How to add clickable links to a field in Django admin?

Question:

I have this admin.py

class LawyerAdmin(admin.ModelAdmin):
    fieldsets = [
        ('Name',   {'fields': ['last', 'first', 'firm_name', 'firm_url', 'school', 'year_graduated']}),
    ]
    list_display = ('last', 'first', 'school', 'year_graduated', 'firm_name', 'firm_url')
    list_filter = ['school', 'year_graduated']
    search_fields = ['last', 'school', 'firm_name']

and I want to make “firm_url” fields clickable with each of the url listed in the field. How can I do this? Thank you.

Asked By: Zeynel

||

Answers:

Define a custom method in your LawyerAdmin class that returns the link as HTML:

def show_firm_url(self, obj):
    return '<a href="%s">%s</a>' % (obj.firm_url, obj.firm_url)
show_firm_url.allow_tags = True

See the documentation.

Answered By: Daniel Roseman

add show_firm_url to list_display

Answered By: diegueus9

But it overrides the text display specified in my models and displays “show firm url” on the head of the column

You can change it by assigning short_description property:

show_firm_url.short_description = "Firm URL"

Use the format_html utility. This will escape any html from parameters and mark the string as safe to use in templates. The allow_tags method attribute has been deprecated in Django 1.9.

from django.utils.html import format_html
from django.contrib import admin

@admin.display(description="Firm URL")
class LawyerAdmin(admin.ModelAdmin):
    list_display = ['show_firm_url', ...]
    ...

    def show_firm_url(self, obj):
        return format_html("<a href='{url}'>{url}</a>", url=obj.firm_url)
    

Now your admin users are safe even in the case of:

firm_url == 'http://a.aa/<script>eval(...);</script>'

See the documentation for more info.

Answered By: Seppo Erviälä

You can handle it in the model if you prefer:

In models.py :

class Foo(models.Model):
...


     def full_url(self):
        url = 'http://google.com'
        from django.utils.html import format_html
        return format_html("<a href='%s'>%s</a>" % (url, url))

admin.py:

    list_display = ('full_url', ... )
Answered By: Mesh

its more easy..

#in your admin.py add

from django.utils.html import format_html

##Code:

#url_file = "your field contain de url of file" example: http://127.0.0.1:8000
/backdown/respaldo.zip

list_display = ["url_file","download_content"]
#Link download
def download_content(self, obj):
    return format_html('<a href="%s">%s</a>' % (obj.url_file, "Download"))

download_content.allow_tags = True
download_content.short_description = "Download Content File"

##in your html see like this..

<a href="http://127.0.0.1:8000/backdown/respaldo.zip">http://127.0.0.1:8000/backdown/respaldo.zip</a>
Answered By: Areliszxz
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