I am new to Django. I have read the official documentations and managed to create models in Django. I have a task to create an estimator tool that could provide the estimate(cost) to the user when he selects the hardware/softwares etc. This could be something similar to this http://calculator.s3.amazonaws.com/index.html. This is only for reference but I want to add details dynamically.
Basically I am curious to know about:
How to add rows in Django?
How is the 'Monthly cost' working in amazon calculator?
I want to achieve something similar to this dynamic estimations.
Any pointers/articles/suggestions will be helpful in my learning and analysis.
You have to override the Django admin template for that.
How to add button next to Add User button in Django Admin Site
for auto-selecting the group value you can use
https://github.com/crucialfelix/django-ajax-selects
Related
I know how to make it possible to create/edit/delete a multiple-row database model through django admin page. But let's say I want to make it possible to modify the single banner image of my site through admin page, what am I supposed to do then?
Should I create an entire database model for one item? Or is there a better way to implement this?
You can either create your own model for these types of settings, or you can use one of the many packages designed to handle this type of situation:
https://djangopackages.org/grids/g/live-setting/
First sorry for my poor English.
I came from asp.net mvc. Now I use django with nanoboxio
In asp, I can create sections like below.
Create model and add it to dbcontext
Right click controller folder and create new controller with views.
Modify as you wish
For now, I know Django can create admin interface for your model. I try and happy with it. but i want to develop it.
I want to for example;
Create a post model.
Create a admin interface for it.
Copy generated admin interface controller and views to another app
Modify it
How can I do that?
Django's MVC is quite different to ASP.
Django's MVC pattern is less strict so you sort of combine the view and the controller in the views.py. However, if you want to change the Admin, the Django docs are quite nice here: docs.djangoproject.com
If you want to create a custom admin functionality the docs should give you a first idea and if you're planning to create a blog, I would advice you to use an existing plugin such as Zinnia. There, you can find the desired functionalities and modify them instead of building them from scratch.
Also, there are a couple of tutorials on how to build reusable apps and they usually include a detailed guideline how to implement admin functionalities there. Just look it up on google.
I hope that helps you!
We have a project in which we have to use 3 different backend frameworks/programming language and integrate them together to form one correlated website: Java(Spring MVC) for the customer end website, PHP(CodeIgniter) for the service provider website, and Python(Django) for the admin site.
My question is, can I use the admin site of Django alone? I will only use it for accepting user registrations, deactivation, and tracking transactions. So it's all just going to be basic CRUD functions for the database. Or is there an alternative framework for me to use? Thank you.
You'd need to define Models for the objects / tables used by other frameworks (which should be possible using Model Meta attributes to map it to an existing table name: https://docs.djangoproject.com/en/1.11/ref/models/options/).
The options you'll definitely want to use are db_table='existing table name' and managed=False (to tell Django not to prepare migrations for this table).
But as long as you can get a working Django Model that can interact with the tables for the other backends you should be able to read / update those records with the Admin interface just as if they were native Django models.
I am new to Django as well as Python. I am trying to build a Employee management System using django framework.
Till now I was able to build models and show the models in admin module. Now I want to show user a user profile and able to update their details.
I am still in starting phase of the project. What I want to do is, I want show the profile page in the below format i.e similar to admin page look.
But till now I was able to get like below
Is it possible to make profile page of each employee look just like admin page look? which has collapsible options as well
You will need to write your views as well as create you template folder with the relevant template tag.
I believe that is what will get you what you want.
The models you create shows the database columns and rows you want Django to create.
Please check the link below assuming you are using Django 1.6:
https://docs.djangoproject.com/en/1.6/intro/tutorial03/
The collapsible part of Django admin, is made with JavaScript (jQuery) & CSS.
If you want your page to be dynamic, use JavaScript.
If you want to make it more beautiful, use CSS.
Is there any ready apps for django admin, that allows to edit model in popup?
I want next functionallity:
View edit form for model in popup.
On model save - update row in list of models.
Motivation: reduce page reloads.
Also, if there any solutions oriented on massive manual data updates for django? I've taken a look at django grappelli - it improves view of data, but edit data is still not usable.
P.S.: If such kind of app is not available - I'll start open source project.
If you want to open a popup, simply create a link to your 'add' view with the following attribute on that link onclick='return showAddAnotherPopup(this);'
You can do most of what you ask there (at least point 1 and 2) using the django built in admin customisations.
Have a look at https://docs.djangoproject.com/en/dev/ref/contrib/admin/
The django admin itself already uses some things similar to this, pay special attention to the django _popup=1 variable in the request URI.
You will have to add a custom modelname_change_list.html file to provide some javascript and in the ModelAdmin override the delet_view, change_view, response_add and potentially response_change.