Django community: RSS
This page, updated regularly, aggregates Django Q&A from the Django community.
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Updating multiple objects at once with a filter in django
The query below filters a queryset of Credentials objects based on these creds and updates the password of a Credentials object with the password that the user put in the form. This works and it updates the password, the problem with this is that it only updates one object, since no more than one object can have the same pcol, user, pcol_user, data_source, and username. creds = Credentials.objects.filter(pcol=pcol, user=user, pcol_user=pcol_user, data_source=data_source,username=username).update(password=password) The query below is what Im trying to do, filter only on these creds since I want to filter multiple objects. If I were to create two objects with the same pcol, user, pcol_user, username, and data_source I wouldn't be able to since they have the same exact creds. In contrast, if I created an object where the user, pcol_user, username, and data source are the same as another object, and the protocol is different it doesn't care and I am able to create the object with no problem. My objective is to update an object based on the user, pcol_user, username, and data source that the user enters. I have tried the query below on two sample objects, where the user, pcol_user, username, and data_source of those two … -
How to show paginated data from 3 different models?
I have three Django models. Person: Fields first_name, last_name, date_of_birth, gender, profile_pic (Nullable), created_at, modified_at PersonStatus: Fields person (fk Person), status_text, created_at, modified_at FormulaFields: formula(TextField that is able to store python code of up to 2000 characters), column_number (Integer >=4), created_at, modified_at. I need to show a paginated table with the same data, Person ID First Name Last Name Person Status: Latest Status of Person Full Name Is Adult (Adult if person > 18 years of age) Profile Pic (Icon) if available For populating columns 5 and 6, I need to add the code in the FormulaFields model. For displaying Person Status, I need to use Subquery and OuterRef (maybe my tech lead is asking so that I learn?) I did pagination with one model and it works, but how do I show the data which are in three different models? Here is my view. I've been stuck on this, please do help. class FileExamListView(generic.ListView): model = Person template_name = "exam_list.html" def get_context_data(self, **kwargs): context = super(FileExamListView, self).get_context_data(**kwargs) list_exam = Person.objects.all().order_by('date_of_birth') paginator = Paginator(list_exam, self.paginate_by) page = self.request.GET.get('page') try: file_exams = paginator.page(page) except PageNotAnInteger: file_exams = paginator.page(1) except EmptyPage: file_exams = paginator.page(paginator.num_pages) context['person'] = file_exams return context I also looked … -
Blog post with image Django3
Hi i'm trying to add some funcionality to my blog post app, i'd like to paste image into post content so i figured is this waw that ive created new model just for images and set it as OneToOne with my Post model i wonder if ther is any way to set this image to content field in the post model models.py class Post(models.Model): title = models.CharField(max_length=100) content = models.TextField() date_posted = models.DateTimeField(default=timezone.now) author = models.ForeignKey(User, on_delete=models.CASCADE) def __str__(self): return self.title def get_absolute_url(self): return reverse('post-detail', kwargs={'pk': self.pk}) class PostImage(models.Model): post = models.OneToOneField(Post, on_delete=models.CASCADE) image = models.ImageField(default=None, upload_to='post_pics', blank=True) def save(self, *args, **kwargs): super().save(*args, **kwargs) img = Image.open(self.image.path) if img.height > 500 or img.width > 500: output_size = (500, 500) img.thumbnail(output_size) img.save(self.image.path) vievs.py def home(request): context = { 'posts': Post.objects.all(), 'user_posts': "active", } return render(request, 'blog/home.html', context) class PostListView(ListView): model = Post template_name = 'blog/home.html' context_object_name = 'posts' ordering = ['-date_posted'] paginate_by = 5 post_template.html {% extends "blog/base.html" %} {% load static %} {% load crispy_forms_tags %} {% block content %} <div class="content-section"> <form method="POST"> {% csrf_token %} <fieldset class="form-group"> <legend class="border-bottom mb-4">Blog Post</legend> {{form|crispy }} </fieldset> <div class="form-group"> <button class="btn btn-outline-info" type="submit">Post</button> </div> </form> </div> {% endblock content %} -
Weasyprint/Cairo loses SVG text labels when converting to PDF
