Reputation: 5952
basically I want to do something quite simple: I want to create a form for deleting entries in a database.
the template is creating a html table with all entries without any trouble. My problem now is: how to convert this to a form with a link in every row.
Of course I could do the manual way of writing html code with a link. But is there a more "flaskish" way? I'm already using wtforms and sqlalchemy
My route:
@app.route('/admin', methods=['GET', 'POST'])
@htpasswd.required
def admin(user):
orders = Order.query.all()
return render_template(
'admin.html', title="AdminPanel", orders=orders
)
The model:
class Order(db.Model):
id = db.Column(db.Integer, primary_key=True)
email = db.Column(db.String(120), index=True, unique=True)
The template:
{% for order in orders %}
<tr>
<td>{{order.email}}</td>
<td><a href="{{ url_for('admin',order_id=order.id, action='delete') }}"><i class="fa fa-trash" aria-hidden="true"></i></a></td>
</tr>
{% endfor %}
Upvotes: 1
Views: 303
Reputation: 719
You should use a different route for deletion and not the same one you are using to render the template. Also you do not need a form for the deletion task. You can use get parameters for that and links like you have tried
add this to your routes:
from flask import redirect, url_for
from app import db # you should import your db or db session instance so you can commit the deletion change this line to wherever your db or db session instance is
@app.route('/delete/<order_id>', methods=['GET', 'POST'])
@htpasswd.required
def delete(order_id):
orders = Order.query.filter(Order.id == order_id).delete()
db.commit()
return redirect(url_for('admin'))
Basically you will perform a delete and then redirect back to the admin route with the above code
Your template file should be changed to:
{% for order in orders %}
<tr>
<td>{{order.email}}</td>
<td><a href="{{ url_for('delete',order_id=order.id) }}"><i class="fa fa-trash" aria-hidden="true"></i></a></td>
</tr>
{% endfor %}
Upvotes: 2