I'm trying to get a string out of an ArrayField in my database but it only prints by characters, not the full string. For example, the ArrayField is named words and in the database it shows as {word1, word2, word3} so in the HTML I put {{ object.words.0 }} and { is rendered on the screen.
How can I get it to render word1?
I have added django.contrib.postgres to INSTALLED_APPS.
This is what it looks like in my model:
from django.db import models
from django.contrib.postgres.fields import ArrayField
class WordArr(models.Model):
words = ArrayField(models.CharField(max_length=200, null=True))
I fixed the issue and it was with my database. The ArrayField in my model was previously a CharField that I changed. I thought the migrations were correct however the database was still reading it as a CharField, so the output was the first character of a string. I migrated to a new database and everything worked fine. My code is below for further reference:
models.py
from django.db import models
from django.contrib.postgres.fields import ArrayField
# Create your models here.
class Word(models.Model):
first = models.CharField(max_length=50)
last = models.CharField(max_length=50)
words = ArrayField(models.CharField(max_length=50, null=True))
def __str__(self):
return self.first
views.py
from django.shortcuts import render
from django.views.generic.detail import DetailView
from django.views.generic import CreateView
from .models import Word
# Create your views here.
class ArrayCreateView(CreateView):
model = Word
fields = ['first', 'last']
success_url = '/'
def form_valid(self, form):
w = form.save(commit=False)
random_words = ["word1", "word2", "word3"]
w.words = random_words
return super().form_valid(form)
class ArrayDetailView(DetailView):
model = Word
word_form.html
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<meta http-equiv="X-UA-Compatible" content="ie=edge">
<title>Document</title>
</head>
<body>
<h1>Create Array</h1>
<div>
<form method="POST">
{% csrf_token %}
{{ form.as_p }}
<button type="submit">Submit</button>
</form>
</div>
</body>
</html>
word_detail.html
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<meta http-equiv="X-UA-Compatible" content="ie=edge">
<title>Document</title>
</head>
<body>
<h1>{{ object.words.0 }}</h1>
</body>
</html>
settings.py
INSTALLED_APPS = [
...
'django.contrib.postgres',
'arr',
]
urls.py
from django.contrib import admin
from django.urls import path
from arr.views import ArrayCreateView, ArrayDetailView
urlpatterns = [
path('admin/', admin.site.urls),
path('create/new/', ArrayCreateView.as_view(), name='create'),
path('detail/<int:pk>/', ArrayDetailView.as_view(), name='detail')
]
Related
I am following this answer for the multi-selecting dropdown. I am using django-multiselectfield to collect data in the DB model. I want to show value in bootstrap multi-selecting dropdown but getting the below result
I am seeking this result
These are my code snippet
model.py
from django.db import models
from multiselectfield import MultiSelectField
class Country(models.Model):
name = models.CharField(max_length=40)
def __str__(self):
return self.name
class Person(models.Model):
obj=Country.objects.all()
TEAM=tuple((ob,ob) for ob in obj)
name = models.CharField(max_length=124)
southasia=MultiSelectField(max_length=200, choices=TEAM)
def __str__(self):
return self.name
views.py
from django.http import JsonResponse
from django.shortcuts import render, redirect, get_object_or_404
from .forms import PersonCreationForm
from .models import Person
def person_create_view(request):
form = PersonCreationForm()
if request.method == 'POST':
form = PersonCreationForm(request.POST)
if form.is_valid():
form.save()
return redirect('person_add')
return render(request, 'miracle/index.html', {'form': form})
forms.py
from django import forms
from .models import Person
class PersonCreationForm(forms.ModelForm):
class Meta:
model = Person
fields = '__all__'
def __init__(self, *args, **kwargs):
super().__init__(*args, **kwargs)
template
{% load crispy_forms_tags %}
<!doctype html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta name="viewport"
content="width=device-width, user-scalable=no, initial-scale=1.0, maximum-scale=1.0, minimum-scale=1.0">
<meta http-equiv="X-UA-Compatible" content="ie=edge">
<link href="https://cdn.jsdelivr.net/npm/select2#4.1.0-rc.0/dist/css/select2.min.css" rel="stylesheet" />
<script src="https://cdn.jsdelivr.net/npm/select2#4.1.0-rc.0/dist/js/select2.min.js"></script>
<title>Dependent Dropdown in Django</title>
<link rel="stylesheet" href="https://stackpath.bootstrapcdn.com/bootstrap/4.1.1/css/bootstrap.min.css">
<link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/bootstrap-select/1.13.1/css/bootstrap-select.css" />
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script src="https://stackpath.bootstrapcdn.com/bootstrap/4.1.1/js/bootstrap.bundle.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/bootstrap-select/1.13.1/js/bootstrap-select.min.js"></script>
</head>
<body>
<h2>Person Form</h2>
<form method="post">
{% csrf_token %}
{{ form.name|as_crispy_field }}
<select class="selectpicker" multiple data-live-search="true">
{{ form.southasia|as_crispy_field }}
</select>
<input type="submit" value="Submit">
</form>
<script>
$('select').selectpicker();
</script>
</body>
</html>
How can do show all values in dropdown multiselect instead of normal checkbox select?
