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How to attach a python file to each row(i.e. each data entry) in database formed using Django?

Ive used default django superuser to create an admin and created a DB using it. But now in that i want to add a python script to each data entry i.e. row. How do i do this???
Nothing special. Used basic Django.

There doesn't seem to be anything particularly complex here. You're just asking if you can use a field on the model to choose something; of course you can.
For instance:
# actions.py
def thing1(obj):
# do something with obj
def thing2(obj):
# do something else with obj
# models.py
from . import actions
ACTIONS = {
"do_thing_1": actions.thing1,
"do_thing_2": actions.thing2,
}
ACTION_CHOICES = [(action, action) for action in ACTIONS]
class MyModel(models.Model):
action = models.CharField(max_length=20, choices=ACTION_CHOICES)
def do_action(self):
return ACTIONS[self.action](self)

Extend django-import-export's import form to specify fixed value for each imported row

I am using django-import-export 1.0.1 with admin integration in Django 2.1.1. I have two models
from django.db import models
class Sector(models.Model):
code = models.CharField(max_length=30, primary_key=True)
class Location(models.Model):
code = models.CharField(max_length=30, primary_key=True)
sector = ForeignKey(Sector, on_delete=models.CASCADE, related_name='locations')
and they can be imported/exported just fine using model resources
from import_export import resources
from import_export.fields import Field
from import_export.widgets import ForeignKeyWidget
class SectorResource(resources.ModelResource):
code = Field(attribute='code', column_name='Sector')
class Meta:
model = Sector
import_id_fields = ('code',)
class LocationResource(resources.ModelResource):
code = Field(attribute='code', column_name='Location')
sector = Field(attribute='sector', column_name='Sector',
widget=ForeignKeyWidget(Sector, 'code'))
class Meta:
model = Location
import_id_fields = ('code',)
and import/export actions can be integrated into the admin by
from django.contrib import admin
from import_export.admin import ImportExportModelAdmin
class SectorAdmin(ImportExportModelAdmin):
resource_class = SectorResource
class LocationAdmin(ImportExportModelAdmin):
resource_class = LocationResource
admin.site.register(Sector, SectorAdmin)
admin.site.register(Location, LocationAdmin)
For Reasons™, I would like to change this set-up so that a spreadsheet of Locations which does not contain a Sector column can be imported; the value of sector (for each imported row) should be taken from an extra field on the ImportForm in the admin.
Such a field can indeed be added by overriding import_action on the ModelAdmin as described in Extending the admin import form for django import_export. The next step, to use this value for all imported rows, is missing there, and I have not been able to figure out how to do it.

EDIT(2): Solved through the use of sessions. Having a get_confirm_import_form hook would still really help here, but even better would be having the existing ConfirmImportForm carry across all the submitted fields & values from the initial import form.
EDIT: I'm sorry, I thought I had this nailed, but my own code wasn't working as well as I thought it was. This doesn't solve the problem of passing along the sector form field in the ConfirmImportForm, which is necessary for the import to complete. Currently looking for a solution which doesn't involve pasting the whole of import_action() into an ImportMixin subclass. Having a get_confirm_import_form() hook would help a lot here.
Still working on a solution for myself, and when I have one I'll update this too.
Don't override import_action. It's a big complicated method that you don't want to replicate. More importantly, as I discovered today: there are easier ways of doing this.
First (as you mentioned), make a custom import form for Location that allows the user to choose a Sector:
class LocationImportForm(ImportForm):
sector = forms.ModelChoiceField(required=True, queryset=Sector.objects.all())
In the Resource API, there's a before_import_row() hook that is called once per row. So, implement that in your LocationResource class, and use it to add the Sector column:
def before_import_row(self, row, **kwargs):
sector = self.request.POST.get('sector', None)
if contract:
self.request.session['import_context_sector'] = sector
else:
# if this raises a KeyError, we want to know about it.
# It means that we got to a point of importing data without
# contract context, and we don't want to continue.
try:
sector = self.request.session['import_context_sector']
except KeyError as e:
raise Exception("Sector context failure on row import, " +
f"check resources.py for more info: {e}")
row['sector'] = sector
(Note: This code uses Django sessions to carry the sector value from the import form to the import confirmation screen. If you're not using sessions, you'll need to find another way to do it.)
This is all you need to get the extra data in, and it works for both the dry-run preview and the actual import.
Note that self.request doesn't exist in the default ModelResource - we have to install it by giving LocationResource a custom constructor:
def __init__(self, request=None):
super()
self.request = request
(Don't worry about self.request sticking around. Each LocationResource instance doesn't persist beyond a single request.)
The request isn't usually passed to the ModelResource constructor, so we need to add it to the kwargs dict for that call. Fortunately, Django Import/Export has a dedicated hook for that. Override ImportExportModelAdmin's get_resource_kwargs method in LocationAdmin:
def get_resource_kwargs(self, request, *args, **kwargs):
rk = super().get_resource_kwargs(request, *args, **kwargs)
rk['request'] = request
return rk
And that's all you need.

