instead of the list class?

I agree it looks funny.


Historical note. When I first learned Python, I expected join to be a method of a list, which would take the delimiter as an argument. Lots of people feel the same way, and there’s a story behind the join method. Prior to Python 1.6, strings didn’t have all these useful methods. There was a separate string module which contained all the string functions; each function took a string as its first argument. The functions were deemed important enough to put onto the strings themselves, which made sense for functions like lower, upper, and split. But many hard-core Python programmers objected to the new join method, arguing that it should be a method of the list instead, or that it shouldn’t move at all but simply stay a part of the old string module (which still has lots of useful stuff in it). I use the new join method exclusively, but you will see code written either way, and if it really bothers you, you can use the old string.join function instead.

--- Mark Pilgrim, Dive into Python


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