Flask send_file done with JavaScript [duplicate] - javascript

This question already has answers here:
Return a download and rendered page in one Flask response
(2 answers)
Closed 6 months ago.
I have this flask app route:
#app.route('/generatecleanbudgetfile', methods=['GET', 'POST'])
def clean_budget():
file = request.files.get('data_file')
app.logger.info('Budget Formatting request has started')
try:
if request.method == 'POST':
file = request.files.get('data_file')
file.seek(0)
buffer = budget_cleaner(file)
buffer.seek(0)
app.logger.info('Conversion Complete')
return send_file(
buffer,
as_attachment=True,
attachment_filename=f'stripped_budget_{dt.today().strftime("%m.%d.%Y")}.xlsx',
mimetype='application/vnd.openxmlformats-officedocument.spreadsheetml.sheet'
)
except:
app.logger.error(f"Budget formatting failed on {file}")
return render_template('error_type.html', title='Unable to process uploaded budget')
My html:
{% extends "base.html" %}
{% block head %}
{% endblock %}
{% block content %}
<div id="vo_budget_file_settings">
{# Upload Final CRO Budget File #}
<p id="uploadPara">Please upload the final CRO budget File</p>
<form class="" action="/generatecleanbudgetfile" method=POST enctype=multipart/form-data>
<input type="file" name="data_file" accept=".xls, .xlsx, .xlsm"/>
<input type="submit" value="Begin Format" onclick="loading();"/>
</form>
</div>
<!-- funtion to show css spinner on button click -->
<script type="text/javascript">
function loading(){
$(".loader").show();
}
</script>
I understand that flask cannot both render_template and send_file since it can only return a single item.
Question:
How would I go about downloading a file via JavaScript instead so I could use my return to render a new template?
I'm wanting to replace this piece:
return send_file(
buffer,
as_attachment=True,
attachment_filename=f'stripped_budget_{dt.today().strftime("%m.%d.%Y")}.xlsx',
mimetype='application/vnd.openxmlformats-officedocument.spreadsheetml.sheet'
)

To download the file with JS, you could encode it and include it in your template as a string literal inside a JS Blob object, then save that Blob as soon as the page renders. Something like this in your template for success, where you pass your encoded file contents into the file_contents template variable:
<body>
Page content here
<script>
const myBlob = new Blob(["{{file_contents}}"], {type:"your-file's-mime-type"})
// then google "how to save js blob to local file".
// This other post might help: https://stackoverflow.com/questions/25547475/save-to-local-file-from-blob
</script>
</body>
To further illustrate as an example, making a blob that has the plaintext contents "Hello World" would go like new Blob(["Hello World"], {type:"text/plain"}). Your case might be more complicated if you file type isn't plaintext or has a weird encoding, but this is the general idea.
Another (probably less hack-y) idea would be do the same Blob thing except use JS's native fetch API to get the file contents from your server to the Blob instead of passing it through the template. This is similar to the other answer that suggested saving the file on the server, except instead of providing an a tag you just do the download in pure JS.

Flask can't do both things(render & send file) in a single request.
But you can have an alternative solution↓
In your generatecleanbudgetfile request:
Generate a file and save it in server
Provide a file link in your html
just render_template in this request
So, when the user get the response, they can click on download link to retrieve the file generated in the first step.

Related

Connect views.py to a Javascript file

I'm working on a Django project showing PDF files from a dummy database. I have the function below in views.py to show the PDF files.
def pdf_view(request):
try:
with open ('static/someData/PDF', 'rb') as pdf:
response = HttpResponse(pdf.read(), content_type="application/pdf")
response['Content-Disposition'] = 'filename=test1.pdf'
return response
pdf.closed
except ValueError:
HttpResponse("PDF not found")
This function needs get connected to another function located in a javascript file.
How do we connect views.py to another Javascript file?
Finally, I added an "Iframe tag" in the JavaScript file. Then there was no need to use the "def pdf_view(request)" function.

