Django - Render a variable which stored in database field

Question

In my application, I will ask user to input some text with variable and store it into database. like this...

Url

id |     url                           |
---+-----------------------------------+
1  | http://example.com/?aaa={{para1}} |
2  | http://example.com/?aaa={{para2}} |
3  | http://example.com/?aaa={{para2}} |
4  | http://example.com/?aaa={{para1}} |

And there is a table with the actual value of those parameter.

Parameter

id |    para   |
---+-----------+
1  | 932580234 |
2  | 865234932 |

I want to render a html and replace the {{para1}} or {{para2}} on it.

views.py

def userpage(request):
    url_obj = Url.objects.filter(id=..not_important...)[0]
    url = url_obj.url

    para1_obj = Parameter.objects.filter(..not_important..)[0]
    para1 = para1_obj.para

    para2_obj = Parameter.objects.filter(..not_important..)[0]
    para2 = para2_obj.para

    return render(request, 'userpage.html', {'url':url, 'para1':para1, 'para2':para2})

templates/userpage.html

You URL is {{ url }}

My expectation is

You URL is http://example.com/?aaa=932580234

However, the actual html is

You URL is http://example.com/?aaa={{para1}}

How can I "pre render" the field in database before "return render...."? Or is there any other solution for this?

I'm using python 3.4 and django 2.0

Thanks!!

willeM_ Van Onsem · Accepted Answer · 2018-05-17 10:08:03Z

1

We can first construct the url as a template:

url_template = Template(url)

then we render it with the given parameters:

context = Context({'para1':para1, 'para2':para2})
our_url = url_template.render(context)

then our_url is rendered correctly, and we can insert it into our real template rendering:

return render(request, 'userpage.html', {'url': our_url})

So that means we get something like:

def userpage(request):
    url_obj = Url.objects.filter(id=..not_important...)[0]
    url = url_obj.url

    para1_obj = Parameter.objects.filter(..not_important..)[0]
    para1 = para1_obj.para

    para2_obj = Parameter.objects.filter(..not_important..)[0]
    para2 = para2_obj.para

    url_template = Template(url)
    context = Context({'para1':para1, 'para2':para2})
    our_url = url_template.render(context)
    return render(request, 'userpage.html', {'url': our_url})

Note that if you have a rather large table of parameters, you better construct the dictionary dynamically with:

para_dict = {
    'para{}'.format(para.id): para.para
    for para in Parameter.objects.all()
}

answered May 17, 2018 at 10:08

willeM_ Van Onsem

482k33 gold badges483 silver badges624 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

user3114168 Over a year ago

Just added to my existing code. It is exactly what I want! Thanks for your explanation and it is clear and let me understand more about django. I'm now enjoying to use django to write my own web application!

Collectives™ on Stack Overflow

Django - Render a variable which stored in database field

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related