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Django smart_str 'NoneType' object has no attribute

In django, I want to download the data to excel using the smart_str, but some record of user dont have data or blank some field like in the error below, how can i do with smart_str which should be unnecessary or error free if no data can be detected?
def downloadcv(request):
response = HttpResponse(content_type='text/csv')
# decide the file name
response['Content-Disposition'] = 'attachment; filename="StudentEnrollmentRecord.csv"'
writer = csv.writer(response, csv.excel)
response.write(u'\ufeff'.encode('utf8'))
writer.writerow([
smart_str(u"Guardians_Fullname"),
smart_str(u"Guardians_Address"),
smart_str(u"Contact_Number"),
])
reports = StudentsEnrollmentRecord.objects.filter(School_Year=yearid).order_by('Education_Levels','-Date_Time')
for report in reports:
last_name = report.Student_Users.Parent_Users.Guardian_LastName or ""
first_name = report.Student_Users.Parent_Users.Guardian_FirstName or ""
initial = report.Student_Users.Parent_Users.Guardian_Middle_Initial or ""
writer.writerow([
smart_str(last_name, first_name, initial),
return response if response is not None else 'None'
this is the error i get

You can make use of a try-catch exception handler
Give this a try
try:
last_name = report.Student_Users.Parent_Users.Guardian_LastName
except AttributeError:
last_name = ""
NB: You may have to do the same for the rest of fields also

Convert uploaded csv file into list in Django

I'm having a problem when I'm trying to convert the the first column of the uploaded csv file in django into a list. Originally, my code was like this without using django:
with open("export.csv") as f:
my_list = [row["BASE_NAME"] for row in DictReader(f)]
But when developing the user interface using django, I'm not having the required my_list like above, what am I doing wrong? I tried many methods that you can see in the commented codes:
Here is my view.py:
def handle_csv_data(csv_file):
logging.warning('Watch out!') # will print a message to the console
#print(csv_file)
# with open(csv_file) as f:
# my_list = [row["BASE_NAME"] for row in DictReader(f)]
users = []
for row in csv_file:
users.append(row)
return (users)
def home(request):
if request.method=="POST":
img = UploadForm(request.POST, request.FILES)
if img.is_valid():
logging.warning('Watch out!') # will print a message to the console
# paramFile = request.FILES['pic']
paramFile =TextIOWrapper(request.FILES['pic'].file).read()
portfolio1 = csv.DictReader(paramFile)
# portfolio = csv.DictReader(request.FILES['pic'].file)
# csv_file = request.FILES['pic'].file
# with open(default_storage.path('images/' + "500 clusters.csv"), 'wb+') as destination:
# for chunk in csv_file.chunks():
# destination.write(chunk)
# ifile = open("500 clusters.csv", "r")
# data = [row for row in DictReader(csv_file.read().splitlines())]
# print(users)
# paramFile = csv_file.read()
# portfolio1 = csv.DictReader(paramFile)
#ifile = open('sample.csv', "r")
#read = csv.reader(ifile)
#for row in read:
# print(row)
#data = [row for row in DictReader(csv_file.read().splitlines())]
# for row in portfolio:
# my_list = [row["BASE_NAME"]]
#print(my_list)
portfolio= handle_csv_data(portfolio1)
print(portfolio)
# my_list = portfolio
# return HttpResponseRedirect(reverse('portfolio'))
return render(request, 'home.html', {'portfolio': portfolio})
else:
img=UploadForm()
images=Upload.objects.all()
return render(request,'home.html',{'form':img,'images':images})
Here is my model.py:
from django.db import models
from django.forms import ModelForm
class Upload(models.Model):
pic = models.FileField("pic", upload_to="images/")
upload_date=models.DateTimeField(auto_now_add =True)
# FileUpload form class.
class UploadForm(ModelForm):
class Meta:
model = Upload
fields = '__all__'
I appreciate the help.

The problem with your code is here:
paramFile =TextIOWrapper(request.FILES['pic'].file).read()
portfolio1 = csv.DictReader(paramFile)
DictReader expects an iterable of strings (usually a file object), where each string represents a line. Because you called .read(), paramFileis a string, which is an iterable of strings but each yielded string is a character, not a line. Remove the .read().

paramFile =io.TextIOWrapper(request.FILES['pic'].file)
portfolio1 = csv.DictReader(paramFile)
print(type(paramFile))
users = []
users = [row["BASE_NAME"] for row in portfolio1]
print(len(users))
Somehow I was able to fix it with the above, by looking at the local variables in the local host page.