Flask returning a generator and handling it in JavaScript [duplicate]

I have a view that generates data and streams it in real time. I can't figure out how to send this data to a variable that I can use in my HTML template. My current solution just outputs the data to a blank page as it arrives, which works, but I want to include it in a larger page with formatting. How do I update, format, and display the data as it is streamed to the page?
import flask
import time, math
app = flask.Flask(__name__)
#app.route('/')
def index():
def inner():
# simulate a long process to watch
for i in range(500):
j = math.sqrt(i)
time.sleep(1)
# this value should be inserted into an HTML template
yield str(i) + '<br/>\n'
return flask.Response(inner(), mimetype='text/html')
app.run(debug=True)
You can stream data in a response, but you can't dynamically update a template the way you describe. The template is rendered once on the server side, then sent to the client.
One solution is to use JavaScript to read the streamed response and output the data on the client side. Use XMLHttpRequest to make a request to the endpoint that will stream the data. Then periodically read from the stream until it's done.
This introduces complexity, but allows updating the page directly and gives complete control over what the output looks like. The following example demonstrates that by displaying both the current value and the log of all values.
This example assumes a very simple message format: a single line of data, followed by a newline. This can be as complex as needed, as long as there's a way to identify each message. For example, each loop could return a JSON object which the client decodes.
from math import sqrt
from time import sleep
from flask import Flask, render_template
app = Flask(__name__)
#app.route("/")
def index():
return render_template("index.html")
#app.route("/stream")
def stream():
def generate():
for i in range(500):
yield "{}\n".format(sqrt(i))
sleep(1)
return app.response_class(generate(), mimetype="text/plain")
<p>This is the latest output: <span id="latest"></span></p>
<p>This is all the output:</p>
<ul id="output"></ul>
<script>
var latest = document.getElementById('latest');
var output = document.getElementById('output');
var xhr = new XMLHttpRequest();
xhr.open('GET', '{{ url_for('stream') }}');
xhr.send();
var position = 0;
function handleNewData() {
// the response text include the entire response so far
// split the messages, then take the messages that haven't been handled yet
// position tracks how many messages have been handled
// messages end with a newline, so split will always show one extra empty message at the end
var messages = xhr.responseText.split('\n');
messages.slice(position, -1).forEach(function(value) {
latest.textContent = value; // update the latest value in place
// build and append a new item to a list to log all output
var item = document.createElement('li');
item.textContent = value;
output.appendChild(item);
});
position = messages.length - 1;
}
var timer;
timer = setInterval(function() {
// check the response for new data
handleNewData();
// stop checking once the response has ended
if (xhr.readyState == XMLHttpRequest.DONE) {
clearInterval(timer);
latest.textContent = 'Done';
}
}, 1000);
</script>
An <iframe> can be used to display streamed HTML output, but it has some downsides. The frame is a separate document, which increases resource usage. Since it's only displaying the streamed data, it might not be easy to style it like the rest of the page. It can only append data, so long output will render below the visible scroll area. It can't modify other parts of the page in response to each event.
index.html renders the page with a frame pointed at the stream endpoint. The frame has fairly small default dimensions, so you may want to to style it further. Use render_template_string, which knows to escape variables, to render the HTML for each item (or use render_template with a more complex template file). An initial line can be yielded to load CSS in the frame first.
from flask import render_template_string, stream_with_context
#app.route("/stream")
def stream():
#stream_with_context
def generate():
yield render_template_string('<link rel=stylesheet href="{{ url_for("static", filename="stream.css") }}">')
for i in range(500):
yield render_template_string("<p>{{ i }}: {{ s }}</p>\n", i=i, s=sqrt(i))
sleep(1)
return app.response_class(generate())
<p>This is all the output:</p>
<iframe src="{{ url_for("stream") }}"></iframe>
5 years late, but this actually can be done the way you were initially trying to do it, javascript is totally unnecessary (Edit: the author of the accepted answer added the iframe section after I wrote this). You just have to include embed the output as an <iframe>:
from flask import Flask, render_template, Response
import time, math
app = Flask(__name__)
#app.route('/content')
def content():
"""
Render the content a url different from index
"""
def inner():
# simulate a long process to watch
for i in range(500):
j = math.sqrt(i)
time.sleep(1)
# this value should be inserted into an HTML template
yield str(i) + '<br/>\n'
return Response(inner(), mimetype='text/html')
#app.route('/')
def index():
"""
Render a template at the index. The content will be embedded in this template
"""
return render_template('index.html.jinja')
app.run(debug=True)
Then the 'index.html.jinja' file will include an <iframe> with the content url as the src, which would something like:
<!doctype html>
<head>
<title>Title</title>
</head>
<body>
<div>
<iframe frameborder="0"
onresize="noresize"
style='background: transparent; width: 100%; height:100%;'
src="{{ url_for('content')}}">
</iframe>
</div>
</body>
When rendering user-provided data render_template_string() should be used to render the content to avoid injection attacks. However, I left this out of the example because it adds additional complexity, is outside the scope of the question, isn't relevant to the OP since he isn't streaming user-provided data, and won't be relevant for the vast majority of people seeing this post since streaming user-provided data is a far edge case that few if any people will ever have to do.
Originally I had a similar problem to the one posted here where a model is being trained and the update should be stationary and formatted in Html. The following answer is for future reference or people trying to solve the same problem and need inspiration.
A good solution to achieve this is to use an EventSource in Javascript, as described here. This listener can be started using a context variable, such as from a form or other source. The listener is stopped by sending a stop command. A sleep command is used for visualization without doing any real work in this example. Lastly, Html formatting can be achieved using Javascript DOM-Manipulation.
Flask Application
import flask
import time
app = flask.Flask(__name__)
#app.route('/learn')
def learn():
def update():
yield 'data: Prepare for learning\n\n'
# Preapre model
time.sleep(1.0)
for i in range(1, 101):
# Perform update
time.sleep(0.1)
yield f'data: {i}%\n\n'
yield 'data: close\n\n'
return flask.Response(update(), mimetype='text/event-stream')
#app.route('/', methods=['GET', 'POST'])
def index():
train_model = False
if flask.request.method == 'POST':
if 'train_model' in list(flask.request.form):
train_model = True
return flask.render_template('index.html', train_model=train_model)
app.run(threaded=True)
HTML Template
<form action="/" method="post">
<input name="train_model" type="submit" value="Train Model" />
</form>
<p id="learn_output"></p>
{% if train_model %}
<script>
var target_output = document.getElementById("learn_output");
var learn_update = new EventSource("/learn");
learn_update.onmessage = function (e) {
if (e.data == "close") {
learn_update.close();
} else {
target_output.innerHTML = "Status: " + e.data;
}
};
</script>
{% endif %}

