Create the index2. Create the index3. Modify the views. Three functions had been defined in the script to return the current URL value in three different formats to the template. The geturl1 function has been defined to retrieve the domain name of the current URL and send it to the index. The geturl2 function has been defined to retrieve the domain name with the path of the current URL and send it to the index2.
On the other hand, request. The geturl3 function has been defined to retrieve the domain name with the http and the path of the current URL and send it to the index3. NOTE: you don't have to include request in settings. The other answers were incorrect, at least in my case. I did not find any proper solution, so I ended up hardcoding the constant part of the url in the View before concatenating it with request.
This is an old question but it can be summed up as easily as this if you're using django-registration. In your Log In and Log Out link lets say in your page header add the next parameter to the link which will go to login or logout. Your link should look like this. That's simply it, nothing else needs to be done, upon logout they will immediately be redirected to the page they are at, for log in, they will fill out the form and it will then redirect to the page that they were on.
Even if they incorrectly try to log in it still works. I am pasting the part of the HTML code snippet that you can see below to understand the use of request.
In Django 3, you want to use url template tag:. For an example, see the documentation. I add the below code in the template file. Stack Overflow for Teams — Collaborate and share knowledge with a private group.
Create a free Team What is Teams? Collectives on Stack Overflow. Learn more. How to get the current URL within a Django template? Ask Question. Asked 11 years, 8 months ago. Active 1 month ago. Viewed k times. Firstly, you have to create a Django template where you will render your results.
To pass a URL parameter to a Django template, you have to edit the views. A sample view. This is the function that will be executed when the specified URL endpoint is hit. This function is taking the URL parameter and passes it to a Django template as a dictionary.
In the Django template, you can receive the parameter and display the results accordingly. For example, you can see in the below sample Django template how to fetch the parameter. As you have seen the steps to get URL parameters in Django, now is the time to see an example. You will get a clearer image of the process by understanding it with an example. Read: Python Django vs Flask. In this section, I will be using my localhost address for the demonstration purpose.
I have created a path to the products endpoint. I have defined the path converter as int to specify that the URL parameter will be an integer. Do not get confused with the path named index. It is just a path pointing to my base URL. The views. The above function will receive the parameter from the request. I have passed this parameter to a Django template named result.
This Django template will be rendered when the request is made. Now let us have a look at the output:. How to Create an App in Django? Skip to content. Change Language. Related Articles. Table of Contents. Improve Article. Save Article.
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