django multiselect will give you check box view. If you want drop-down multiple list one of the work around is you can create another model for the choices and use manytomany relation . Django rendering should automatically detect it as droplist view and gives you opttion to select multiple choices.
Like this.
Class model1(model.models)
Choices = models.charfield(choice=choice)
Class person(models.model)
Choice = manytomany (model1).
1.views.py
from django.shortcuts import render
from . models import User
from .forms import Userform
def Home(request):
if request.method == "POST":
form = Userform(request.POST or None,request.FILES or None)
if form.is_valid():
form.save()
else:
form = Userform()
return render(request,"home.html",{"form":form})
def read(request):
read = User.objects.all()
return render(request,"read.html",{"read":read})
2.models.py
from django.db import models
class User(models.Model):
name = models.CharField(max_length=12)
rollno = models.IntegerField()
# files = models.FileField()
3.form.py
from django import forms
from .models import User
class Userform(forms.ModelForm):
class Meta:
model = User
fields = ('name','rollno')
urls.py
from django.contrib import admin
from django.urls import path
from app import views
from django.conf import settings
from django.conf.urls.static import static
urlpatterns = [
path('admin/', admin.site.urls),
path("",views.Home,name="Home"),
path("read/",views.read, name="read"),
]
urlpatterns+=static(settings.MEDIA_URL,document_root=settings.MEDIA_ROOT)
4.home.html
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Document</title>
</head>
<body>
<form method="POST" action="read/" enctype="multipart/form-data">
{%csrf_token%}
{{form}}
<button type=submit>click to submit</button>
</form>
</body>
</html>
read.html
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Document</title>
</head>
<body>
{%block containt%}
{%for i in read%}
{{i.name}}
{{i.rollno}}
{%endfor%}
{%endblock%}
</body>
</html>
i want to add data to django admin using a form but data is not uploading at django admin
Not able to upload data into django admin hence cant read it
please help
i want to add data to django admin using a form but data is not uploading at django admin
Not able to upload data into django admin hence cant read it
please help
Problem is in the action of your form, you have entered the wrong URL in the action
You have to put home URL instead of read/, because you are handling your form submission in your home function so replace your form action by this
<form method="POST" action="{% url 'Home' %}" enctype="multipart/form-data">
{%csrf_token%}
{{form}}
<button type=submit>click to submit</button>
</form>
I am building a blog website where I can upload audio/video files to the database and then run a for loop to output the files in HTML audio/video tags. I have succeeded in saving the files to the database and also output them on the website but for some reason I can't get them to play.
I am still pretty new to Python/Django and I would appreciate it if anybody can help me figure this one out. Thanks in anticipation
Here is my models.py
from django.db import models
class Track(models.Model):
track_title = models.CharField(max_length=250)
track = models.FileField(upload_to='album/')
def __str__(self):
return self.track_title
Here is my index.html
{% load static %}
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<meta http-equiv="X-UA-Compatible" content="ie=edge">
<title>Document</title>
</head>
<body>
{% for track in tracks %}
<figure>
<figcaption>Audio</figcaption>
<audio controls>
<source src="{{ track.track.url }}" type="audio/mp3">
</audio>
</figure>
{% endfor %}
This is my views.py
from django.shortcuts import render
from .models import Track
def songView(request):
tracks = Track.objects.all()
context = {
'tracks':tracks
}
return render(request, 'album/index.html', context)
And my urls.py
from django.urls import path
from . import views
app_name = 'album'
urlpatterns = [
path('', views.songView, name='song'),
]
I am trying to create a user form and then post the input to the database model.However I keep getting this error message
Page not found (404)
Request Method: GET
Request URL: http://127.0.0.1:8000/hiresite/Recruiter
Using the URLconf defined in recruitment.urls, Django tried these URL patterns, in this order:
^hiresite ^$ [name='index']
^hiresite ^Recruiter$ [name='Recruiter']
^admin/
The current URL, hiresite/Recruiter, didn't match any of these.
