Define static file directory per route (URL) in flask
You have to create the static folder and then you have to add all the folders with the image. You can follow this directory structure:
PROJECT NAME>> static >> 1-directory >> 1.jpg >> 2-directory >> 2.jpg >> 3-directory >> 3.jpg >> 4-directory >> 4.jpg>> templates >> 1.html >> 2.html >> 3.html >> 4.html >> show.html>> venv folder>> app.pyYou can use this code for app.py:
import flaskapp = flask.Flask(__name__)@app.route('/docs/show/<string:i>', methods=['GET'])def show(i): with open('templates/' + str(i) + '.html', 'r') as f: content = f.read() return flask.render_template('show.html', content = content)if __name__=='__main__': app.run('0.0.0.0',port=<your_port_name>)You can keep the 1.html as below:
<h1>Hello</h1><img src="/static/1-directory/1.jpg" />You can keep the 2.html as below:
<h1>Hello</h1><img src="/static/2-directory/2.jpg" />You can keep the 3.html as below:
<h1>Hello</h1><img src="/static/3-directory/3.jpg" />You can keep the 4.html as below:
<h1>Hello</h1><img src="/static/4-directory/4.jpg" />And, you can display your code in show.html (with the same way you have shown it):
<html><body>{{content | safe}}</body></html>EDIT (As per your comments):
I have created the file structure in the following way as per your comments:
PROJECT NAME>> templates >> 1-directory >> 1.jpg >> 2-directory >> 2.jpg >> 3-directory >> 3.jpg >> 4-directory >> 4.jpg >> 1.html >> 2.html >> 3.html >> 4.html >> show.html>> venv folder>> app.pyYou can code for app.py like this:
@app.route('/docs/show/<string:i>', methods=['GET'])def show(i): internal_html = flask.render_template(str(i)+'.html', file_name = str(i)+'-directory/'+str(i)+'.jpg') return flask.render_template('show.html', content = internal_html)@app.route('/serve_file/<path:filename>')def serve_file(filename): return flask.send_from_directory('templates/', filename)All your HTML files will be as shown below:
<h1>Hello</h1><img src="{{ url_for('serve_file', filename=file_name) }}" />Also, your show.html will be the same as the above code.Here, we are using this send_from_directory, because the flask does not serve other than /static folder. Thus, for external use of the folder, we have to use send_from_directory.
I'm not aware of any way that this can be done "in flask". I can think of a couple of ways you would want to do this by hand anyways (validation, authorization, security).
But to answer the question, there are several ways you could go about it, depending on how "magical" you want the solution to be. (I assume that if you want flask to handle it, then you want magic)
One solution would be a decorator that injects the loaded snippet into the view function as a new parameter.
import functoolsdef inject_snippet(func): @functools.wraps(func) def wrapper(i, *args, **kwargs): filepath = f"{i}-directory/{i}.html" with open(filepath) as f: snippet = f.read() return func(snippet, *args, **kwargs) return wrapper@app.route('/docs/show/<int:i>')@inject_snippetdef show(snippet): return render_template('show.html', content=snippet)