How to make HTML imported Campaign Monitor templates editable? - python

During the import procedure I made sure to mark the imgs editable, and put several content sections into multiline constructs. The imported templates show up in the client's template set, but when I try to edit them, I only presented with the zip/HTML upload option instead of the actual editor:
From my code base, my own template object has the exporting code:
class EmailBodyTemplate(models.Model):
...
def migrate_email_template(self, cm_client_id, cm_token):
cm_template = Template(cm_token)
# Basically I expose a specially tailored HTML version of my template
# for CM: adding editable to imgs, introducing multiline wrappers and
# adding unsubscribe, view in browser and other special links
html_url = settings.HOME_URL + reverse('email_import_view', args=(self.id,))
template_id = cm_template.create(cm_client_id, self.name, html_url, None)
self.template_id = template_id
self.save()

I reached out to Campaign Monitor support on the matter, and got the info that their editor is not applicable for imported HTML templates from an external source. In that case you need to keep editing those outside of the system and re-upload after edit.

Related

Django - How to dynamically add scripts/classes to the code?

I have a Django project where users can register and add their XML/CSV feeds.
What the project does is (everyday morning):
download the feed
parse the feed and store it to the model Product
generate XML/CSV export from these products in a different format
Now the problem is that users can register anytime during the project lifetime and many feeds have to be either downloaded, parsed or exported in a specific way.
So I need to be able to react quickly when the feed is added and write custom functions/classes/scripts to download/parse or export these sources.
Let's say it would be enough to add or extend some base classes like Downloader,Parser or Exporter
I'm looking for the most common way to do this.
I tried multiple approaches. For example, I've created a package parsers and there is a __init__.py with CHOICES attribute. When a user adds new feed, I create a parser__sourcename.py file which I add into parsers package which contains Parser class which extends BaseParser class with parse method where is the custom code.
Then I add import into __init__.py file so it looks like this:
from feeds_core.parsers.parser__source1 import Parser as ParserForSource1
from feeds_core.parsers.parser__source2 import Parser as ParserForSource2
from feeds_core.parsers.parser__source3 import Parser as ParserForSource3
PARSER__SOURCE1 = 'source1'
PARSER__SOURCE2 = 'source2'
PARSER__SOURCE3 = 'source3'
CHOICES = {
PARSER__SOURCE1: ParserForSource1,
PARSER__SOURCE2: ParserForSource2,
PARSER__SOURCE3: ParserForSource3,
}
def get_parser(choice):
return CHOICES[choice]
Then I have a model Source with this field:
PARSER_CHOICES = [(x, x) for x in CHOICES.keys()]
parser = models.CharField(max_length=128, choices=PARSER_CHOICES, null=True, blank=True)
def get_parser(self):
...
elif self.parser:
return self.get_parser_class()(self.get_last_filepaths(), self.user_eshop, self)
else:
raise self.NoParserDefinedException
def parse(self): # TODO raise spec exc
self.parse_started()
self.get_parser().parse()
self.parse_ended()
def get_parser_class(self) -> BaseParser:
return get_parser(self.parser)
And when there is a new feed, I create the file, modify the __init__.py and choose the parser for this source in the Django admin interface.
But it is little bit complicated and moreover, I'm afraid I have to restart production server every time.
Do you have any ideas/experiences? Is there some best practice how to do such things?
A very basic implementation where you would add a file to a directory (settings.PARSER_FILE_DIR), it would then be an available choice for Source.parser_file, then using importlib you would load the file and extract the Profile class out of it. settings.PARSER_FILE_DIR would probably have to be in your PYTHONPATH
This is an extremely basic example and you would need a lot of error handling and would need to make sure it was very secure
import importlib
class Source(models.Model):
parser_file = models.FilePathField(path=settings.PARSER_FILE_DIR, match='.*\.py$')
def get_parser_class(self):
return importlib.import_module(self.parser_file.replace('.py', '')).Parser