I have an SVG image generated by Bokeh (link here) with labels on the left and bottom of the chart. The image itself shows the text within the chart just fine, however, when converting to PDF through Weasyprint, the text is lost. Has anyone else encountered a similar issue? I'm not sure how to debug this as there are no errors when converting. A snippet of my export function: html = HTML(string=html_string) result = html.write_pdf('./tmp/example.pdf', stylesheets=[css], font_config=font_config) Within the HTML template, I use css to embed the image as a background image as so: html body article#columns section#linechart{ width: 100%; background: url(./tmp/linechart.svg) no-repeat center; background-size: contain; margin: 0; height: 500px; overflow: visible; } Thanks in advance! Current version info: CairoSVG = 2.4.2 Weasyprint = 51 Python = 3.73 -
How can I create redirect button with post method in Django template
I'm trying to create a button in the Django template which will redirect to another URL. But getting error 404 since Django can't recognize URL path rescribed in the urls.py. HTML part <form method="post" action='sts'> {% csrf_token %} <button class="btn btn-outline-success my-2 my-sm-0" type="submit" name="cts_link">cts</button> </form> urls.py from django.conf.urls import include, url from django.contrib import admin from rtRegRes.views import units from rtRegRes.views import spartan urlpatterns = [ url(r'^admin/', admin.site.urls), url(r'^units/$', units), url(r'^units/sts/?$', spartan), ] views.py from django.shortcuts import render, redirect, reverse, render_to_response from .models import rt_reg_res from django.http import HttpResponse, JsonResponse def units(request): """Return main webpage""" return render_to_response('runtime.html') def spartan(request): """Link to the other unit webpages""" table = rt_reg_res.objects.all() if request.method == 'POST': qatables = request.POST.get("cts_link") if qatables: return render(request, 'cts.html', {'table': table}) Clicking the button following error message appears: enter image description here Could somebody point me what is wrong in my code Thanks -
I'm trying to connect customer to User but getting this error: AttributeError at /user/' User' object has no attribute 'Customer
I'm trying to get the user data from Customer model. For this I also connect the Customer to User which I import from django.contrib.auth.models. But getting this error: AttributeError at /user/ 'User' object has no attribute 'Customer' at views page: @login_required(login_url='login') def user_page(request): orders=request.user.Customer.order_set.all() context={'orders':orders} return render(request, 'blog/user_page.html', context) at models page from django.db import models from django.contrib.auth.models import User # Create your models here. class Customer(models.Model): User=models.OneToOneField(User, null=True, on_delete=models.CASCADE) name=models.CharField(max_length=200, null=True) email=models.EmailField() phone=models.IntegerField(null=True) date_created=models.DateTimeField(auto_now_add=True, null=True) def __str__(self): return self.name -
Styling HTML for model property with verbose name in Django admin
I have a Django model like this: class MyModel (models.Model): value = IntegerField() def custom_field(self): return self.value + 1 Then in my admin.py file, I have a model: class MyModelAdmin(admin.ModelAdmin): list_display = ('id', 'custom_field') def custom_field(self, obj): if obj.total_counter_actual != obj.total_counter: color = 'red' return format_html( '<b style="color:{};">{}</b>', color, obj.custom_field ) else: return obj.custom_field custom_field.short_description='My custom column name' Although column name (implying analogue of verbose name) and styling both work as exepcted, instead of values I see something like <bound method MyModel.custom_field of <MyModel: TEST_VALUE>>. Any way to fix this? -
linking miniconda environment to wsgi.py django apache windows deployment?
I was following this guide to deploy Django app on windows. I'm using miniconda environment as my environment how should I link it in my windows_wsgi.py file i tried activate_this = 'C:/ProgramData/Miniconda3/envs/sentizer/python.exe' # execfile(activate_this, dict(__file__=activate_this)) exec(open(activate_this).read(),dict(__file__=activate_this)) but I get this error in the logs Traceback (most recent call last):\r File "C:/Users/Administrator/Eyelizer/Eyelizer/wsgi_windows.py", line 3, in <module>\r exec(open(activate_this).read(),dict(__file__=activate_this))\r File "c:\\programdata\\miniconda3\\envs\\sentizer\\lib\\encodings\\cp1252.py", line 23, in decode\r return codecs.charmap_decode(input,self.errors,decoding_table)[0]\r UnicodeDecodeError: 'charmap' codec can't decode byte 0x90 in position 2: character maps to <undefined>\r mod_wsgi (pid=4268): Failed to exec Python script file 'C:/Users/Administrator/Eyelizer/Eyelizer/wsgi_windows.py'. mod_wsgi (pid=4268): Exception occurred processing WSGI script 'C:/Users/Administrator/Eyelizer/Eyelizer/wsgi_windows.py'. -
Why is my login_required redirect not working?