Download SQL data in xlsx format with Django

I have to write a function which on input has a name of a table and conditionals/filters for that query, on output it returns a link which should be automatically be download by the client (browser).
How to implement this task using python/django?
E.g. I've written a small piece of code, but I'm not sure that it works correctly and there is no implementation of query conditionals parsing (I don't know how to implement it):
direct_db.py:
from django.db import connection
class DirectSQL:
def __init__(self,in_sql):
self.sql=in_sql
self.cursor = connection.cursor()
self.cursor.execute(in_sql)
def getDescription(self):
columns = [desc[0] for desc in self.cursor.description]
return columns
def getResult(self):
row = self.cursor.fetchall()
return row
def getResultAsDict(self):
desc = self.cursor.description
return [dict(zip([col[0].lower() for col in desc], row)) for row in self.cursor.fetchall()]
excel.py:
from ecc.direct_db import DirectSQL
import pandas as ps
class Excel:
def __init__(self, table_name):
self.table_name = table_name
def convert(in_args):
q = DirectSQL("select * from self.table_name" ) # where... order by... like...
columns = [desc[0] for desc in q.getDescription()]
data = q.getResults()
df = ps.DataFrame(list(data), columns)
writer = ps.ExcelWriter('converted.xlsx')
df.to_excel(writer, sheet_name='converted')
writer.save()

I've worked in something like this before, I used xlsxwriter, you can check its docs to find out how to create a xlsx and how set data into it. Then you should need some view:
from django.views.generic import View
from django.http import HttpResponse
class CreateReport(View):
def get_data(self):
# Query your data here, probably using self.request to get query string
...
return data
def generate_report(self):
# Here you will create xlsx doc and populate with data according to docs linked before
...
return workbook
def get(self, request, *args, **kwargs):
document = self.generate_report()
_file = open(document.filename, 'r')
response(HttpResponse(_file, content_type='application/vnd.openxmlformats-officedocument.spreadsheetml.sheet'))
response['Content-Disposition'] = 'attachment; filename=%s' % document.filename.split('/')[-1] # Here will return a full path, that's why probably you will need a split to get only the filename
add_never_cache_headers(response=response) # To avoid download the same file with out of date data.
return response
Then you will need an url
from myapp.views import CreateReport
url(r'^create_report/(?P<some_param_if_needed>[-\w]+)',
CreateReport.as_view(),
name='create_report'),
and finally in template
Download Report
EDIT
Here's a more complete example for get_data() method.
get_data(self):
# Let's supose you have a `MyElements` model
elements = MyElements.objects.all()
# And let's supose you want to filter data with some GET parameter
filter = self.request.GET.get('filter_name', None)
if filter is not None:
elements = elements.filter(filter_field=filter)
return elements

Exporting items from a model to CSV Django / Python

I'm fairly new to django and Python and want to be able to export a list of items in my model i.e products. I'm looking at the documentation here - https://docs.djangoproject.com/en/dev/howto/outputting-csv/
I'm persuming I need will need to create a variable that stores all the data that I want. But not sure where it would within the snippet of code on the link above.
Apologies as this is a very noobish question but would really Any help at all.
Here is the code to my script so far:
import csv
from products.models import Product
from django.http import HttpResponse
def export_to_csv(request):
response = HttpResponse(content_type='text/csv')
response['Content-Disposition'] = 'attachment; filename="mytest.csv"'

Have a look at the python csv module.
You'll probably want to get the models fields with
def get_model_fields(model):
return model._meta.fields
Then use
getattr(instance, field.name)
to get the field values (as in this question).
Then you'll want something like
with open('your.csv', 'wb') as csvfile:
writer = csv.writer(csvfile)
# write your header first
for obj in YourModel.objects.all():
row = ""
for field in fields:
row += getattr(obj, field.name) + ","
writer.writerow(row)
It's a bit verbose (and untested), but it should give you an idea. (Oh and don't forget to close your file)