Passing parameter(s) to Javascript file from Python Flask app through HTML

I am building a Python/Flask based web app. The python script produces a dictionary of words and their corresponding weights. I have a javascript file (let's call it custom.js), which I call from the output.html. The way this javascript works is that it takes this dictionary and then uses d3.v3.min.js and d3.layout.cloud.js to create a wordcloud. When the dictionary is hard-coded into custom.js, the output file shows the wordcloud. However, the dictionary values will change depending on other parameters in the python script. Therefore, I would like to pass this dictionary from Python to custom.js. I am not sure how to do that.
I know that the parameters could be passed to HTML using {{ params |safe }}, but I am trying to figure out how to do that so that custom.js will receive the parameters (dictionary of words and weights, in this case) and word clouds can be rendered dynamically.
Thank you in advance!
If I understood you correctly you need to create a view function (a route) in the flask backend with url like this /get_dictionary. This function can look like this:
from flask import request, jsonify
...
#app.route('/get_dictionary'):
def get_dictionary():
...
your_dictionary = []
# Fill in your_dictionary with data
...
render_template('your_template.html', your_dictionary=your_dictionary)
EDIT:
You can pass the data from flask to script section of the html template using standard jinja2 notation:
<html>
<head>
<script>
your_dictionary = {{ your_dictionary | tojson }}
<!-- Do what you need with your_dictionary -->
</script>
...
</head>
...
you can try define a var in your template html, like this:
<script>
var your_var = '{{ value }}'
</script>
then use "your_var" in external js file. But please make sure above definition is at ahead of your js file refer.