Update row (SQLAlchemy) with data from marshmallow

I'm using Flask, Flask-SQLAlchemy, Flask-Marshmallow + marshmallow-sqlalchemy, trying to implement REST api PUT method. I haven't found any tutorial using SQLA and Marshmallow implementing update.
Here is the code:
class NodeSchema(ma.Schema):
# ...
class NodeAPI(MethodView):
decorators = [login_required, ]
model = Node
def get_queryset(self):
if g.user.is_admin:
return self.model.query
return self.model.query.filter(self.model.owner == g.user)
def put(self, node_id):
json_data = request.get_json()
if not json_data:
return jsonify({'message': 'Invalid request'}), 400
# Here is part which I can't make it work for me
data, errors = node_schema.load(json_data)
if errors:
return jsonify(errors), 422
queryset = self.get_queryset()
node = queryset.filter(Node.id == node_id).first_or_404()
# Here I need some way to update this object
node.update(data) #=> raises AttributeError: 'Node' object has no attribute 'update'
# Also tried:
# node = queryset.filter(Node.id == node_id)
# node.update(data) <-- It doesn't if know there is any object
# Wrote testcase, when user1 tries to modify node of user2. Node doesn't change (OK), but user1 gets status code 200 (NOT OK).
db.session.commit()
return jsonify(), 200

UPDATED, 2022-12-08
Extending the ModelSchema from marshmallow-sqlalchemy instead of Flask-Marshmallow you can use the load method, which is defined like this:
load(data, *, session=None, instance=None, transient=False, **kwargs)
Putting that to use, it should look like that (or similar query):
node_schema.load(json_data, session= current_app.session, instance=Node().query.get(node_id))
And if you want to load without all required fields of Model, you can add the partial=True argument, like this:
node_schema.load(json_data, instance=Node().query.get(node_id), partial=True)
See the docs for more info (does not include definition of ModelSchema.load).
See the code for the load definition.

