How can I output raw unescaped html from this Rails route/helpers? - html

I am trying to use the following bit of code to generate a link on my page
%h2= link_to '{{ label }}', product_path('{{ id }}'.html_safe)
This is actually part of a larger HTML block which will serve as a javascript template, and I will later parse using Underscore.js to fill in the {{ id }} and {{ label }} placeholders. So I would like rails to output something to my HTML like
/products/{{ id }}
However, it keeps escaping the spaces and brackeets, and giving me
{{ label }}
So the url_helper is escaping my string, even though I don't want it to. How can I force it to not do this?
I've tried
%h2= link_to '{{ label }}', product_path('{{ id }}'.html_safe)
%h2= link_to '{{ label }}', product_path(raw '{{ id }}')
%h2= link_to '{{ label }}', raw(product_path('{{ id }}'))
and
%h2=raw( link_to '{{ label }}', product_path('{{ id }}'.html_safe))
But none of them work
EDIT:
Another way to play with this is from rails console,
include ActionController::UrlWriter
ruby-1.9.2-p0 :010 > product_path '{{ id }}'.html_safe
=> "/products/%7B%7B%20id%20%7D%7D"
Any help appreciated... thanks
Thanks

What about CGI::unescape(product_path('{{ id }}') ? (with the require 'cgi' that goes with it.)
I believe this is Ruby 1.9.2 only but it seems to be the version you're using.

Related

Jekyll Collection Filter

I'm trying to grab a specific item from a collection called 'content' based on an id using where_exp, but for the life of me I can't get it to work. Here's the code:
filter:
{% assign var = site.content | where_exp:"content", "content.id == 'testId'" | first %}
frontmatter for post in collection:
---
layout: content
title: "This is the title"
image: "assets/photos/image.jpg"
id: "testId"
---
html:
<img class="full-width-poto" src="{{ var.image }}">
I can't figure out what I'm doing wrong.
Note, I've been referring to this post: Getting a specific item from a collection in Jekyll and https://riptutorial.com/jekyll/example/28446/accessing-a-specific-collection-item
Ok, I figured out my problem, just in case someone comes across this. For some reason I can't use the key 'id' for this...it must be hardcoded for something else.
I swapped in 'myid' and it works fine now...

Sorted Map Iteration in Go Templates?

I'm building a website in Go, using the Hugo static site generator. What I'm trying to do is build a dynamic navigation bar for my web pages.
Here's what I'm doing:
In my config.yml file, I've defined a Map of links that I'd like to appear in my navbar -- here's what this file looks like:
baseurl: "https://www.rdegges.com/"
languageCode: "en-us"
title: "Randall Degges"
params:
navLinks: {"Twitter": "https://twitter.com/rdegges", "Facebook": "https://www.facebook.com/rdegges", "Google+": "https://plus.google.com/109157194342162880262", "Github": "https://github.com/rdegges"}
So, I've also got an index.html template in Hugo that contains a navbar which looks like this:
<nav>
<ul>
{{ range sort $title, $link := .Site.Params.navLinks }}
<li>{{ $title }}</li>
{{ end }}
</ul>
</nav>
This above code works correctly, with one exception: I'd like to order the results of my links instead of having them randomly ordered each time.
I know that Maps are not inherently structured in Go -- but is there a way to retain the original ordering of my navigation elements in some way?
Thanks for the help!
Go templates sort maps by key. If you want to force a specific order, then use a slice:
Here's the YAML:
baseurl: "https://www.rdegges.com/"
languageCode: "en-us"
title: "Randall Degges"
params:
navLinks:
- title: Twitter
url: https://twitter.com/rdegges
- title: Facebook
url: https://www.facebook.com/rdegges
... and the template:
<nav>
<ul>
{{ range $link := .Site.Params.navLinks }}
<li>{{ $link.title }}</li>
{{ end }}
</ul>
</nav>

webapp2 changes html tags in pure text

I want to make my GAE application webapp2 compatible.
This code worked great with webapp:
insert = '<p><font color="red"><b>some text</b></font></p>'
template_values = {
'insert': insert,
...
}
path = ...
self.response.out.write(template.render(path,template_values))
The content of the variable insert was just put into the web page output by webapp. Now the content of the variable is "analyzed" by webapp2 and the content is changed when it is inserted in the webpage.
webapp2 inserts this:
<p><font color="red"><b>some text</b></font></p>
How can I go back to the old behavior?
Thanks for any help.
Have a look at
safe : https://docs.djangoproject.com/en/dev/ref/templates/builtins/#safe &
autoescape : https://docs.djangoproject.com/en/dev/ref/templates/builtins/#autoescape.
Eg:
{{ insertHTML|safe }} OR
{% autoescape off %}{{ inserHTML }}{% endautoescape %}

Django - Shorten text content retrieved in html template

I would like to know how to shorten text content retrieved to get a preview instead of getting a block of text in the html template
From: aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
To:
aaaaaaaaaaaaaaaaaaaaaaaaaaa....
The code that is used to retrieve the contents is :
<a class="list_content" href="/qna/view" onclick="document.forms['q_title_{{ q.id }}'].submit(); return false" title="{{ q.content }}">{{ q.content }} </a><br/>
Many thanks!
If you are using Django 1.4, truncatechars filter will be the best way to go:
{{ q.content|truncatechars:25 }}
Assuming that "q.content" is a string you could use the slice command:
{{ q.content|slice:":255" }}
truncatechars template filter for django 1.4

Create hyperlink in django template of object that has a space

I am trying to create a dynamic hyperlink that depends on a value passed from a function:
{% for item in field_list %}
<a href={% url index_view %}{{ item }}/> {{ item }} </a> <br>
{% endfor %}
The problem is that one of the items in field_list is "Hockey Player". The link for some reason is dropping everything after the space, so it creates the hyperlink on the entire "Hockey Player", but the address is
http://126.0.0.1:8000/Hockey
How can I get it to go to
http://126.0.0.1:8000/Hockey Player/
instead?
Use the urlencode filter.
{{ item|urlencode }}
But why are you taking the name? You should be passing the appropriate view and PK or slug to url which will create a suitable URL on its own.
Since spaces are illegal in URLs,
http://126.0.0.1:8000/Hockey Player/
is unacceptable. The urlencode filter will simply replace the space with %20, which is ugly/inelegant, even if it does kind of get the job done. A much better solution is to use a "slug" field on your model that represents a cleaned-up version of the title field (I'll assume it's called the title field). You want to end up with a clean URL like:
http://126.0.0.1:8000/hockey_player/
To make that happen, use something like this in your model:
class Player(models.Model):
title = models.CharField(max_length=60)
slug = models.SlugField()
...
If you want the slug field to be pre-populated in the admin, use something like this in your admin.py:
class PlayerAdmin(admin.ModelAdmin):
prepopulated_fields = {"slug": ("title",)}
....
admin.site.register(Player,PlayerAdmin)
Now when you enter a new Player in the admin, if you type "Hockey Player" for the Title, the Slug field will become "hockey_player" automatically.
In the template you would then use:
{% for item in field_list %}
<a href={% url index_view %}{{ item.slug }}/> {{ item }} </a> <br>
{% endfor %}
There is this builtin filter .
http://docs.djangoproject.com/en/dev/ref/templates/builtins/#urlencode
Although you should be using one of these
http://docs.djangoproject.com/en/dev/ref/models/fields/#slugfield