I want to create a back button to a page which can be accessed with a link from another page. To do this, I want to put the first page's path to the link, and on the next page I can put it to the back button.
I have this anchor tag:
Go to page
When I try to go to the site I get the following error:
Reverse for 'page' with keyword arguments '{'path': '/my_site/'}' not found. 1 pattern(s) tried: ['notifications\\/(?P<path>[^/]+)\\Z']
If you know the url name of the first page,
path('first_page/', views.first_page, name='first_page'),
then you can simply use:
Go the page
Otherwise, you must add the url of the previous page to the context of the second page's template in order to use that in that template. How to do that depends on your style. For instance, you can somehow keep a log of the visited pages on the client side or server side, and use that collection to determine the previous page.
Related
I am trying to use HTTP_REFERER to create a link to the previous page. However, let's say I have a 'page1' that calls the 'page2' get method. It will show a form at the 'page2' where I can do a post request. When I post on the 'page2', the HTTP_REFERER is referencing the same page (page2), therefore I can't to back to the 'page1'. Is there anything I can do to go back to the 'page1'?
You will need to add a hidden next field to your form . This will capture the previous page by making use of the HTTP_REFERER. After your post request you can then perform a redirect to the page you were on before the form page.
Edit:
More information can also be found here.
Right now. I have a search function in my page to search for item id. When I click search, I will render the same page with the result items and show item. And in other pages where I also display the item id, I want to add a link to the id to go to the same page where I search for that id.
Example: id: 123, I want the same page when:
1. search '123' in my search page(my search only accept exact match)
2. In other pages, click '123', go to the search page with results
How should I achieve this, I have tried many ways which don't wok.
You need to make use of the GET method that HTML forms provide. When you perform a search from the first page, you must make sure that you are doing so using the GET method in the form. This will append the form data into the URL.
E.g. If you have a 'name' field in your form which has 'John' inputted. The submission of this form will compose a URL like so:
http://someurl.com/?name=John
This can then be accessed using the Django request object:
name = request.GET['name']
You've probably done something similar already for displaying your search results. So, all you need to do is create a link in your second page that redirects to the search page with GET request variables appended.
E.g.
<a href="{% url 'search_page' %}?searchterm=232> Item 232 </a>
I am very new to Django and I'm nearing the end of the django girls tutorial. I have added "#login_required" above my post_detail in views (view for clicking on a specific post) and added a login.html template. So when I click on a post title I get redirected to my login page (so far, so good) and the url is: http://127.0.0.1:8000/accounts/login/?next=/post/11/ (trying this on my computer atm.)
Then I type in my admin name/password and automatically get redirected to http://127.0.0.1:8000/accounts/profile/ and of course get a "Page not found (404)" (since I have no url/view/template for that). I thought "Dang, I just wanted to be redirected to /post/11/"!
Looked around on stack overflow and found this question:
Signing in leads to "/accounts/profile/" in Django (sounds about right)
and got the answer
Change the value of LOGIN_REDIRECT_URL in your settings.py.
So I looked up LOGIN_REDIRECT_URL in the Django documentation:
Default: '/accounts/profile/'
The URL where requests are redirected after login when the contrib.auth.login view gets no next parameter.
This is used by the login_required() decorator, for example.
This setting also accepts named URL patterns which can be used to reduce configuration duplication since you don’t have to define the URL in two places (settings and URLconf).
Deprecated since version 1.8: The setting may also be a dotted Python path to a view function. Support for this will be removed in Django 1.10.
But doesn't my contrib.auth.login get a next parameter? (looking at my url that say "?next=/post/11/" at the end) Please help me out here, I'm lost for what the problem could be here :(
You can view the page at:
http://finbel.pythonanywhere.com/
And the source code at:
https://github.com/Finbel/my-first-blog
UPDATE (1):
So I now know that the LOGIN_REDIRECT_URL is the thing that's deciding where I end up next, which must mean that it ignores the next-parameter in the url. I googled further on the problem and found this question which was very similar to my problem, i.e.
Documentation states that I need to use the "next" parameter and context processors. I have the {{next}} in my template, but I'm confused on how to actually pass the "/gallery/(username)". Any help would be greatly appreciated.
(I don't even have the {{next}} in my template, where/how should I add it?)
The preferred answer to that question seemed to be:
Django's login view django.contrib.auth.views.login accepts a dictionary named extra_context. The values in the dictionary are directly passed to the template. So you can use that to set the next parameter. Once that is done, you can set a hidden field with name next and value {{ next }} so that it gets rendered in the template.
But I'm not sure how to interpret this. While writing this edit I got an answer on this post (by kacperd) and will read it through now)
The problem is that contrib.auth.login doesn't get the next parameter.
When you try to get the login_required view without credentials your request is redirect to login view, and the template you created is rendered. The next parameter is present in this view, but when you perform the login action which is submitting the form, you are not including next in your request so contrib.auth.login doesn't get it and redirects to default page.
The solution to your problem is to include the next param and pass it forward. You can do this by modifying your login template. Simply add ?next={{ request.GET.next }} to form action attribute.
<form method="post" action="{% url 'django.contrib.auth.views.login' %}?next={{ request.GET.next }}">
I originally followed this tutorial (https://django-haystack.readthedocs.org/en/latest/tutorial.html), and have so far been able to highlight my query within my returned results. However, I want to highlight this same query when visiting the next page that I load with a separate template. Is there any way to save/access this query so that I can highlight the same results within this other template?
Whenever I try and include a statement like this, I get an error, which I'm thinking is because I'm not trying to access the query properly.
{% highlight section.body with query html_tag "span" css_class "highlighted" %}
You have to send to the next page, the information that you use to highlight the results in the first page. You can use the request.session to store the data and call it in the next page, or you can send the sqs by the url to the next page.
If you want to know how to manage the search query set, and how to edit that kind of stuff, I recommend you to read the views.py forms.py and the elasticsearch_backend in the haystack folder at: "/usr/local/lib/python2.7/dist-packages/haystack"
This is the url for the documentation of Django Session: Django Session
This is the url for the documentation to pass parameters trhough url: URL dispatcher
I have a question: how to get the current path of the url. Let's say, I have 3 navigation bars, about , blog and contact page. In each page, I have facebook, twitter and a manual email a friend button. When I clicked the email a friend button, and the current URL is www.example.com/about, the current URL is now already www.example.com/emailafriend. How can I get the www.example/about? Also in blog and contact. Please help me. Thanks.
How does your email a friend button work? Is it a django view that takes the current URL and emails it? If so, you don't want the "current" URL, which, as you note, is actually the email a friend URL. What you want to do is pass the URL you want to share as a URL parameter, ie:
/share?url=http://www.example.com/blog
Adding more info based on comments:
When I was referencing URL above, I was not referring to your django URL configuration. Let's take a step back.
On your About page you have a link to email a friend, right? That link is probably generated in your template, but it's the same on every page. Something like:
Email a friend
Instead of this, try this:
Email a friend
Now you need to make your email_a_friend view handle this. It can get the url via
request.get('url', '').
Some additional information:
You might want to escape the {{ request.get_full_path }} function so that it's escaped and URL safe, then you'll have to unescape it in your view. Once you get the URL back to your view, you can do as you please with it.
{{ request.get_full_path|urlencode }}
Try using Relative URLs like for example From www.example.com/about to get to www.example.com/email use /email. Using relative urls is the simplest solution .
Take a look at this.
Absolute vs relative URLs
It sounds like your want to get the referring URL (the URL that sent you to the current page). That is available to you in the request object, although it is not 100% reliable:
request.META['HTTP_REFERER']
See the documentation on HttpRequest objects for more information.