I seem to be doing everything correctly, yet I am still receiving a 404 when I try to login into a page that is login_required only rather than being redirected to the login page. Settings.py/Login_url LOGIN_URL = '/dating_app/login/' dating_app/urls/login path('login/', LoginView.as_view(template_name = 'dating_app/login.html'), name='login'), project_urls/dating_app path('', include('dating_app.urls', namespace= 'dating_app')), project_directory . ├── 11_env │ ├── bin │ ├── include │ ├── lib │ └── pyvenv.cfg ├── dating_app │ ├── __init__.py │ ├── __pycache__ │ ├── admin.py │ ├── apps.py │ ├── chat.html │ ├── forms.py │ ├── media │ ├── migrations │ ├── models.py │ ├── static │ ├── tag.py │ ├── templates │ ├── templatetags │ ├── tests.py │ ├── urls.py │ └── views.py ├── dating_project │ ├── __init__.py │ ├── __pycache__ │ ├── asgi.py │ ├── settings.py │ ├── urls.py │ └── wsgi.py ├── db.sqlite3 └── manage.py -
How to return a token with graphql jwt upon user sign up
I'm wondering how can I return a token instead of a user. I'm using graphql-jwt for signing in and I don't want to have to sign up a user then sign-in them in the fronted I want a token to be returned upon successful signup from django.contrib.auth import get_user_model import graphene from graphene_django import DjangoObjectType class UserType(DjangoObjectType): class Meta: model = get_user_model() class CreateUser(graphene.Mutation): user = graphene.Field(UserType) class Arguments: username = graphene.String(required=True) password = graphene.String(required=True) email = graphene.String(required=True) def mutate(self, info, username, password, email): user = get_user_model()( username=username, email=email, ) user.set_password(password) user.save() return CreateUser(user=user) class Mutation(graphene.ObjectType): create_user = CreateUser.Field() -
Deploying a Django app to Heroku: H14 - No web dynos running
I have deployed my app successfully on heroku following the "Getting Started" documentation. After deployment ended, I entered: heroku ps:scale web=1. But in return I got : Scaling dynos... ! ▸ Couldn't find that process type (web). I printed my logs and I got: 2020-05-07T16:55:45.691324+00:00 heroku[router]: at=error code=H14 desc="No web processes running" method=GET path="/favicon.ico" host=thawing-retreat-70669.herokuapp.com request_id=08bf5ac4-dc36-404d-98a3-dd3434084a44 fwd="90.2.155.70" dyno= connect= service= status=503 bytes= protocol=https And heroku ps Free dyno hours quota remaining this month: 548h 53m (99%) Free dyno usage for this app: 0h 0m (0%) For more information on dyno sleeping and how to upgrade, see: https://devcenter.heroku.com/articles/dyno-sleeping No dynos on ⬢ thawing-retreat-70669 I've read that it may be my Procfile; I have one that reads: web: gunicorn pur_beurre.wsgi --log-file- I've checked the builpacks and I got: heroku/python -
Add request user to InlineFormSet: TypeError: __init__() got an unexpected keyword argument 'user'
I'm using InlineFormSetView from extra_views to create a custom inlineform. On form save I need to make some changes that's relevant to the request user. So the request user is required in the form. From what I found, I came up with this solution but it seems that get_formset_kwargs may be used elsewhere. Causing an exception to occur. I'm not quite sure what is the cause. May be there is another solution to adding request user to an inline form. View: class MyModelSetsView(InlineFormSetView): model = MyModel inline_model = MySubModel form_class = MySubModelSetForm template_name = "update_sets.html" success_message = "Updated successfully." permission_required = [] factory_kwargs = { 'extra': 1, } def get_success_url(self): return self.object.get_absolute_url() def get_formset_kwargs(self): kwargs = super(MyModelSetsView, self).get_formset_kwargs() kwargs.update({'user': self.request.user}) return kwargs Form: class MySubModelSetForm(forms.ModelForm): def __init__(self, *args, **kwargs): self.user = kwargs.pop('user', None) super(PeakPeriodSetForm, self).