How to loop over custom plugin data on the template in django cms 3.5.3

I am trying to implement a website that uses RSS Feeds.
My use case is to be able to display one RSS feed at a time, then loop over to the next maybe after 5 seconds. These RSS feeds need to be displayed on the same placeholder.
I am new to django cms in case you need more info please let me know.
I have googled a lot but all I can see is how to add plugins from the frontend and they will all automatically show inside the placeholder.
Or modify the render method of the custom plugin class to display what you want.
But I want to display everything but one at a time in a continuous manner
#plugin_pool.register_plugin
class ExternalArticlePlugin(CMSPluginBase):
model = ExternalArticle
name = _("External article")
render_template = "external_article.html"
cache = False
def render(self, context, instance, placeholder):
context = super(ExternalArticlePlugin, self).render(
context, instance, placeholder
)
return context
I expect to display one RSS feed at a time inside my placeholder.
Those feeds are links to the actual webpage with more info.
one way would be to write a function random_rss_feed_url() in ExternalArticle Model which renders a random rss instance.
model.py
class ExternalArticle(models.Model):
def random_rss_feed_link(self):
return ExternalArticle.objects.order_by('?')[0].link
then you do in plugins external_article.html:
{{ instance.random_rss_feed_link }}
Edited:
if you want to change automatically without page reload, then you need something like this in javascript in your template:
var rss_links = ['link_1', 'link_2', 'link_3'];
setInterval(function() {
// take random rss link
var random_link = rss_links[Math.floor(Math.random()*rss_links.length)];
// change the link
document.getElementById('rss_link_element').href = random_link;
}, 5000);

Django custom admin action for FeinCMS actions column

I'm making an admin panel for a Django-Mptt tree structure using the FeinCMS TreeEditor interface. This interface provides an 'actions column' per-node for things like adding or moving nodes quickly without using the typical Django admin action select box.
What I am trying to do is add a custom admin action to this collection which passes the pk of the node to a celery task which will then add a collection of nodes as children. Existing functions are simply href links to the URL for that task(add/delete/move), so thus far I have simply mimicked this.
My solution currently involves:
Define the action as a function on the model
Create a view which uses this function and redirects back to the changelist
Add this view to the admin URLs
Super the TreeEditor actions column into the ModelAdmin class
Add an action to the collection which calls this URL
Surely there must be a better method than this? It works, but it feels massively convoluted and un-DRY, and I'm sure it'll break in odd ways.
Unfortunately I'm only a month or two into working with Django so there's probably some obvious functions I could be using. I suspect that I might be able to do something with get_urls() and defining the function directly in the ModelAdmin, or use a codeblock within the injected HTML to call the function directly, though I'm not sure how and whether it's considered a better option.
Code:
I've renamed everything to a simpler library <> books example to remove the unrelated functionality from the above example image.
models.py
class Library(models.Model):
def get_books(self):
# Celery task; file omitted for brevity
get_books_in_library.delay(self.pk)
views.py
def get_books_in_library(request, library_id):
this_library = Library.objects.get(pk=library_id)
this_library.get_books_in_library()
messages.add_message(request, messages.SUCCESS, 'Library "{0}" books requested.'.format(this_library.name))
redirect_url = urlresolvers.reverse('admin:myapp_library_changelist')
return HttpResponseRedirect(redirect_url)
urls.py
urlpatterns = [
url(r'^admin/myapp/library/(?P<library_id>[0-9]+)/get_books/$', get_books_in_library, name='get books in library'),
url(r'^admin/', include(admin.site.urls)),
]
admin.py
class LibraryAdmin(TreeEditor):
model = Library
def _actions_column(self, obj):
actions = super(LibraryAdmin, self)._actions_column(obj)
actions.insert(
0, u'<a title="{0}" href="{1}/get_books"><img src="{2}admin/img/icon_addlink.gif" alt="{0}" /></a>'.format(
_('Get Books'),
obj.pk,
settings.STATIC_URL
)
)
return actions
Note that I may have broken something in renaming things and removing the extraneous cruft if you try to execute this code, I think it should adequately illustrate what I'm trying to do here however.
After digging around today and simply trying various other solutions, I've put together one that uses get_urls and a view defined directly into the admin interface which feels tidier though it's effectively just moving the code from multiple django files into the admin interface - though it does make use of the admin wrapper to stop unauthenticated users, which is an improvement.
I'll leave a copy of the working code here for anyone who finds this in future, as I've seen very few examples of TreeEditor et al. being used in newer versions of Django.
class NodeAdmin(TreeEditor):
model = Node
# < ... > Other details removed for brevity
def get_urls(self):
urls = super(NodeAdmin, self).get_urls()
my_urls = [
url(r'^(?P<node_id>[0-9]+)/get_suggestions/$', self.admin_site.admin_view(self.get_suggestions)),
]
return my_urls + urls
def get_suggestions(self, request, node_id):
this_node = Node.objects.get(pk=node_id)
get_suggestions(this_node.pk)
messages.add_message(request, messages.SUCCESS, 'Requested suggestions for {0}'.format(this_node.term))
redirect_url = urlresolvers.reverse('admin:trinket_node_changelist')
return HttpResponseRedirect(redirect_url)
def _actions_column(self, obj):
actions = super(NodeAdmin, self)._actions_column(obj)
# Adds an 'get suggestions' action to the Node editor using a search icon
actions.insert(
0, u'<a title="{0}" href="{1}/get_suggestions"><img src="{2}admin/img/selector-search.gif" alt="{0}" /></a>'.format(
_('Get Suggestions'),
obj.pk,
settings.STATIC_URL,
)
)
# Adds an 'add child' action to the Node editor using a plus icon
actions.insert(
0, u'<a title="{0}" href="add/?{1}={2}"><img src="{3}admin/img/icon_addlink.gif" alt="{0}" /></a>'.format(
_('Add child'),
getattr(self.model._meta,'parent_attr', 'parent'),
obj.pk,
settings.STATIC_URL
)
)
return actions

