I am using Flask as a web framework, and I am trying to implement the first example from the book Getting Started with D3, by Mike Dewar. I have a Python script named run.py and two directories, templates/ and static/, containing index.html and service_status.json, respectively. Unfortunately, my code is not rendering the data at all, nor is it producing any glaring errors.
This is what I have in run.py:
#!/usr/bin/env python
from flask import Flask, render_template, url_for
app = Flask(__name__)
#app.route('/')
def index():
return render_template('index.html')
if __name__=="__main__":
port = 5000
app.debug = True
app.run( port=port )
This is what I have in templates/index.html:
<!DOCTYPE HTML>
<HTML>
<HEAD>
<META CHARSET="utf-8">
<SCRIPT SRC="http://d3js.org/d3.v3.min.js"></SCRIPT>
<SCRIPT>
function draw(data) {
"use strict";
d3.select("body")
.append("ul")
.selectAll("li")
.data(data)
.enter()
.append("li")
.text( function(d){
return d.name + ": " + d.status;
}
);
}
</SCRIPT>
<TITLE>MTA Data</TITLE>
</HEAD>
<BODY>
<H1>MTA Availability Data</H1>
<SCRIPT>
d3.json("{{ url_for( 'static', filename='service_status.json') }}",draw); // <---- BIG PROBLEM
</SCRIPT>
</BODY>
</HTML>
I am using Windows 7, Google Chrome, and Python 2.7.
If the JSON file is not going to change, then you should put it in the static directory and use
from flask import url_for
url_for('static', filename='service_status.json')
For this to work, also change the path in the JavaScript to '/static/service_status.json'
Static files like your json document, are by default served from a different directory from the templates - by default 'static'
You dont need to use the url_for call in your view, you can use it in your template:
d3.json("{{ url_for('static', filename='service_status.json') }}",draw);
So to summarise: 1) Move your json document in to the static folder (a folder called static along side your templates folder, by default), and 2) use the url_for call in your template to get the correct URI for your json document.
If you want to use a folder other than static, you can change that by passing static_folder to the Flask object contructor
You seem to be getting a 304 status code as you mentioned in earlier comments. I see that your JSON has the following date/time:
"Date": [
"12/15/2011"
],
"Time": [
" 7:35AM"
],
I am not 100% sure but this might help:
http://www.w3.org/Protocols/HTTP/HTRQ_Headers.html#if-modified-since
Basically, it says that
"This request header is used with GET method to make it conditional: if the requested document has not changed since the time specified in this field the document will not be sent, but instead a Not Modified 304 reply.
Format of this field is the same as for Date:"
So, may be you can check the timestamp on the JSON and probably just do a fresh save ?
Related
I am trying to create a very simple one-page Flask application for a python script that I have. The script requires multiple user inputs in a for-loop with the number of loops being user input as well.
Here is the code in my script to make it more clear:
def shared_books():
import requests as re
from bs4 import BeautifulSoup
import time
num_lists = int(input('Enter the number of lists you would like to search:'))
urls = []
page_counts = []
for i in range(num_lists):
urls.append(input(f'Enter the url for list {i + 1}:'))
page_counts.append(int(input(f'Enter the number of pages for list {i + 1}:')))
I want a simple HTML that will ask the user for the number of lists, then the URL and page count for each list as is shown in my function. Then it will run the entire function.
The HTML code I have right now is super simple and I don't want much else outside of the input parts:
<html>
<head>
<title>Goodreads-App</title>
</head>
<body>
<h1>Welcome to my app!</h1>
<<p>This app will allow you to see books that are
shared between multiple lists on goodreads</p>
</body>
</html>
Please let me know how I can set up this application!
Firstly, I suggest you take a look at the Flask docs. You are doing it right in terms of having a view function, but the input() python keyword doesn't work like that in Flask. Instead, you should render an html template which you can then put your form input field into. Here is an example:
from flask import Flask, render_template
#flask initialising stuff, read docs for info
#app.route("/home")
def home():
return render_template("home.html")
Flask runs on your computer's local server "localhost", which is not publicly accessible. It conventionally runs on port 5000, which gives the name "localhost:5000".
When someone visits "localhost:5000/home", flask will look for a file called "home.html" in a pre-designated templates folder – the default is a directory called "templates" which you should put your html files into.
So if this is your "home.html" file:
<html>
<head>
<title>Goodreads-App</title>
</head>
<body>
<h1>Welcome to my app!</h1>
<p>This app will allow you to see books that are
shared between multiple lists on goodreads</p>
</body>
</html>
When you load the page associated with a specific function, it will return a template which is rendered as html. The above should look something like this:
And that is how to start.