__init__(*args, **kwargs) Exception: File "..\venv\lib\site-packages\extra_views\formsets.py", line 268, in get formset = self.construct_formset() File "...venv\lib\site-packages\extra_views\formsets.py", line 36, in construct_formset return formset_class(**self.get_formset_kwargs()) File "...\venv\lib\site-packages\django\forms\models.py", line 897, in __init__ super().__init__(data, files, prefix=prefix, queryset=qs, **kwargs) File "...\venv\lib\site-packages\django\forms\models.py", line 569, in __init__ super().__init__(**{'data': data, 'files': files, 'auto_id': auto_id, 'prefix': prefix, **kwargs}) TypeError: __init__() got an unexpected keyword argument 'user' -
Django 3: AttributeError: 'AdminSite' object has no attribute 'Register'
AttributeError: 'AdminSite' object has no attribute 'Register' when i go to make migrations and open my admin section to add so dummy info and image I receive the error. admin.site.Register(Bet, BetAdmin) What I have done so far is delete and hit return to see if it was indentation mistake on the class. I also installed pylint to see more details on errors based on the one suggested answers. Class has no objects member pip install pylint-django then I added to my setttings.json { "terminal.integrated.shell.windows": "C:\\WINDOWS\\Sysnative\\WindowsPowerShell\\v1.0\\powershell.exe", "workbench.colorTheme": "PowerShell ISE", "python.pythonPath": "C:\\Users\\taylo\\AppData\\Local\\Programs\\Python\\Python38-32\\python.exe", "json.schemas": [ ], "python.linting.pylintArgs": [ "--load-plugins=pylint_django" ], "[python]": { } } but it didn't seam to solve anything related to issue. FYI: I did a bunch of searches and view on stack and but solutions seen where for spelling errors ased on the word being spelled incorrectly with q like reqister instead of register is and was not spelled wrong. the full terminal error (venv) PS C:\Users\taylo\django\pybet_project> python manage.py makemigrations Traceback (most recent call last): File "manage.py", line 21, in <module> main() File "manage.py", line 17, in main execute_from_command_line(sys.argv) File "C:\Users\taylo\django\pybet_project\venv\lib\site-packages\django\core\management\__init__.py", line 401, in execute_from_command_line utility.execute() File "C:\Users\taylo\django\pybet_project\venv\lib\site-packages\django\core\management\__init__.py", line 377, in execute django.setup() File "C:\Users\taylo\django\pybet_project\venv\lib\site-packages\django\__init__.py", line 24, in setup apps.populate(settings.INSTALLED_APPS) … -
Django get template id tags values from view
In my project i have a template for example like this: ... <div class="row no-gutters row_app" id="first_id"> <div class="col-12 bg-066AB3" id="second_val"><img src="{% static "img/Icona_Medico.png" %}"></div> <div class="col-12 col-sm-12 col-md-6 col-lg-6 col-xl-6 bg-BCBCBB"> <div class="row no-gutters" id="Other_val"> <div class="col-12 d-xl-flex justify-content-xl-center align-items-xl-center"> <div class="width_100 box_testo"> ... well, i would then when in my views.py all start a python function that check every values of ids in my template and put its in a list: def index(request, **kwargs): ids_val = [] #Here a function that return the template ids value and put into my ids_val list context_dict = {} response = render(request, 'base_home.html', context_dict) return response Someone know how is possible extract id template tags values from my django template? So many thanks in advance -
How do I make Django form fields required conditionally?