I wrestled with this issue for some time, and in consequence came back again and again to this post. In the end what made my situation difficult was that there was a confounding issue involving SQLAlchemy sessions. I figure this is common enough to Flask, Flask-SQLAlchemy, SQLAlchemy, and Marshmallow, to put down a discussion. I certainly, do not claim to be an expert on this, and yet I believe what I state below is essentially correct.
The db.session is, in fact, closely tied to the process of updating the DB with Marshmallow, and because of that decided to to give the details, but first the short of it.
Short Answer
Here is the answer I arrived at for updating the database using Marshmallow. It is a different approach from the very helpful post of Jair Perrut. I did look at the Marshmallow API and yet was unable to get his solution working in the code presented, because at the time I was experimenting with his solution I was not managing my SQLAlchemy sessions properly. To go a bit further, one might say that I wasn't managing them at all. The model can be updated in the following way:
user_model = user_schema.load(user)
db.session.add(user_model.data)
db.session.commit()
Give the session.add() a model with primary key and it will assume an update, leave the primary key out and a new record is created instead. This isn't all that surprising since MySQL has an ON DUPLICATE KEY UPDATE clause which performs an update if the key is present and creates if not.
Details
SQLAlchemy sessions are handled by Flask-SQLAlchemy during a request to the application. At the beginning of the request the session is opened, and when the request is closed that session is also closed. Flask provides hooks for setting up and tearing down the application where code for managing sessions and connections may be found. In the end, though, the SQLAlchemy session is managed by the developer, and Flask-SQLAlchemy just helps. Here is a particular case that illustrates the management of sessions.
Consider a function that gets a user dictionary as an argument and uses that with Marshmallow() to load the dictionary into a model. In this case, what is required is not the creation of a new object, but the update of an existing object. There are 2 things to keep in mind at the start:
The model classes are defined in a python module separate from any code, and these models require the session. Often the developer (Flask documentation) will put a line db = SQLAlchemy() at the head of this file to meet this requirement. This in fact, creates a session for the model.
from flask_sqlalchemy import SQLAlchemy
db = SQLAlchemy()
In some other separate file there may be a need for a SQLAlchemy session as well. For example, the code may need to update the model, or create a new entry, by calling a function there. Here is where one might find db.session.add(user_model) and db.session.commit(). This session is created in the same way as in the bullet point above.
There are 2 SQLAlchemy sessions created. The model sits in one (SignallingSession) and the module uses its own (scoped_session). In fact, there are 3. The Marshmallow UserSchema has sqla_session = db.session: a session is attached to it. This then is the third, and the details are found in the code below:
from marshmallow_sqlalchemy import ModelSchema
from donate_api.models.donation import UserModel
from flask_sqlalchemy import SQLAlchemy
db = SQLAlchemy()
class UserSchema(ModelSchema):
class Meta(object):
model = UserModel
strict = True
sqla_session = db.session
def some_function(user):
user_schema = UserSchema()
user['customer_id'] = '654321'
user_model = user_schema.load(user)
# Debug code:
user_model_query = UserModel.query.filter_by(id=3255161).first()
print db.session.object_session(user_model_query)
print db.session.object_session(user_model.data)
print db.session
db.session.add(user_model.data)
db.session.commit()
return
At the head of this module the model is imported, which creates its session, and then the module will create its own. Of course, as pointed out there is also the Marshmallow session. This is entirely acceptable to some degree because SQLAlchemy allows the developer to manage the sessions. Consider what happens when some_function(user) is called where user['id'] is assigned some value that exists in the database.
Since the user includes a valid primary key then db.session.add(user_model.data) knows that it is not creating a new row, but updating an existing one. This behavior should not be surprising, and is to be at least somewhat expected since from the MySQL documentation:
13.2.5.2 INSERT ... ON DUPLICATE KEY UPDATE Syntax
If you specify an ON DUPLICATE KEY UPDATE clause and a row to be inserted would cause a duplicate value in a UNIQUE index or PRIMARY KEY, an UPDATE of the old row occurs.
The snippet of code is then seen to be updating the customer_id on the dictionary for the user with primary key 32155161. The new customer_id is '654321'. The dictionary is loaded with Marshmallow and a commit done to the database. Examining the database it can be found that it was indeed updated. You might try two ways of verifying this:
In the code: db.session.query(UserModel).filter_by(id=325516).first()
In MySQL: select * from user
If you were to consider the following:
In the code: UserModel.query.filter_by(id=3255161).customer_id
You would find that the query brings back None. The model is not synchronized with the database. I have failed to manage our SQLAlchemy sessions correctly. In an attempt to bring clarity to this consider the output of the print statements when separate imports are made:
<sqlalchemy.orm.session.SignallingSession object at 0x7f81b9107b90>
<sqlalchemy.orm.session.SignallingSession object at 0x7f81b90a6150>
<sqlalchemy.orm.scoping.scoped_session object at 0x7f81b95eac50>
In this case the UserModel.query session is different from the Marshmallow session. The Marshmallow session is what gets loaded and added. This means that querying the model will not show our changes. In fact, if we do:
db.session.object_session(user_model.data).commit()
The model query will now bring back the updated customer_id! Consider the second alternative where the imports are done through flask_essentials:
from flask_sqlalchemy import SQLAlchemy
from flask_marshmallow import Marshmallow
db = SQLAlchemy()
ma = Marshmallow()
<sqlalchemy.orm.session.SignallingSession object at 0x7f00fe227910>
<sqlalchemy.orm.session.SignallingSession object at 0x7f00fe227910>
<sqlalchemy.orm.scoping.scoped_session object at 0x7f00fed38710>
And the UserModel.query session is now the same as the user_model.data (Marshmallow) session. Now the UserModel.query does reflect the change in the database: the Marshmallow and UserModel.query sessions are the same.
A note: the signalling session is the default session that Flask-SQLAlchemy uses. It extends the default session system with bind selection and modification tracking.

