CODEWITHSUNDEEP
pythondjangolist
MacPython
MacPython

Reputation: 18281

Python Lists Beginner

I created a list in Python:

mylist=os.listdir("/User/Me/Folder")

now I have a list of files in a List.

What I would like to do is:

Take one file name after the other and add a URL to it:

/myurl/ + each item in mylist

And then I would like to write the result in a html template from Django.

So that it display all the images in that folder in a list of html

How can I achieve this?

Thanks

Upvotes: 4

Views: 216

Answers (3)

Joubert Nel
Joubert Nel

Reputation: 3214

Using list comprehensions, you can transform your original list, "mylist", into a list with the URL prefix like so:

urllist = ['/myurl/%s' % the_file for the_file in mylist]

Analysis:

a) the expression in the square brackets is the list comprehension. it says: iterate over each item in "mylist", temporarily calling the iterated item "the_file" and transform it using the subexpression: '/myurl/%s' % the_file

b) the transformation expression says, create a string where %s is replaced by the string represented by the value of "the_file"

Upvotes: 5

Ignacio Vazquez-Abrams
Ignacio Vazquez-Abrams

Reputation: 799130

For contrast, if you feel like doing this in the view (even though there's no good reason to do so):

mylist[:] = [url + x for x in mylist]

If you don't need to modify the existing list then you can drop the [:].

Upvotes: 1

Ignacio Vazquez-Abrams
Ignacio Vazquez-Abrams

Reputation: 799130

{% for filename in listoffilenames %}
  /myurl/{{ filename }}
{% endfor %}

Upvotes: 3

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