I am new to Django and I am trying learn about creating web apps. I am trying to delete a record without the form fields being required. I only want them to be required when creating a new record. Here is a screen shot: enter image description here Here is my form.py: from django import forms class ExpenseForm(forms.Form): name = forms.CharField( max_length=255, widget=forms.TextInput(attrs={ "class": "form-control", "placeholder": "Expense Name" }) ) amount = forms.DecimalField( max_digits=16, decimal_places=2, widget=forms.NumberInput(attrs={ "class": "form-control", "placeholder": "$400.00" }) ) Here is my view.py: from django.shortcuts import render from budget.models import Expense from .forms import ExpenseForm from django.http import HttpResponse, HttpResponseRedirect from django.db.models import Sum def budget_view(request): expenses = Expense.objects.all() total = Expense.objects.aggregate(total=Sum('amount'))['total'] form = ExpenseForm() if request.method == 'POST': form = ExpenseForm(request.POST) if form.is_valid() and 'create-expense' in request.POST: expense = Expense( name=form.cleaned_data["name"], amount=form.cleaned_data["amount"] ) expense.save() return HttpResponseRedirect('/') elif form.is_valid() and 'edit-expense' in request.POST: return HttpResponseRedirect('/') else: expense = Expense.objects.get(id=request.POST.get("delete-expense")) expense.delete() return HttpResponseRedirect('/') context = { "form": form, "expenses": expenses, "total": total, } return render(request, "budget_view.html", context) I tried adding form.fields[<field_name>].required() = False to the else block, but to no avail. Do I need to create a custom clean() in the form.py to perform validation on each field … -
Getting python usage errors using Django to display a table of files in a directory
So I am trying to get Django to display a template that shows a list of files on a local directory. My views.py looks like this: from django.shortcuts import render from django.http import HttpResponse from os import listdir def index(request): files = listdir("/path/to/directory/") filelen = len(files) return render(request, 'path/to/template.html') My html template looks like this: {%for i in range(1, filelen)%} <tr> <td>{{i}}</td> <td>{{files[i]}}</td> {% if "a" in files[i] %} <td>✓</td> {% else %} <td>x</td> {% endif %} {% if "b" in files[i] %} <td>✓</td> {% else %} <td>x</td> {% endif %} But when I try to run it I get the following error: Error at line 39 'for' statements should use the format 'for x in y': for i in range(1, len_files) Anybody know how to get this to work? I have tried replacing filelen with {{filelen}} but that gave me the same error. -
NoReverseMatch at /blog/ - Django by example code list and detail view issue
I'm working on the code in chapter 1 of the django by example. Followed all instructions and reviewed many feedbacks on this site and other areas but not to my luck. I had earlier attempted get_absolute_url and return reverse method unsuccessfully. I was trying an instructed approach and land exactly the same issues in listview and detailview. I'm a beginner so I guess I'm missing something may be fundamental. Is there any version dependency. I've installed latest django and python. Need advice and thanks for any help! This is the error message models.py from django.db import models from django.utils import timezone from django.contrib.auth.models import User from django.urls import reverse # Create your models here. class PublishedManager(models.Manager): def get_queryset(self): return super().get_queryset().filter(status='published') class Post(models.Model): STATUS_CHOICES = ( ('draft', 'Draft'), ('published', 'Published'), ) title = models.CharField(max_length=250) slug = models.SlugField(max_length=250, unique_for_date='publish_date') author = models.ForeignKey(User, on_delete=models.CASCADE, related_name='blog_posts') body = models.TextField() publish_date = models.DateTimeField(default=timezone.now) created_date = models.DateTimeField(auto_now_add=True) updated_date = models.DateTimeField(auto_now_add=True) status = models.CharField(max_length=10, choices=STATUS_CHOICES, default='draft') objects = models.Manager() # default Manager published = PublishedManager() # our custom manager class Meta: ordering = ['-publish_date',] def __str__(self): return self.title def get_absolute_url(self): return reverse( 'blog:post_detail', args=[ self.publish_date.year, self.publish_date.month, self.publish_date.day, self.slug, ] ) views.py from django.shortcuts import render, get_object_or_404 from … -
Django Built in PasswordChangeView keeps redirecting me to the admin password change after declaring my templates
urls.py from django.urls import path, include from django.contrib.auth import views as auth_views from .views import ( dashboardview ) urlpatterns = [ `path('dashboard/', dashboardview, name='dashboard'), path('login/', auth_views.LoginView.as_view(), name='login'), path('logout/', auth_views.LogoutView.as_view(), name='logout'), path('change-password/', auth_views.PasswordChangeView.as_view(template_name='registration/password_change_form.html'), name='password_change'), path('password/change/done', auth_views.PasswordChangeDoneView.as_view(), name='password_change_done'), ` ] -
