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Flask upload example
<VirtualHost *>
ServerName example.com
WSGIDaemonProcess www user=max group=max threads=5
WSGIScriptAlias / /home/max/Projekte/flask-upload/flask-upload.wsgi
<Directory /home/max/Projekte/flask-upload>
WSGIProcessGroup www
WSGIApplicationGroup %{GLOBAL}
Order deny,allow
Allow from all
</Directory>
</VirtualHost>
activate_this = '/home/max/.virtualenvs/flask/bin/activate_this.py'
execfile(activate_this, dict(__file__=activate_this))
import sys
sys.path.insert(0, '/home/max/Projekte/flask-upload')
from server import app as application
import os
from flask import Flask, request, redirect, url_for
from werkzeug import secure_filename
UPLOAD_FOLDER = '/tmp/'
ALLOWED_EXTENSIONS = set(['txt'])
app = Flask(__name__)
app.config['UPLOAD_FOLDER'] = UPLOAD_FOLDER
def allowed_file(filename):
return '.' in filename and \
filename.rsplit('.', 1)[1] in ALLOWED_EXTENSIONS
@app.route("/", methods=['GET', 'POST'])
def index():
if request.method == 'POST':
file = request.files['file']
if file and allowed_file(file.filename):
filename = secure_filename(file.filename)
file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
return redirect(url_for('index'))
return """
<!doctype html>
<title>Upload new File</title>
<h1>Upload new File</h1>
<form action="" method=post enctype=multipart/form-data>
<p><input type=file name=file>
<input type=submit value=Upload>
</form>
<p>%s</p>
""" % "<br>".join(os.listdir(app.config['UPLOAD_FOLDER'],))
if __name__ == "__main__":
app.run(host='0.0.0.0', port=5001, debug=True)
@swdevbali

I love this gist. Clearly show how file uploading works in Flask.
Thanks @dAnjou!

@casatwy
casatwy commented Jun 18, 2014

thanks! @dAnjou

@bhadley
bhadley commented Jul 29, 2014

Thanks!

@wangyu190810

Thanks

@Thuruv
Thuruv commented May 29, 2015

Extremely awesome. Thanks a bunch.

@jzp113
jzp113 commented Sep 7, 2015

I fellow your code ,when I upload file is normal but I can not found the file in dir

@mrmagcore

How do you get form data? When I add a line like this:

item_name = request.form['name']

flask throws a 400

@bakztfuture

@mrmagcore I was able to get the form data by putting name='name' in the html form, as opposed to id='name'. In flask, I was able to access the POST data via name = request.form.getlist('name')

@d33tah
d33tah commented Feb 21, 2016

Thanks! A coding style suggestion: could you change the pattern:

if allowed_filename(...):
    # code

To:

if not allowed_filename(...):
    return error

# code goes unindented here
@gaoning777

thanks a lot, having a hard time to do this

@maratmkhitaryan

fanks you!

@dardan
dardan commented Oct 27, 2016 edited

What if we want to rename a file if it allreade exists ;), here is How

import os
from flask import Flask, request, redirect, url_for
from werkzeug import secure_filename

UPLOAD_FOLDER = '/tmp/'
ALLOWED_EXTENSIONS = set(['txt'])

app = Flask(name)
app.config['UPLOAD_FOLDER'] = UPLOAD_FOLDER

def allowed_file(filename):
return '.' in filename and
filename.rsplit('.', 1)[1] in ALLOWED_EXTENSIONS

def filenamechanger(filename):
if os.path.isfile(os.path.join(app.config['UPLOAD_FOLDER'], filename)):
myfilename = filename.split('.')
if len(myfilename[0].split(''))>2:
a = myfilename[0].split('
')
a.pop()
b = []
for xk in range(0,len(a)):
if xk < len(a)-1:
b.append(''.join([a[xk],'']))
else:
b.append(a[xk])
myfilename[0] = ''.join(b)
print(myfilename)
i = 0
sexer = filename
while os.path.isfile(os.path.join(app.config['UPLOAD_FOLDER'], sexer)):
sexer = ''.join([myfilename[0],'
',str(i),'.',myfilename[1]])
print(sexer)
i+= 1
filename = sexer
return filename
return filename

@app.route("/", methods=['GET', 'POST'])
def index():
if request.method == 'POST':
file = request.files['file']
if file and allowed_file(file.filename):
filename = secure_filename(file.filename)
filename = filenamechanger(filename)
file.save(os.path.join(app.config['UPLOAD_FOLDER'],filename))
print('My Filename: ',filename)
return redirect(url_for('index'))
return """
<!doctype html> <title>Upload new File</title> <h1>Upload new File</h1> <form action="" method=post enctype=multipart/form-data> <p><input type=file name=file> <input type=submit value=Upload> </form> <p>%s</p>""" %<br> "".join(os.listdir(app.config['UPLOAD_FOLDER'],))

if name == 'main':
app.run(
host="0.0.0.0",
port=int("80"),
debug=True
)

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