With Flask, how can I serve robots.txt and sitemap.xml as static files?
Question:
I’ve read on quiet a few places that serving static files should be left to the server, for example in a couple of the answers on this SO question. But I use the OpenShift PaaS, and can’t figure out how to modify the .htaccess file there.
I came across this piece of code that serves the sitemap from a template. I did that on my app for both the sitemap, and robots.txt, like so –
@app.route("/sitemap.xml")
def sitemap_xml():
response= make_response(render_template("sitemap.xml"))
response.headers['Content-Type'] = 'application/xml'
return response
@app.route("/robots.txt")
def robots_txt():
return render_template("robots.txt")
Is there any harm in this, or is my approach okay?
Answers:
Flask has built in support for serving static files.
Make a /static directory and put your files there. Then, when you instantiate Flask, specify the static_url_path parameter:
app = Flask(__name__, static_url_path='/')
The default is to serve static files from the /static/ path, but you want them served from / so they are where expected.
See the Flask API Docs for more info.
In addition to overhead and unnecessary code, the problem with your approach is if / when one of the files you want to serve contains something that looks like a template tag to render_template — you can cause a rendering error. If you were to read the file into memory (once, not inside the method) then use that string as the body of the response without calling render_template, you would at least avoid that problem.
Put robots.txt and sitemap.xml into your app’s static directory and define this view:
from flask import Flask, request, send_from_directory
@app.route('/robots.txt')
@app.route('/sitemap.xml')
def static_from_root():
return send_from_directory(app.static_folder, request.path[1:])
The best way is to set static_url_path to root url
from flask import Flask
app = Flask(__name__, static_folder='static', static_url_path='')
I’ve read on quiet a few places that serving static files should be left to the server, for example in a couple of the answers on this SO question. But I use the OpenShift PaaS, and can’t figure out how to modify the .htaccess file there.
I came across this piece of code that serves the sitemap from a template. I did that on my app for both the sitemap, and robots.txt, like so –
@app.route("/sitemap.xml")
def sitemap_xml():
response= make_response(render_template("sitemap.xml"))
response.headers['Content-Type'] = 'application/xml'
return response
@app.route("/robots.txt")
def robots_txt():
return render_template("robots.txt")
Is there any harm in this, or is my approach okay?
Flask has built in support for serving static files.
Make a /static directory and put your files there. Then, when you instantiate Flask, specify the static_url_path parameter:
app = Flask(__name__, static_url_path='/')
The default is to serve static files from the /static/ path, but you want them served from / so they are where expected.
See the Flask API Docs for more info.
In addition to overhead and unnecessary code, the problem with your approach is if / when one of the files you want to serve contains something that looks like a template tag to render_template — you can cause a rendering error. If you were to read the file into memory (once, not inside the method) then use that string as the body of the response without calling render_template, you would at least avoid that problem.
Put robots.txt and sitemap.xml into your app’s static directory and define this view:
from flask import Flask, request, send_from_directory
@app.route('/robots.txt')
@app.route('/sitemap.xml')
def static_from_root():
return send_from_directory(app.static_folder, request.path[1:])
The best way is to set static_url_path to root url
from flask import Flask
app = Flask(__name__, static_folder='static', static_url_path='')