Parse value in html from dictionary in django

Question

I have a dictionary like this in django

my_dict = {0: 'Dog - Wikipedia',
 1: 'https://en.wikipedia.org/wiki/Dog',
 2: 'Funny babies annoying dogs - Cute dog & baby compilation ...',
 3: 'https://www.youtube.com/watch?v=M1djO19aSFQ'}

What I want is to have title as title but the link to be appeared as a link in html template...Can u guys please help me sort this out?i am new in this case.. :(

ps: how do i iterate over the dictionary to create one as a title and the next one as a link?

• Alex Hall: This is a very strange format for storing pairs of items. How did the dictionary come about and can we change its format at the source?

Answer 1

That will be something like this:

{% for k, v in my_dict.items %)
 {% if k == 0 %} <h1>{{ v }}</h1> {% endif %}
 {% elif k == 1 %} <a href="{{ v }}">Some link text</a> {% endif %}
{% endfor %}

Learn more about Django Template Tags.

• Chiefir: I will appreciate if you accept the answer if it helped :)
• Chiefir: So you can mark this answer as the right answer - just press that grey tick below arrows to the left from this answer ))
• Chiefir: Yes, that worked, thank you :) that points help users to do more things on this site :)

Answer 2

To be clear, you want to take that dictionary and make a new dictionary that contains the title as the key value and the url as the value of the key? If so, the following should do the trick.

new_dict = {}
for i in range(len(my_dict.values())):
 if not i%2:
 new_dict[my_dict[i]] = my_dict[i+1]

Or you can write it pythonically with a dict comprehension.

newer_dict = { my_dict[i]:my_dict[i+1] for i in range(len(my_dict.values())) if not i%2 }