I have rolled out own solution. Hope it helps someone else. Solution implements update method on Node model.
Solution:
class Node(db.Model):
# ...
def update(self, **kwargs):
# py2 & py3 compatibility do:
# from six import iteritems
# for key, value in six.iteritems(kwargs):
for key, value in kwargs.items():
setattr(self, key, value)
class NodeAPI(MethodView):
decorators = [login_required, ]
model = Node
def get_queryset(self):
if g.user.is_admin:
return self.model.query
return self.model.query.filter(self.model.owner == g.user)
def put(self, node_id):
json_data = request.get_json()
if not json_data:
abort(400)
data, errors = node_schema.load(json_data) # validate with marshmallow
if errors:
return jsonify(errors), 422
queryset = self.get_queryset()
node = queryset.filter(self.model.id == node_id).first_or_404()
node.update(**data)
db.session.commit()
return jsonify(message='Successfuly updated'), 200

Latest Update [2020]:
You might facing the issue of mapping keys to the database models. Your request body have only updated fields so, you want to change only those without affecting others. There is an option to write multiple if conditions but that's not a good approach.
Solution
You can implement patch or put methods using sqlalchemy library only.
For example:
YourModelName.query.filter_by(
your_model_column_id = 12 #change 12: where condition to find particular row
).update(request_data)
request_data should be dict object. For ex.
{
"your_model_column_name_1": "Hello",
"your_model_column_name_2": "World",
}
In above case, only two columns will be updated that is: your_model_column_name_1 and your_model_column_name_2
Update function maps request_data to the database models and creates update query for you. Checkout this: https://docs.sqlalchemy.org/en/13/core/dml.html#sqlalchemy.sql.expression.update

Previous answer seems to be outdated as ModelSchema is now deprecated.
You should instead SQLAlchemyAutoSchema with the proper options.
class NodeSchema(SQLAlchemyAutoSchema):
class Meta:
model = Node
load_instance = True
sqla_session = db.session
node_schema = NodeSchema()
# then when you need to update a Node orm instance :
node_schema.load(node_data, instance=node, partial=True)
db.session.update()

Below is my solution with Flask-Marshmallow + marshmallow-sqlalchemy bundle as the author requested initially.
schemas.py
from flask import current_app
from flask_marshmallow import Marshmallow
from app.models import Node
ma = Marshmallow(current_app)
class NodeSchema(ma.SQLAlchemyAutoSchema):
class Meta:
model = Node
load_instance = True
load_instance is a key point here to make an update further.
routes.py
from flask import jsonify, request
from marshmallow import ValidationError
from app import db
#bp.route("/node/<node_uuid>/edit", methods=["POST"])
def edit_node(node_uuid):
json_data = request.get_json(force=True, silent=True)
node = Node.query.filter_by(
node_uuid=node_uuid
).first()
if node:
try:
schema = NodeSchema()
json_data["node_uuid"] = node_uuid
node = schema.load(json_data, instance=node)
db.session.commit()
return schema.jsonify(node)
except ValidationError as err:
return jsonify(err.messages), 422
else:
return jsonify("Not found"), 404
You have to check for existence of Node first, otherwise the new instance will be created.

Django Admin import_export module used for custom IMPORT

I was trying to follow official doc of import-export:
https://django-import-export.readthedocs.org/en/latest/import_workflow.html#import-data-method-workflow
But I still do not know how to glue it to my admin assuming that:
I want only subset of fields (I created Resource model with listed fields, but It crashes while importing anyway with: KeyError full stack bellow.
Where - in which method - in my admin class (inheriting of course ImportExportModelAdmin and using defined resource_class) should i place the code responsible for some custom actions I want to happen after validating, that import data are correct but before inserting them into database.
I am not very advanced in Django and will be thankful for some hints.
Example of working implementation will be appreciated, so if you know something similar on github - share.

you can override it as
to create a new instance
def get_instance(self, instance_loader, row):
return False
your custom save
def save_instance(self, instance, real_dry_run):
if not real_dry_run:
try:
obj = YourModel.objects.get(some_val=instance.some_val)
# extra logic if object already exist
except NFCTag.DoesNotExist:
# create new object
obj = YourModel(some_val=instance.some_val)
obj.save()
def before_import(self, dataset, dry_run):
if dataset.headers:
dataset.headers = [str(header).lower().strip() for header in dataset.headers]
# if id column not in headers in your file
if 'id' not in dataset.headers:
dataset.headers.append('id')

Django models: Only permit one entry in a model?