Dropzone.js prevents Flask from rendering template

I am using Dropzone.js to allow drag and drop upload of CSV files via a Flask web site. The upload process works great. I save the uploaded file to my specified folder and can then use df.to_html() to convert the dataframe into HTML code, which I then pass to my template. It gets to that point in the code, but it doesn't render the template and no errors are thrown. So my question is why is Dropzone.js preventing the render from happening?
I have also tried just return the HTML code from the table and not using render_template, but this also does not work.
init.py
import os
from flask import Flask, render_template, request
import pandas as pd
app = Flask(__name__)
# get the current folder
APP_ROOT = os.path.dirname(os.path.abspath(__file__))
#app.route('/')
def index():
return render_template('upload1.html')
#app.route('/upload', methods=['POST'])
def upload():
# set the target save path
target = os.path.join(APP_ROOT, 'uploads/')
# loop over files since we allow multiple files
for file in request.files.getlist("file"):
# get the filename
filename = file.filename
# combine filename and path
destination = "/".join([target, filename])
# save the file
file.save(destination)
#upload the file
df = pd.read_csv(destination)
table += df.to_html()
return render_template('complete.html', table=table)
if __name__ == '__main__':
app.run(port=4555, debug=True)
upload1.html
<!DOCTYPE html>
<meta charset="utf-8">
<script src="https://rawgit.com/enyo/dropzone/master/dist/dropzone.js"></script>
<link rel="stylesheet" href="https://rawgit.com/enyo/dropzone/master/dist/dropzone.css">
<table width="500">
<tr>
<td>
<form action="{{ url_for('upload') }}", method="POST" class="dropzone"></form>
</td>
</tr>
</table>
EDIT
Here is the sample csv data I am uploading:
Person,Count
A,10
B,12
C,13
Complete.html
<html>
<body>
{{table | safe }}
</body>
</html>
Update: Now you can use Flask-Dropzone, a Flask extension that integrates Dropzone.js with Flask. For this issue, you can set DROPZONE_REDIRECT_VIEW to the view you want to redirect when uploading complete.
Dropzone.js use AJAX to post data, that's why it will not give back the control to your view function.
There are two methods to redirect (or render template) when all files were complete uploading.
You can add a button to redirect.
Upload Complete
You can add an event listener to automatic redirect page (use jQuery).
<script>
Dropzone.autoDiscover = false;
$(function() {
var myDropzone = new Dropzone("#my-dropzone");
myDropzone.on("queuecomplete", function(file) {
// Called when all files in the queue finish uploading.
window.location = "{{ url_for('upload') }}";
});
})
</script>
In view function, add an if statement to check whether the HTTP method was POST:
import os
from flask import Flask, render_template, request
app = Flask(__name__)
app.config['UPLOADED_PATH'] = 'the/path/to/upload'
#app.route('/')
def index():
# render upload page
return render_template('index.html')
#app.route('/upload', methods=['GET', 'POST'])
def upload():
if request.method == 'POST':
for f in request.files.getlist('file'):
f.save(os.path.join('the/path/to/upload', f.filename))
return render_template('your template to render')
Your code does work. Your template will be rendered and returned.
Dropzone will upload files you drag and drop into your browser 'in the background'.
It will consume the response from the server and leave the page as is. It uses the response from the server to know if the upload was successful.
To see this in action:
Navigate to your page
Open up your favourite browser dev tools; (in firefox press CTRL+SHIFT+K)
Select the network tab
Drag your csv into the dropzone pane and note that the request shows in the dev tools network table
Here is a screen shot from my browser. I copied your code as is from your question.
To actually see the rendered complete.html you will need to add another flask endpoint and have a way to navigate to that.
For example:
in upload1.html add:
Click here when you have finished uploading
in init.py change and add:
def upload():
...
# you do not need to read_csv in upload()
#upload the file
#df = pd.read_csv(destination)
#table += df.to_html()
return "OK"
# simply returning HTTP 200 is enough for dropzone to treat it as successful
# return render_template('complete.html', table=table)
# add the new upload_complete endpoint
# this is for example only, it is not suitable for production use
#app.route('/upload-complete')
def upload_complete():
target = os.path.join(APP_ROOT, 'uploads/')
table=""
for file_name in os.listdir(target):
df = pd.read_csv(file_name)
table += df.to_html()
return render_template('complete.html', table=table)
If you are using Flask-Dropzone then:
{{ dropzone.config(redirect_url=url_for('endpoint',foo=bar)) }}

Accessing Django dictionary in Javascript

I'm passing a dictionary from my Django view and want to access the dictionary in my code.
view code:
res.responses is a dictionary
def index(request):
import pprint
pprint.pprint(res.responses)
print 'type = ', type(res.responses)
return render_to_response("deploy/index.html", {"responses":res.responses})
Javascript code:
$(document).ready(function(){
//{% for each in responses%}
// console.log(Hi)
//{% endfor %}
var response = "{{responses}}"
console.log(response)
I tried accessing the variable directly using for loop and also accessing the variable directly. Both throw me an error. Please provide some suggestion.
You can do both.
Option 1:
Provide the script via a template that will send the code with the values. Will look ugly but work. Your javascript file or even the html must be parsed by the django template engine. They can't be static
Option 2:
Provide a new view with a json response, that is accessed from your javascript code (ie: via JQuery)

Categories

Resources