Return server.log file on server using Flask [duplicate]
0
This question already has an answer here:
How to serve static files in Flask
13 answers
Display the contents of a log file as it is updated
2 answers
I am using python 3 and flask server. This server can browse a file and upload it to a folder called 'uploads'
After all processing the result is saved to 'server.log'
I want to display result to android app. For that, I want to return server.log to server so it can be accessed by an android app.
Any kind of help is appreciated. Thanks in advance.
Here is the code:
import os
from flask import Flask, request, redirect, url_for, send_from_directory, jsonify
from werkzeug import secure_filename
UPLOAD_FOLDER = 'uploads'
ALLOWED_EXTENSIONS = set(['txt', 'pdf', 'png', 'jpg', 'jpeg', 'mp4'])
app = Flask(__name__)
app.config['UPLOAD_FOLDER'] = UPLOAD_FOLDER
def allowed_file(filename):
# this has changed from the original example because the original did not work for me
return filename[-3:].lower() in ALLOWED_EXTENSIONS
@app.route('/', methods=['GET', 'POST'])
def upload_file():
print("Printing the request {}".format(request.headers))
print("Printing the request {}".format(request))
print("Printing the request {}".format(request.data))
print("Printing the request {}".format(request.form))
if request.method == 'POST':
file = request.files['file']
if file and allowed_file(file.filename):
print('**found file', file.filename)
filename = secure_filename(file.filename)
file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
# for browser, add 'redirect' function on top of 'url_for'
return jsonify({"success" : url_for('uploaded_file',
filename=filename)})
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>
'''
@app.route('/uploads/<filename>') #this part of server.py is missing from video attendance output.py
def uploaded_file(filename):
return send_from_directory(app.config['UPLOAD_FOLDER'], filename)
if __name__ == '__main__':
app.run(host= '192.168.0.102', debug=True)
python-3.x flask
asked Nov 23 '18 at 6:06
Afshan Anwarali
224
marked as duplicate by davidism flask
Users with the flask badge can single-handedly close flask questions as duplicates and reopen them as needed.
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Nov 23 '18 at 14:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
0
This question already has an answer here:
How to serve static files in Flask
13 answers
Display the contents of a log file as it is updated
2 answers
I am using python 3 and flask server. This server can browse a file and upload it to a folder called 'uploads'
After all processing the result is saved to 'server.log'
I want to display result to android app. For that, I want to return server.log to server so it can be accessed by an android app.
Any kind of help is appreciated. Thanks in advance.
Here is the code:
import os
from flask import Flask, request, redirect, url_for, send_from_directory, jsonify
from werkzeug import secure_filename
UPLOAD_FOLDER = 'uploads'
ALLOWED_EXTENSIONS = set(['txt', 'pdf', 'png', 'jpg', 'jpeg', 'mp4'])
app = Flask(__name__)
app.config['UPLOAD_FOLDER'] = UPLOAD_FOLDER
def allowed_file(filename):
# this has changed from the original example because the original did not work for me
return filename[-3:].lower() in ALLOWED_EXTENSIONS
@app.route('/', methods=['GET', 'POST'])
def upload_file():
print("Printing the request {}".format(request.headers))
print("Printing the request {}".format(request))
print("Printing the request {}".format(request.data))
print("Printing the request {}".format(request.form))
if request.method == 'POST':
file = request.files['file']
if file and allowed_file(file.filename):
print('**found file', file.filename)
filename = secure_filename(file.filename)
file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
# for browser, add 'redirect' function on top of 'url_for'
return jsonify({"success" : url_for('uploaded_file',
filename=filename)})
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>
'''
@app.route('/uploads/<filename>') #this part of server.py is missing from video attendance output.py
def uploaded_file(filename):
return send_from_directory(app.config['UPLOAD_FOLDER'], filename)
if __name__ == '__main__':
app.run(host= '192.168.0.102', debug=True)
python-3.x flask
asked Nov 23 '18 at 6:06
Afshan Anwarali
224
marked as duplicate by davidism flask
Users with the flask badge can single-handedly close flask questions as duplicates and reopen them as needed.
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Nov 23 '18 at 14:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
0
0
0
This question already has an answer here:
How to serve static files in Flask
13 answers
Display the contents of a log file as it is updated
2 answers
I am using python 3 and flask server. This server can browse a file and upload it to a folder called 'uploads'
After all processing the result is saved to 'server.log'
I want to display result to android app. For that, I want to return server.log to server so it can be accessed by an android app.