Depending on the scenario - you may want to have a CSV of your model. If you have access to the Django Admin site, you can plug in a generic action for any model displayed as a list (google: django admin actions)
http://djangosnippets.org/snippets/790/
If you're operating with a console (python manage.py ...), you can use such a script, which I just used:
(place it in: yourapp/management/commands/model2csv.py)
"""
Prints CSV of all fields of a model.
"""
from django.core.management.base import BaseCommand, CommandError
import csv
import sys
class Command(BaseCommand):
help = ("Output the specified model as CSV")
args = '[appname.ModelName]'
def handle(self, *app_labels, **options):
from django.db.models import get_model
app_name, model_name = app_labels[0].split('.')
model = get_model(app_name, model_name)
field_names = [f.name for f in model._meta.fields]
writer = csv.writer(sys.stdout, quoting=csv.QUOTE_ALL)
writer.writerow(field_names)
for instance in model.objects.all():
writer.writerow([unicode(getattr(instance, f)).encode('utf-8') for f in field_names])
This does not catch any exceptions etc., but as an Admin you won't cause them to be raised, right?
Use it like:
./manage.py model2csv my_ecommerce.Product > products.csv

You can also make a template to assist in formatting!
The template is a common Django template
from django.template import loader
def export_to_csv(request):
response = HttpResponse(mimetype='text/csv')
response['Content-Disposition'] = 'attachment; filename="products-list.csv"'
template = loader.get_template('templates/products_template.csb')
response.write(template.render(Context({'products': Products.objects.all()})))
return response

Using django.db.models.query.QuerySet.values results in more optimised queries for my use case.
import csv
from datetime import datetime
from django.http import HttpResponse
# Populate this list with your model's fields
# Replace MODEL with your model
fields = [f.name for f in MODEL._meta.fields]
# The following code will live inside your view
timestamp = datetime.now().isoformat()
response = HttpResponse(content_type="text/csv")
response[
"Content-Disposition"
] = f"attachment; filename={timestamp}.csv"
writer = csv.writer(response)
# Write the header row
writer.writerow(fields)
# Replace MODEL with your model
for row in MODEL.objects.values(*fields):
writer.writerow([row[field] for field in fields])
return response

I use this on my code. A function called from view.
It automatically get model fields to make columns.
You can also customize the field list you want to export.
Function
import csv
from django.http import HttpResponse
from .models import Books
def export_qs_to_csv(model_class = None, qs = None, field_names = None):
if model_class and not qs:
qs = model_class.objects.all()
if qs and not model_class:
model_class = qs.model
meta = model_class._meta
if not field_names:
field_names = [field.name for field in meta.fields]
response = HttpResponse(content_type='text/csv')
response['Content-Disposition'] = 'attachment; filename={}.csv'.format(meta)
writer = csv.writer(response)
writer.writerow(field_names)
for obj in qs:
row = writer.writerow([getattr(obj, field) for field in field_names])
return response
Usage
#user_passes_test(lambda u: u.is_superuser)
def export_books(request):
return export_qs_to_csv(model_class = Books)
# or
return export_qs_to_csv(qs = Books.objects.filter(published = True))
# or
return export_qs_to_csv(
qs = Books.objects.filter(published = True),
field_names = [
"title",
"price",
"publishing_date",
]
)
Original answer
It works, and it needs only to define model class in model_class variable.
This Django view let use downloads CSV. CSV name is Django_app.model_name.csv.
import csv
from django.http import HttpResponse
from .models import Trade
def export_to_csv(request):
# The only line to customize
model_class = Trade
meta = model_class._meta
field_names = [field.name for field in meta.fields]
response = HttpResponse(content_type='text/csv')
response['Content-Disposition'] = 'attachment; filename={}.csv'.format(meta)
writer = csv.writer(response)
writer.writerow(field_names)
for obj in model_class.objects.all():
row = writer.writerow([getattr(obj, field) for field in field_names])
return response