I want to make some of my Django global settings configurable through the admin interface.
To that end, I've decided to set them as database fields, rather than in settings.py.
These are the settings I care about:
class ManagementEmail(models.Model):
librarian_email = models.EmailField()
intro_text = models.CharField(max_length=1000)
signoff_text = models.CharField(max_length=1000)
These are one-off global settings, so I only ever want there to be a single librarian_email, intro_text etc floating around the system.
Is there a way I can prevent admin users from adding new records here, without preventing them from editing the existing record?
I guess I can do this by writing a custom admin template for this model, but I'd like to know if there's a neater way to configure this.
Could I use something other than class, for example?
Thanks!

Please see this question on "keep[ing] settings in database", where the answer seems to be django-dbsettings
Update
Just thought of another option: you can create the following model:
from django.contrib.sites.models import Site
class ManagementEmail(models.Model):
site = models.OneToOneField(Site)
librarian_email = models.EmailField()
intro_text = models.CharField(max_length=1000)
signoff_text = models.CharField(max_length=1000)
Because of the OneToOneField field, you can only have one ManagementEmail record per site. Then, just make sure you're using sites and then you can pull the settings thusly:
from django.contrib.sites.models import Site
managementemail = Site.objects.get_current().managementemail
Note that what everyone else is telling you is true; if your goal is to store settings, adding them one by one as fields to a model is not the best implementation. Adding settings over time is going to be a headache: you have to add the field to your model, update the database structure, and modify the code that is calling that setting.
That's why I'd recommend using the django app I mentioned above, since it does exactly what you want -- provide for user-editable settings -- without making you do any extra, unnecessary work.

I think the easiest way you can do this is using has_add_permissions function of the ModelAdmin:
class ContactUsAdmin(admin.ModelAdmin):
form = ContactUsForm
def has_add_permission(self, request):
return False if self.model.objects.count() > 0 else super().has_add_permission(request)
You can set the above to be any number you like, see the django docs.
If you need more granularity than that, and make the class a singleton at the model level, see django-solo. There are many singleton implementations also that I came across.
For StackedInline, you can use max_num = 1.

Try django-constance.
Here are some useful links:
https://github.com/jezdez/django-constance
http://django-constance.readthedocs.org/en/latest/

I'd take a page out of wordpress and create a Model that support settings.
class Settings(models.Model):
option_name = models.CharField(max_length=1000)
option_value = models.CharField(max_length=25000)
option_meta = models.CharField(max_length=1000)
Then you can just pickle (serialize) objects into the fields and you'll be solid.
Build a little api, and you can be as crafty as wordpress and call. AdminOptions.get_option(opt_name)
Then you can just load the custom settings into the runtime, keeping the settings.py module separate, but equal. A good place to write this would be in an __init__.py file.

Just set up an GlobalSettings app or something with a Key and Value field.
You could easily prevent admin users from changing values by not giving them permission to edit the GlobalSettings app.
class GlobalSettingsManager(models.Manager):
def get_setting(self, key):
try:
setting = GlobalSettings.objects.get(key=key)
except:
raise MyExceptionOrWhatever
return setting
class GlobalSettings(models.Model):
key = models.CharField(unique=True, max_length=255)
value = models.CharField(max_length=255)
objects = GlobalSettingsManager()
>>> APP_SETTING = GlobalSettings.objects.get_setting('APP_SETTING')
There are apps for this but I prefer looking at them and writing my own.

You can prevent users from adding/deleting an object by overriding this method on your admin class:
ModelAdmin.has_add_permission(self, request)
ModelAdmin.has_delete_permission(self, request, obj=None)

Modification of #radtek answer to prevent deleting if only one entry is left
class SendgridEmailQuotaAdmin(admin.ModelAdmin):
list_display = ('quota','used')
def has_add_permission(self, request):
return False if self.model.objects.count() > 0 else True
def has_delete_permission(self, request, obj=None):
return False if self.model.objects.count() <= 1 else True
def get_actions(self, request):
actions = super(SendgridEmailQuotaAdmin, self).get_actions(request)
if(self.model.objects.count() <= 1):
del actions['delete_selected']
return actions

I had basically the same problem as the original poster describes, and it's easily fixed by overriding modelAdmin classes. Something similar to this in an admin.py file easily prevents adding a new object but allows editing the current one:
class TitleAdmin(admin.ModelAdmin):
def has_delete_permission(self, request, obj=TestModel.title):
return False
def has_add_permission(self, request):
return False
def has_change_permission(self, request, obj=TestModel.title):
return True
This doesn't prevent a user from posting a form that edits data, but keeps things from happening in the Admin site. Depending on whether or not you feel it's necessary for your needs you can enable deletion and addition of a record with a minimum of coding.