Remove href params from address bar.

Heya,

To achieve this in Flask, you don’t typically pass the template name through the URL as you’re doing. Instead, you should separate the routing logic from the template rendering.

Here’s a simple refactoring of your situation:

1. Flask Route

In your Flask app, you’d define a route for ‘pictures’ that will render the desired template:

from flask import Flask, render_template

app = Flask(__name__)

@app.route('/pictures')
def pictures():
    return render_template('pictures.html')

This sets up a route /pictures which will render the pictures.html template when accessed.

2. Navbar Link

In your HTML (probably within a base template if you’re using one), your link to the pictures page would look like:

<li class="nav-item">
    <a class="nav-link" href="{{ url_for('pictures') }}">Pictures</a>
</li>

By doing this, the resulting link will be http://127.0.0.1:5000/pictures without any extra query parameters.

3. General Tips

• The Flask render_template function looks for templates in a folder named templates by default. So you typically don’t include templates/ in the template name.
• url_for takes the name of the function (the view function in Flask terminology) as its argument, not the template filename. Hence url_for('pictures') corresponds to the pictures function in your Flask app.
• Ensure you have a templates folder in the same directory as your Flask script, and that’s where you put your pictures.html file.
• Always separate routing logic (which view function to execute) from rendering logic (which template to use) for cleaner and more maintainable code.