How to do POST using requests module with Flask server?
Question:
I am having trouble uploading a file to my Flask server using the Requests module for Python.
import os
from flask import Flask, request, redirect, url_for
from werkzeug import secure_filename
UPLOAD_FOLDER = '/Upload/'
app = Flask(__name__)
app.config['UPLOAD_FOLDER'] = UPLOAD_FOLDER
@app.route("/", methods=['GET', 'POST'])
def index():
if request.method == 'POST':
file = request.files['file']
if file:
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', debug=True)
I am able to upload a file via web page, but I wanted to upload file with the requests module like this:
import requests
r = requests.post('http://127.0.0.1:5000', files={'random.txt': open('random.txt', 'rb')})
It keeps returning 400 and saying that “The browser (or proxy) sent a request that this server could not understand”
I feel like I am missing something simple, but I cannot figure it out.
Answers:
You upload the file as the random.txt
field:
files={'random.txt': open('random.txt', 'rb')}
# ^^^^^^^^^^^^ this is the field name
but look for a field named file
instead:
file = request.files['file']
# ^^^^^^ the field name
Make those two match; using file
for the files
dictionary, for example:
files={'file': open('random.txt', 'rb')}
Note that requests
will automatically detect the filename for that open fileobject and include it in the part headers.
Because you have <input>
with name=file
so you need
files={'file': ('random.txt', open('random.txt', 'rb'))}
Examples in requests doc: POST a Multipart-Encoded File
I am having trouble uploading a file to my Flask server using the Requests module for Python.
import os
from flask import Flask, request, redirect, url_for
from werkzeug import secure_filename
UPLOAD_FOLDER = '/Upload/'
app = Flask(__name__)
app.config['UPLOAD_FOLDER'] = UPLOAD_FOLDER
@app.route("/", methods=['GET', 'POST'])
def index():
if request.method == 'POST':
file = request.files['file']
if file:
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', debug=True)
I am able to upload a file via web page, but I wanted to upload file with the requests module like this:
import requests
r = requests.post('http://127.0.0.1:5000', files={'random.txt': open('random.txt', 'rb')})
It keeps returning 400 and saying that “The browser (or proxy) sent a request that this server could not understand”
I feel like I am missing something simple, but I cannot figure it out.
You upload the file as the random.txt
field:
files={'random.txt': open('random.txt', 'rb')}
# ^^^^^^^^^^^^ this is the field name
but look for a field named file
instead:
file = request.files['file']
# ^^^^^^ the field name
Make those two match; using file
for the files
dictionary, for example:
files={'file': open('random.txt', 'rb')}
Note that requests
will automatically detect the filename for that open fileobject and include it in the part headers.
Because you have <input>
with name=file
so you need
files={'file': ('random.txt', open('random.txt', 'rb'))}
Examples in requests doc: POST a Multipart-Encoded File