Any kind of help is appreciated. Thanks in advance.
Here is the code:
import os
from flask import Flask, request, redirect, url_for, send_from_directory, jsonify
from werkzeug import secure_filename
UPLOAD_FOLDER = 'uploads'
ALLOWED_EXTENSIONS = set(['txt', 'pdf', 'png', 'jpg', 'jpeg', 'mp4'])
app = Flask(__name__)
app.config['UPLOAD_FOLDER'] = UPLOAD_FOLDER
def allowed_file(filename):
# this has changed from the original example because the original did not work for me
return filename[-3:].lower() in ALLOWED_EXTENSIONS
@app.route('/', methods=['GET', 'POST'])
def upload_file():
print("Printing the request {}".format(request.headers))
print("Printing the request {}".format(request))
print("Printing the request {}".format(request.data))
print("Printing the request {}".format(request.form))
if request.method == 'POST':
file = request.files['file']
if file and allowed_file(file.filename):
print('**found file', file.filename)
filename = secure_filename(file.filename)
file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
# for browser, add 'redirect' function on top of 'url_for'
return jsonify({"success" : url_for('uploaded_file',
filename=filename)})
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>
'''
@app.route('/uploads/<filename>') #this part of server.py is missing from video attendance output.py
def uploaded_file(filename):
return send_from_directory(app.config['UPLOAD_FOLDER'], filename)
if __name__ == '__main__':
app.run(host= '192.168.0.102', debug=True)
python-3.x flask
asked Nov 23 '18 at 6:06
Afshan Anwarali
224
This question already has an answer here:
How to serve static files in Flask
13 answers
Display the contents of a log file as it is updated
2 answers
I am using python 3 and flask server. This server can browse a file and upload it to a folder called 'uploads'
After all processing the result is saved to 'server.log'
I want to display result to android app. For that, I want to return server.log to server so it can be accessed by an android app.
Any kind of help is appreciated. Thanks in advance.
Here is the code:
import os
from flask import Flask, request, redirect, url_for, send_from_directory, jsonify
from werkzeug import secure_filename
UPLOAD_FOLDER = 'uploads'
ALLOWED_EXTENSIONS = set(['txt', 'pdf', 'png', 'jpg', 'jpeg', 'mp4'])
app = Flask(__name__)
app.config['UPLOAD_FOLDER'] = UPLOAD_FOLDER
def allowed_file(filename):
# this has changed from the original example because the original did not work for me
return filename[-3:].lower() in ALLOWED_EXTENSIONS
@app.route('/', methods=['GET', 'POST'])
def upload_file():
print("Printing the request {}".format(request.headers))
print("Printing the request {}".format(request))
print("Printing the request {}".format(request.data))
print("Printing the request {}".format(request.form))
if request.method == 'POST':
file = request.files['file']
if file and allowed_file(file.filename):
print('**found file', file.filename)
filename = secure_filename(file.filename)
file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
# for browser, add 'redirect' function on top of 'url_for'
return jsonify({"success" : url_for('uploaded_file',
filename=filename)})
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>
'''
@app.route('/uploads/<filename>') #this part of server.py is missing from video attendance output.py
def uploaded_file(filename):
return send_from_directory(app.config['UPLOAD_FOLDER'], filename)
if __name__ == '__main__':
app.run(host= '192.168.0.102', debug=True)
This question already has an answer here:
How to serve static files in Flask
13 answers
Display the contents of a log file as it is updated
2 answers
python-3.x flask
python-3.x flask
asked Nov 23 '18 at 6:06
Afshan Anwarali
224
asked Nov 23 '18 at 6:06
Afshan Anwarali
224
asked Nov 23 '18 at 6:06
Afshan Anwarali
224
asked Nov 23 '18 at 6:06
Afshan Anwarali
224
asked Nov 23 '18 at 6:06
Afshan Anwarali
224
224
marked as duplicate by davidism flask
Users with the flask badge can single-handedly close flask questions as duplicates and reopen them as needed.
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Nov 23 '18 at 14:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by davidism flask
Users with the flask badge can single-handedly close flask questions as duplicates and reopen them as needed.
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Nov 23 '18 at 14:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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