Here is a potential solution, based on #tomasz-gandor 's answer, but updated to 2020:
"""
Prints CSV of all fields of a model.
"""
import csv
from django.core.management.base import BaseCommand, CommandError
class Command(BaseCommand):
help = ("Output the specified model as CSV")
def add_arguments(self, parser):
parser.add_argument('model',
nargs=1,
type=str,
help='Model name to export, like <app.model> or "members.Member"')
parser.add_argument('outfile',
nargs=1,
type=str,
help='Save path, like </path/to/outfile.csv> or "/data/members.csv"')
def handle(self, *app_labels, **options):
from django.apps import apps
app_name, model_name = options['model'][0].split('.')
model = apps.get_model(app_name, model_name)
field_names = [f.name for f in model._meta.fields]
writer = csv.writer(open(options['outfile'][0], 'w'), quoting=csv.QUOTE_ALL, delimiter=',')
writer.writerow(field_names)
for instance in model.objects.all():
writer.writerow([str(getattr(instance, f)) for f in field_names])
Can easily be used with:
python manage.py model2csv members.Member /data/members_export.csv

If you don't care about fieldnames and want all the fields, just do this.
with open('file_name.csv', 'w') as csvfile:
writer = csv.writer(csvfile)
for obj in YourModel.objects.values_list():
row = list(obj)
writer.writerow(row)

I combined some of the previous answers, because I needed to import some data from production and change some of it along the way. So here is my solution, which you can use to override some field values while writing the CSV file.
Export some queryset data into CSV file:
import csv
from myapp.models import MyModel
from user.models import User
# Make some queryset manually using Django shell:
user = User.objects.get(username='peterhil')
queryset = MyModel.objects.filter(user=user)
def query_to_csv(queryset, filename='items.csv', **override):
field_names = [field.name for field in queryset.model._meta.fields]
def field_value(row, field_name):
if field_name in override.keys():
return override[field_name]
else:
return row[field_name]
with open(filename, 'w') as csvfile:
writer = csv.writer(csvfile, quoting=csv.QUOTE_ALL, delimiter=',')
writer.writerow(field_names) # write the header
for row in queryset.values(*field_names):
writer.writerow([field_value(row, field) for field in field_names])
# Example usage:
query_to_csv(queryset, filename='data.csv', user=1, group=1)

Use this solution for model csv file.might being helpful
# Create the HttpResponse object with the appropriate CSV header.
response = HttpResponse(content_type='text/csv')
response['Content-Disposition'] = 'attachment;
filename="somefilename.csv"'
writer = csv.writer(response);
writer.writerow(["username","Email"]);
for i in User.objects.all():
writer.writerow([i.username,i.email])
return response

I used the django-queryset-csv package.
Follow these steps:
pip install django-queryset-csv
Your views.py:
import djqscsv
from products.models import Product
def get_csv(request):
qs = Product.objects.all()
return djqscsv.render_to_csv_response(qs)

Want to prompt browser to save csv

Want to prompt browser to save csv using pyramid.response.Response searched for clues and found here's a link Django answer but i can't use it with Pyramid wsgi my code looks like this:
from pyramid.response import Response
def get_list_names_emails(request):
session, env = request.db, request.client_env
response = Response(content_type='text/csv')
output = StringIO()
writer = csv.writer(output)
writer.writerow(['SomeName', 'SomeEmail', 'CompanyName])
csv_output = output.getvalue()
return csv_output

As a cleaner way to do that, you can register a renderer.
In your configuration set-up, add:
config.add_renderer(name='csv',
factory='mypackage.renderers.CSVRenderer')
then in mypackage/renderers.py:
class CSVRenderer(object):
def __init__(self, info):
pass
def __call__(self, value, system):
fout = StringIO.StringIO()
writer = csv.writer(fout, delimiter=';', quoting=csv.QUOTE_ALL)
writer.writerow(value['header'])
writer.writerows(value['rows'])
resp = system['request'].response
resp.content_type = 'text/csv'
resp.content_disposition = 'attachment;filename="report.csv"'
return fout.getvalue()
After that, you can decorate your view with the renderer:
#view_config(..., renderer='csv')
def myview(self):
header = ['name', 'surname', 'address']
rows = [
(
row['name'],
row['surname'],
row['address'],
)
for row in query_rows(.....)
]
return {
'header': header,
'rows': rows
}
The advantage of this approach is better testable view code (you just check for the dictionary values, no need to parse anything) and you can also add a XLS or whatever renderer to the same view:
#view_config(..., renderer='xls')
#view_config(..., renderer='csv')
def myview(self):
...

Try adding Content-Disposition:
response['Content-Disposition'] = 'attachment; filename="report.csv"'

It's better to set content type as well
response['Content-type'] = 'text/csv'
response['Content-Disposition'] = 'attachment; filename="report.csv"'