Thank you for the answers! I haven't quite solved the previous issue but have approached it from a different angle which is working now! I will potentially post again if I don't solve it.
I am using flask forms to do what I was trying.
I have written a piece of code that matches if the string has anything of the pattern text.someExtension, In my case, it would be fileName.png in the string, converts it into an img tag and displays on the HTML file using python and flask. Let us take the example string:
"What is the output of this program? e.png"
the code matches e.png and it then replaces e.png by
"<br><img src="{{url_for('static', filename='pics/e.png')}}" /><br>"
The image e.png is put in the folder pics inside the static folder.
If this string is pushed into a flask variable even by adding Markup() to it it isn't rendering the image but showing the following output.
output on html page
why is it so? Any way to make it display the image e.png?
my code is:
import re
from flask import Flask, Markup, render_template
app = Flask(__name__)
def rep(x):
text = re.findall(r'\w+[.]\w+', x)
for i in text:
b ="<img src=\"{{ url_for('static',filename='pics/"+i+"')}}\">"
x=x.replace(i,b)
return x
#app.route('/')
def home():
a = "What is the output of this program? e.png"
a = rep(a)
return render_template('new.html',var=Markup(a))
if __name__ == '__main__':
app.run(debug=True, host='localhost', port=8032)
And the HTML file is,
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Title</title>
</head>
<body>
{{var}}
</body>
</html>
The problem is that the value you're passing to your template is a string and even though the string you're inserting is formatted with the {{ brackets, flask doesn't interpret those. Notice if you look at the html being served, it actually contains the string 'url_for'...
You're also not even importing the url_for function from flask, so that wouldn't have worked anyway.
The solution:
import re
from flask import Flask, url_for, Markup, render_template
app = Flask(__name__)
def rep(x):
text = re.findall(r'\w+[.]\w+', x)
for i in text:
b ="<img src='" + url_for('static', filename='pics/'+i) + "'>"
x=x.replace(i,b)
return x
#app.route('/')
def home():
a = "What is the output of this program? e.png"
a = rep(a)
return render_template('new.html', var=Markup(a))
#app.route('/static/<path:path>')
def static_file(path):
return send_from_directory('static', path)
if __name__ == '__main__':
app.run(debug=True, host='localhost', port=8032)
The html file can remain unchanged. This solution
separately tells the flask server to listen and serve the static files under the /static page.
Creates the html with the image url by string concatenation, instead of trying to use template rendering.
This question already has answers here:
How to serve static files in Flask
(24 answers)
Link to Flask static files with url_for
(2 answers)
Closed 4 years ago.
I'm pretty new to python, even less experienced with flask, and I cannot figure out this issue. I have the following simple web page with jQuery functionality that works great when I double click the file and open it in a browser:
<!DOCTYPE html>
<html>
<head>
<script src="jquery-3.3.1.js"></script>
</head>
<body>
<script type="text/javascript">
$(document).ready(function() {
$("#updateBtn").on("click", function() {
text = "<h2>The div has been updated!</h2>";
$("#jQuery_div").html(text);
});
});
</script>
<div>
<h1>This is a non-jQuery div</h1>
</div>
<div id="jQuery_div">
<h2>This div should update with jQuery</h2>
</div>
<button id="updateBtn">update</button>
</body>
</html>
However, when flask delivers the web page on localhost:5000, the jQuery functionality is no longer present. My python is as follows:
from flask import Flask, render_template
app = Flask(__name__)
#app.route('/')
def render():
return render_template("jquery_test.html")
if __name__ == "__main__":
app.run(port=5000, debug=True)
My app's file tree is:
/AJAX_practice
ajax_practice.py
/templates
jquery-3.3.1.js
jquery_test.html
I was trying to follow this tutorial when I couldn't get the "echo" button to work. In my efforts to debug, I have slowly chipped away and drastically simplified the program to the above code to see why I cannot get my jQuery to work through flask. I am still at a loss. I am running the flask app by pressing F5 in IDLE, with no errors in Python 2.7.13 Shell, and the Terminal (from which I started IDLE with $ sudo idle) showing:
my ip - - [date and time] "GET / HTTP/1.1" 200 -
my ip - - [date and time] "GET /jquery-3.3.1.js HTTP/1.1" 404 -
From this, my best guess is that flask cannot find the jquery.3.3.1.js file, though I have tried putting it everywhere in the file tree with no luck. I cannot use the script src to https for jQuery dependencies, as my server will eventually be on a non-internet connected LAN. Am I on the right track? If so, how does flask find and/or navigate jQuery dependencies? Can anyone point me towards some documentation that might help my fundamental understanding of this issue?
Any help on this matter would be greatly appreciated.
