上传文件，烧瓶时请求（400）错误

Question

I'm trying to do a file upload to my Flask backend

My Python code

@app.route('/new_upload/', methods=['GET', 'POST'])
@login_required
def upload_file():

    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))
        flash("File uploaded: Thanks!", "success")
        return redirect(url_for('upload.html'))
    return render_template('upload.html', filename=filename)

My HTML looks like this:

{% extends "layout.html" %}
{% from "macros.html" import render_field %}

{% block content %}
<form action="" method=post enctype=multipart/form-data>
      <p><input type=file name=file>
         <input type=submit value=Upload>
</form>
{% endblock %}

In home page when I clicked on upload file link browser show me

Bad Request

The browser (or proxy) sent a request that this server could not understand.

For clearance Home page HTML and image are attached bellow

 <div class="main">

        <nav>

          <a href="{{ url_for('index') }}">All</a>

          {% if current_user.is_authenticated %}
          <a href="{{ url_for('stream', username=current_user.username) }}"> Following</a>
          <a href="{{ url_for('post') }}" class="new">Create New Post</a>
          <a href="{{ url_for('upload_file') }}" class="new">Upload file</a>
          {% endif %}

        </nav>

        {% block content %}{% endblock %}

      </div>

Home Page

After click

Please try help me, I am just learning

Answer 1

In this piece of code

return redirect(url_for('upload.html'))

You should change url_for('upload.html') to url_for('upload') or what is suppose to be the name of the function instead of the html template.

Also if you are about to use the same function "def upload_file()" for HTTP GET and HTTP POST requests, then you should specify the piece of code that would be executed on post and another piece that would be executed when only GET request is performed. Something like:

# Import request if you haven't.
from flask import request 

@app.route('/new_upload', methods=['GET', 'POST'])
@login_required
def upload_file():

    if request.method == 'POST':
        # This will be executed on POST request.
        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))
            flash("File uploaded: Thanks!", "success")
            return redirect(url_for('upload_file'))

    # This will be executed on GET request.
    return render_template('upload.html')

I haven't tested the above code, but this should be the approach if you use one function for GET and POST http request. If you do not differentiate the upload functionality (on POST HTTP request) and rendering the template (on GET request) it would try to execute all the code on every request and would fall in loop where would return redirect(url_for('upload_file')) every time and would not get to the return render_template('upload.html') where is suppose to show you page (HTTP request with code 200 instead of code 400).

You can strictly follow this example: http://flask.pocoo.org/docs/0.10/patterns/fileuploads/ to get the overal idea.

You can also have a look at the HTTP Methods: https://en.wikipedia.org/wiki/Hypertext_Transfer_Protocol#Request_methods so you would have better overview what is POST and GET request.