Django如何将模型中的值添加到另一个应用程序的视图中？

Question

I have 2 applications in my "aplikacja" django project:

• articles
• qr

From articles model I would like to get a value "title" from the first article and then put it into qr.views (it prepares for me a pdf file)

Articles models:

from django.db import models

class Article(models.Model):
        title = models.CharField(max_length=150)
        content = models.TextField(verbose_name="Zawartosc")
        published = models.DateTimeField(verbose_name="Data Publikacji")

How to get a "title" value into qr views? I suppose I need to import article from aplikacja.articles.models. But how to get exactly value in test_qr method?

from reportlab.pdfgen import canvas
from django.http import HttpResponse
from reportlab.graphics.shapes import Drawing 
from reportlab.graphics.barcode.qr import QrCodeWidget 
from reportlab.graphics import renderPDF
from django.contrib.auth.models import User
from aplikacja.articles.models import article


def test_qr(request):
    # Create the HttpResponse object with the appropriate PDF headers.
    response = HttpResponse(content_type='application/pdf')
    response['Content-Disposition'] = 'attachment; filename="somefilename.pdf"'

    a= map(lambda x: str(x), User.objects.values_list('id', flat=True).order_by('id'))

    p = canvas.Canvas(response)
    p.drawString(10, 800, a[1])

    qrw = QrCodeWidget(a[1]) 
    b = qrw.getBounds()

    w=b[2]-b[0] 
    h=b[3]-b[1] 

    d = Drawing(200,200,transform=[200./w,0,0,200./h,0,0]) 
    d.add(qrw)

    renderPDF.draw(d, p, 1, 1)

    p.showPage()
    p.save()
    return response

Answer 1

To get the first article use the first() method:

article = article.objects.all().order_by('published').first()
title = article.title if article else None

BTW python lists are zero based so to get the first element of a list you should use a[0] instead of a[1] . But anyway I suggest you to use the same first() method to get the id of the first user:

first_user_id = str(User.objects.values_list('id', flat=True) \
                                .order_by('id').first())