Django CMS how to create nested plugins programatically

I am writing a migration script to parse an old html website into Django CMS pages.
The thing I need is to understand on how to nest plugins programatically.
In particular case I need to have html < a> tags converted into CMS LinkPlugin objects, nested inside text that is edited by standard ckeditor TextPlugin of Django-CMS.
How to programmatically nest plugins inside other plugins of Django CMS. In my case I need to nest a CMS Link plugin inside of the TextPlugin in the text.
I know on how to parse text. I do not understand on how to do it from nested CMS plugins perspective?
I can not interconnect the Link CMS plugin object instance and CMSPlugin object instance that I insert into the ancestor TextPlugin.
More context:
Note I really know how to do this from UI perspective. I need to emulate this in a script.
I have dumped the database into JSON and noticed there are certain things there.
First I have a CMSPlugin class instance that is placed into a page placeholder. (Sotle this part from placeholderadmin.py of the CMS)
position = CMSPlugin.objects.filter(language=lang, parent=parent).count()
plugin = CMSPlugin(
language='en',
position=position,
plugin_type=plugin_type,
placeholder=placeholder,
)
plugin.insert_at(parent, position='last-child', save=False)
plugin.save()
# ?????
plugin.link = Link(
name='Link text',
page_link=target_page,
placeholder=placeholder,
)
plugin.save()
This creates a nested plugin in a proper placeholder and attaches it into a text plugin. However it is added with a blank LinkPlugin instance. I'm later creating an instance of a Link plugin in the CMS.
The thing is I do not know on how to do this properly.
From UI perspective the CMS plugin is added nested but contains no real plugin instance. SO the Admin plugins tree for that placeholder is rendered with empty Link plugins.
CMSPlugins are added Link < Empty>.
I can edit this created Link plugin through admin and add a text and target link. How to do this programatically. E.g. inside of a script? Script must do 1000-s of pages so I can not do it manually
Sorry just to be consistent on this. This is far more complicated in logic then it seems.
I did an article on this.
Django CMS Adding plugins inside plugins programmatically
In general the solution is to mimic the CMS way of doing this.
# Getting an site admin instance
admin_site = AdminSite()
instance, plugin_admin = plugin.get_plugin_instance(admin_site)
plugin_admin.cms_plugin_instance = plugin
plugin_admin.placeholder = plugin.placeholder
# Triggering the Django Admin add view with our request.
# That's how Django-CMS does this action itself.
response = plugin_admin.add_view(request)
Look for the full snippet in article. Hope this helps someone with similar problems.
To add nested plugins you need to do this:
add_plugin(
placeholder=placeholder,
plugin_type='TextPlugin',
language=translation.get_language(),
)
target = placeholder.get_plugins().get(plugin_type='TextPlugin')
add_plugin(
placeholder=placeholder, #same placeholder as the parent plugin
plugin_type='LinkPlugin',
language=translation.get_language(),
target=target, #the parent plugin
#here comes the params from the selected plugin
name='Google',
url='http://www.google.com'
)
This also works with custom plugins.
Did you tried to save Link plugin that you've created?
plugin.link = Link(
name='Link text',
page_link=target_page,
placeholder=placeholder,
)
maybe try to add
plugin.link.save()
I hope that is the case.
To create nested plugins in your cms_plugins.py file
from .models import ParentPlugin, ChildPlugin
#plugin_pool.register_plugin
class ParentCMSPlugin(CMSPluginBase):
render_template = 'parent.html'
name = 'Parent'
model = ParentPlugin
allow_children = True # This enables the parent plugin to accept child plugins
# You can also specify a list of plugins that are accepted as children,
# or leave it away completely to accept all
# child_classes = ['ChildCMSPlugin']
def render(self, context, instance, placeholder):
context = super().render(context, instance, placeholder)
return context
#plugin_pool.register_plugin
class ChildCMSPlugin(CMSPluginBase):
render_template = 'child.html'
name = 'Child'
model = ChildPlugin
require_parent = True # Is it required that this plugin is a child of another plugin?
# You can also specify a list of plugins that are accepted as parents,
# or leave it away completely to accept all
# parent_classes = ['ParentCMSPlugin']
def render(self, context, instance, placeholder):
context = super(ChildCMSPlugin, self).render(context, instance, placeholder)
return context
In your plugin template file of parent plugin
{% load cms_tags %}
<div class="plugin parent">
{% for plugin in instance.child_plugin_instances %}
{% render_plugin plugin %}
{% endfor %}
</div>
For detailed documentation, go through
https://docs.django-cms.org/en/latest/how_to/custom_plugins.html#nested-plugins