You are trying to serve JavaScript file from templates folder. Add a static folder and use that to serve static content.
in your case create a directory structure like "static/js/jquery.min.js"
and then add script reference like this
<script src="{{url_for('static', filename='js/jquery.min.js')}}"></script>
See this :
http://exploreflask.com/en/latest/static.html
If you don't want to keep it in "static" folder and use another local directory you can use send_from_directory as shown in this example :
https://stackoverflow.com/a/20648053/2118215
This has always worked for me with Flask in the past:
<script src="{{ url_for('static', filename='jquery-3.3.1.js') }}"></script>
'static' is the name of the folder it's in (and the 'static' folder is in the root of my project). You can edit this to suit your preferred structure and naming, so change 'static' to 'templates' if that's where you'd rather keep your jquery file, although I would recommend keeping it in a separate folder from your HTML templates, purely in the interests of keeping your project well organised.
I believe the path to jquery should be /templates/jquery-3.3.1.js
On me flask server when i serve jquery it has the full path from the home directory: /static/js/jquery.min.js
I have this issue with Flask when i try running this code with Flask :
#app.route('/viz')
def root():
return render_template('page.html')
The file 'page.html' is in the templates folder, and contains some lines like this :
<link rel="stylesheet" href="static/main.css">
But when I execute my program, the file doen't look for main.css in the static directory, but in the viz/static direcory ( /viz is the route of the view).
How can I solve this issue ?
Thanks and sorry for my english.
You need a leading slash to tell the browser to use an absolute path.
<link rel="stylesheet" href="/static/main.css">
You need to place the static folder within the package or next to your module. See the Flask Quickstart documentation for more details. Without knowing more about your structure, my guess is your tree should look something like this:
- top level directory (e.g. your package "root")
| app.py
| static/
| viz/
I'm prototyping an idea for a website that will use the HTML5 offline application cache for certain purposes. The website will be built with Python and Flask and that's where my main problem comes from: I'm working with those two for the first time, so I'm having a hard time getting the manifest file to work as expected.
The issue is that I'm getting 404's from the static files included in the manifest file. The manifest itself seems to be downloaded correctly, but the files that it points to are not. This is what is spit out in the console when loading the page:
Creating Application Cache with manifest http://127.0.0.1:5000/static/manifest.appcache offline-app:1
Application Cache Checking event offline-app:1
Application Cache Downloading event offline-app:1
Application Cache Progress event (0 of 2) http://127.0.0.1:5000/style.css offline-app:1
Application Cache Error event: Resource fetch failed (404) http://127.0.0.1:5000/style.css
The error is in the last line.
When the appcache fails even once, it stops the process completely and the offline cache doesn't work.
This is how my files are structured:
sandbox
offline-app
offline-app.py
static
manifest.appcache
script.js
style.css
templates
offline-app.html
This is the content of offline-app.py:
from flask import Flask, render_template
app = Flask(__name__)
#app.route('/offline-app')
def offline_app():
return render_template('offline-app.html')
if __name__ == '__main__':
app.run(host='0.0.0.0', debug=True)
This is what I have in offline-app.html:
<!DOCTYPE html>
<html manifest="{{ url_for('static', filename='manifest.appcache') }}">
<head>
<title>Offline App Sandbox - main page</title>
</head>
<body>
<h1>Welcome to the main page for the Offline App Sandbox!</h1>
<p>Some placeholder text</p>
</body>
</html>
This is my manifest.appcache file:
CACHE MANIFEST
/style.css
/script.js
I've tried having the manifest file in all different ways I could think of:
CACHE MANIFEST
/static/style.css
/static/script.js
or
CACHE MANIFEST
/offline-app/static/style.css
/offline-app/static/script.js
None of these worked. The same error was returned every time.
I'm certain the issue here is how the server is serving up the files listed in the manifest. Those files are probably being looked up in the wrong place, I guess. I either should place them somewhere else or I need something different in the cache manifest, but I have no idea what. I couldn't find anything online about having HTML5 offline applications with Flask.
Is anyone able to help me out?
I would have thought this one would work:
CACHE MANIFEST
/static/style.css
/static/script.js
But in any case, you should not hardcode the URLs for your static files. It's best to serve the manifest as a template (moved to the "templates" folder) so that you can use url_for to generate the path to the static files, something like this:
CACHE MANIFEST
{{ url_for('static', filename='style.css') }}
{{ url_for('static', filename='script.js') }}
Then in your HTML template you would have a reference to a route instead of a static file:
<html manifest="{{ url_for('manifest') }}">
And finally, you would add a new view function that returns the manifest:
from flask import make_response
#app.route('/manifest')
def manifest():
res = make_response(render_template('manifest.appcache'), 200)
res.headers["Content-Type"] = "text/cache-manifest"
return res