how to upload 'django-rest-api server', and 'crawling.py'
Hi, guys. i have a problem. i made an android application written by kotlin. this app communicates to 'django-restapi server' to get some data. 'django-restapi server' gets data from 'crawling.py' file(using selenium,headless chrome). app : show data to people django-restapi server : get data from crawling.py, and saving data on database crawling.py(with chromedriver) : crawling data from website, and sending data to server so, what i want to know is, i don't know how to upload them. i want to use aws. what i'm thinking now is, upload 'django-restapi server' and 'crawling.py' to aws lambda seperately. then, i will run 2 lambda. but i think it's bad architecture. can you advice me better answer? -
"cleaned_data" isn't working as my browser shows back - " Exception Value: 'FormName' object has no attribute 'cleaned_data' "
and here is my VIEWS.PY, def form_view(request): form = forms.FormName() if request.method == 'POST': form = forms.FormName(request.POST) if form.is_valid: print("VALIDATION SUCCESS !!") d = form.cleaned_data['name'] print("name:", d) return render(request, 'baleno/form_pg.html', {'form': form}) -
how to complete my django app in the backend
I created a django app , I also set templates and static files . I made migrations my questions is : I want to scrape data from different websites and save it in django to be displayed in my templates what would be the next steps then ? -
I want to know,how does 'category' knows which model to access
template code- <a class="dropdown-item" href="{% url 'home' %}">All Products</a> {% for category in links %} <a class="dropdown-item" href="{{category.get_url}}">{{category}}</a> {% endfor %} context_processor.py def menu_links(request): links=Category.objects.all() return dict(links=links) Category Model- class Category (models.Model): name=models.CharField(max_length=250, unique=True) slug=models.SlugField(max_length=250, unique=True) description=models.TextField(blank=True) image=models.ImageField(upload_to='category',blank=True) def get_url(self): return reverse('products_by_category',args=[self.slug]) def __str__(self): return self.name Product model- class Product(models.Model): name=models.CharField(max_length=250, unique=True) slug=models.SlugField(max_length=250, unique=True) description=models.TextField(blank=True) category=models.ForeignKey(Category, on_delete=models.CASCADE) price=models.DecimalField(max_digits=10,decimal_places=2) image=models.ImageField(upload_to='product',blank=True) stock=models.IntegerField() available=models.BooleanField(default=True) created=models.DateTimeField(auto_now_add=True) updated=models.DateTimeField(auto_now=True) def get_url(self): return reverse('product_detail',args=[self.category.slug,self.slug]) def __str__ (self): return self.name here, as you can see the def get_url(self) is in Category as well as Product model I just want to know how come category in {{category.get_url}} knows to access the method in Category and not in Product model -
Reuse existing Django rest framework serializer
I am currently using Django Rest Framework and I am looking for a way to reuse some attributes from an already defined Serializer. In order to explain, I am going to expose some of the serializers involved: Completed Serializer: class ProductSerializer(serializers.ModelSerializer): subscribed = serializers.SerializerMethodField() other_field = serializers.SerializerMethodField() class Meta: model = Product fields = [ 'id', 'name', 'subscribed', 'other_field', 'x', 'y', 'z' ] def get_subscribed(self, product: Product): return product.is_user_subscribed(self.context['request'].user) Simplified Serializer: class ProductSimplifiedSerializer(serializers.ModelSerializer): subscribed = serializers.SerializerMethodField() class Meta: model = Product fields = [ 'id', 'name', 'subscribed' ] def get_subscribed(self, product: Product): return product.is_user_subscribed(self.context['request'].user) As you can notice above, those serializers are almost the same, but one of them is a simplified version of the object because I don't want to retrieve unnecessary information in some parts. The problem here is that in this scenario we have a method serializer that will need to be maintained two times. Maybe in the future, I would want to add another field, and I will need to add it to both. So, how could be achieved a Based Serializer in which all the fields are included but I could reuse it and extract specific fields from it? I have already thought these options: … -
Django returns error but dosen't check for more Integrity errors afterwards
So a request that violates the check for postive_integer (e.g. positive_integer = -1), django will not check for errors against unique_together (e.g. record with same name and user already exist) and will only return the original error from the positive_integer. class Foo(models.Model): positive_integer = models.PositiveIntegerField() name = models.CharField(max_length=100) user = models.CharField(max_length=100) class Meta: unique_together = (('name', 'user',)) -
Project migration to django data
Can i migraine my project from jam.py to Django? Without any change in the code or rewriting it to suit the framework? Or do you prefer to create my project on Django in the first place