I am a bit confused because I can see the url Recruiter above and yet I get the error message.Your help would be much appreciated .
1.Here is my Urls.py for the app
from django .conf.urls import url
from. import views
urlpatterns = [
url(r'^$', views.index, name='index'),
url(r'^Recruiter$', views.Recruiter, name='Recruiter')
]
2.Here is my Urls.py for the Project
from django.conf.urls import include, url
from django.contrib import admin
urlpatterns = [
url(r'^hiresite', include('hiresite.urls')),
url(r'^admin/', admin.site.urls),
]
3.Here is the view for the Url Recruiter
def Recruiter(request):
if request.method == 'POST':
form = register_job(request.POST)
if form.is_valid():
title = request.POST.get('title', ' ')
description = request.POST.get('description', ' ')
salary = request.POST.get('salary', ' ')
reference = request.POST.get('reference', ' ')
user_obj = jobsearch(title=title, description=description, salary=salary, reference=reference)
user_obj.save()
return render(request, 'hiresite/Recruiter.html', {'user_obj ': user_obj, 'is_registered': True})
else:
form = register_job()
return render(request, 'hiresite/Recruiter.html', {'form': form})
4.Here is the Html template file used in the views.py file for the Url Recruiter
!DOCTYPE html>
<html lang="en">
<head>
<title>Learning Html the Hard way</title>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1">
<meta name="description" content="">
<meta name="author" content="">
<link rel="icon" href="../../favicon.ico">
{% load staticfiles %}
<link rel='stylesheet' href= " {% static 'css/bootstrap.min.css' %}" type = 'text/css'/>
</head>
<body>
<form action="{% url 'hiresite:Recruiter' %}" method="post">
{% csrf_token %}
{{ form }}
<input type="submit" value="Submit">
</form>
</body>
</html>
convert this line :
url(r'^hiresite', include('hiresite.urls')),
To this line :
url(r'^hiresite/', include('hiresite.urls')),
I am new to Django. I want to draw a google graph using django graphos. I have written small code for that and i am getting empty template..any lead would be appritiated..
view.py
from graphos.renderers import gchart
from django.shortcuts import render
from django.shortcuts import render_to_response
from graphos.sources.model import ModelDataSource
from models import MyCityViews
def get_context_data(self):
queryset = MyCityViews.objects.all()
data_source = ModelDataSource(queryset,fields=['city', 'views'])
line_chart = gchart.LineChart(data_source)
context = {"chart": line_chart}
return render_to_response('gchart.html', {'chart': line_chart})
models.py
from django.db import models
# Create your models here.
class MyCityViews(models.Model):
city = models.CharField(max_length = 30, blank = True , null = True)
views = models.IntegerField()
class Meta:
ordering = ['city']
verbose_name = 'city'
verbose_name_plural = 'city'
def __unicode__(self):
return self.city
gchart.html
<!DOCTYPE html>
<html lang="en">
<head>
<meta http-equiv="content-type" content="text/html; charset=UTF-8">
<meta charset="utf-8">
<title>Graphos</title>
{% load static from staticfiles %}
<link rel="stylesheet" href="{% static "css/bootstrap.css" %}">
<script type="text/javascript" src="https://www.google.com/jsapi"></script>
<script type="text/javascript">
google.load("visualization", "1", {packages:["corechart"]});
google.setOnLoadCallback({{line_chart.as_html}});
</script>
</head>
<body>
{{line_chart.as_html}}
<div id="container"></div>
</body>
</html>
Where i am doing wrong...
Did not pass view object to div block thats why i was not getting HTML view.
<body>
<div id="container">
{{line_chart.as_html}}
</div>
</body>