Import excel data into models via django admin

I need the Django Admin interface to accept administrator uploads of Excel files where the data in each Excel file is inserted into my database models. How can I make such an “Upload” button appear on a Django model admin page, where clicking the button asks the administrator to choose an .xls file, whose data then gets added to database once its upload is complete?
I have done this, but I just set up a simple view with a file upload (actually this makes more sense than adding it directly into a Django admin page, as one edit page = one model instance,and I assume that your excel contains multiple models).
in forms.py, a simple form with a file upload field
class ImportExcelForm(forms.Form):
file = forms.FileField(label= "Choose excel to upload")
in views.py, a view to process the upload
def test_flowcell(request):
c = RequestContext(request, {'other_context':'details here'})
if request.method == 'POST': # If the form has been submitted...
form = ImportExcelForm(request.POST, request.FILES) # A form bound to the POST data
if form.is_valid(): # All validation rules pass
excel_parser= ExcelParser()
success, log = excel_parser.read_excel(request.FILES['file'] )
if success:
return redirect(reverse('admin:index') + "pages/flowcell_good/") ## redirects to aliquot page ordered by the most recent
else:
errors = '* Problem with flowcell * <br><br>log details below:<br>' + "<br>".join(log)
c['errors'] = mark_safe(errors)
else:
c['errors'] = form.errors
else:
form = ImportExcelForm() # An unbound form
c['form'] = form
return render_to_response('sequencing/file_upload.html')
and as suggested in the other post use xlrd to read the data in from the excel file. I have a separate file ExcelParser.py for this
import xlrd
class ExcelParser(object, excel_name):
#transaction.commit_on_success
def read_excel(self):
wb = xlrd.open_workbook(excel_name)
...
do your parsing in here.....
...
(May I add, that excel is a terrible, and error prone way to import data. I do a lot of it at my work, and am trying to convince management that there are far better solutions.)
I'm not sure about the Django side of things, but you can use xlrd to read and manipulate Excel files. There is a free PDF which explains this called Working with Excel files in Python
django-import-export could be helpful.
It creates two buttons "import" and "export" for admin objects and permits select many types of extensions, including xls. It also show the data do be imported and asks to be confirmed before execute the execution.
You just need to include it in INSTALLED_APPS and create an import-export resource of the class you want to upload and a subclass of ImportExportModelAdmin related to the resource class created before to show buttons in admin.
more info at:
http://django-import-export.readthedocs.org/en/latest/getting_started.html
https://github.com/bmihelac/django-import-export.

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