flask bootstrap route 404 [duplicate]

Question

im new to flask and boostrap,im trying to make a button that link to another html

Project layout

-Norman
  -args
    -static
      -css
        -bootstrap.min.css
        -style.css
      -img
        -wall.jpg
    -templates
      -index.html
      -form.html
    -__init__.py
    -routes.py
 -.flaskenv 
 -readme.txt   
 -book.py

route.py:

from args import app # init.py
from flask import render_template
@app.route('/')
def index():
return render_template('index.html')
@app.route('/form')
def form():
return render_template('form.html')

__init__.py:

from flask import Flask
from flask_bootstrap import Bootstrap

app = Flask(__name__)
Bootstrap(app)

from args import routes

here is the code from index.html:

<div>
    <button class="submit-btn">Check availability</button>
    <a href="/form.html" class="btn btn-primary pull-right" role="button">check form</a>
</div>

im trying to link the button to open form.html with href but it says the requested url was not found on the server. opening localhost:5000 worked but the second route doesnt working

shmsr · Accepted Answer · 2019-12-18 17:10:13Z

0

In your index.html: Replace /form.html to /form as you have registered the handle /form which when triggered, would render form.html

<div>
    <button class="submit-btn">Check availability</button>
    <a href="/form" class="btn btn-primary pull-right" role="button">check form</a>
</div>

answered Dec 18, 2019 at 17:10

shmsr

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1 Comment

edward wibowo Over a year ago

i tried that but still doesnt work, anyway i just solve the problem by renaming form. not sure what happen

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