I am developing a chatbot using NodeJS and BotBuilder. I have the file chatbot.jpg stored in a folder named image. This sits just beneath the root directory. I am able to display the image (using Kudu to find the URL) in a web browser as follows:
https://mysite.scm.azurewebsites.net/api/vfs/site/wwwroot/images/chatbot.jpg
If I remove the .scm element within the URL, I can no longer display the image.
Even more curiously, if I use the above URL in my bot code, the image doesn't display.
var welcomeCard = new builder.HeroCard(session)
.title("This is the new")
.subtitle('Virtual Assistant')
.images([
new builder.CardImage(session)
.url("http://mysite.scm.azurewebsites.net/api/vfs/site/wwwroot/images/chatbot.jpg")
alt("Virtual Assistant")
]);
session.send(new builder.Message(session)
.addAttachment(welcomeCard));
My question is, how do I find out the regular URL of the image stored in the Azure App Service, so that I can use it in my code?
From the public web, the URL should be https://mysite.azurewebsites.net/images/chatbot.jpg
The wwwroot folder is the root folder served by the app service.
Your code can't just the scm URL as that URL requires you to be logged in to the Azure portal; it's an admin URL.
To accomplish your goal, you need to configure your Restify server to serve static files.
Example Restify config (add to your bot code):
server.get(/\/images\/?.*/, restify.serveStatic({
directory: './images'
}));
Related
I want to make a main URL and that URL will be the main route for all my Django apps.
Suppose I have 3 apps: blog, linux and business. I want to make URLs like this:
127.0.0.1:8000/blog/ (when i will click blog menu)
127.0.0.1:8000/blog/linux ( when i will click linux menu)
127.0.0.1:8000/blog/business (when i will click business menu)
The blog app will be my index home page and all the apps' links will be like this. I don't want to import all of the views and models in the blog app. The project's structure will be the same and I don't want to change it.
All I want is to make blog common for all the apps. If I click a post on a linux page, the URL will be like this:
127.0.0.1:8000/blog/linux/post name
How can I do this?
This is my project's structure:
[]
You can write all the URLs into the urls.py in your project folder or you can make a urls.py in your apps folder. Maybe read this
After Creating the App in django project by following procedure in https://docs.djangoproject.com/en/3.0/intro/tutorial01/
App Folder (blog app folder, found inside project folder)
Add these urls
127.0.0.1:8000/blog/linux
127.0.0.1:8000/blog/business
to the app/urls.py file above the admin.site url path
Project Folder:
Configure the urls.py file in the project folder by adding the 127.0.0.1:8000/blog/ url above the admin.site url path. This configuration is done to load this url when the server starts running.
The django checks for the requested url pattern and will route the first discovered url.
The django cannot identify duplicate url.
I am not able to see my SPA page after deploying it to Azure WebApp from VS Code. Its says
"Hey, Node developers!
Your app service is up and running.
Time to take the next step and deploy your code."
I have seen at so may site that I need to set default document and New physical path. But i don't see any Default Document Tab in Configuration menu of Web App. There are only three tabs. 1- Application Setting 2- General Setting 3- Path Mapping.
The Issue is where to set the Default document and new physical path.
If you deployed to a Node Linux Web App the default document would be hostingstart.html located in /home/site/wwwroot/.
According this document:Things You Should Know: Web Apps and Linux, there is a description about default document in a Node.js app.
When you create a Node.js app, by default, it's going to use hostingstart.html as the default document unless you configure it to look for a different file. You can use a JavaScript file to configure your default document. Create a file called index.js in the root folder of your site and add the following content.
So go to your ssh terminal, navigate to /home/site/wwwroot , create the index.js with the following code:
var express = require('express');
var server = express();
var options = {
index: 'index.html'
};
server.use('/', express.static('/home/site/wwwroot', options));
server.listen(process.env.PORT);
I''m developing an android app, a web server using flask and firebase.
When client app uploads an image file, the web server saves an image url in database.
And then, client app gets the image url and open image from firebase storage.
So my web server, and app don't know file name.
However, to delete a file in storage,
Knowing the file's name is needed.
Only what I can get is file url.
How can my web server delete a file using file url?
The following code is for deleteting file using filename.
I wanna change this code to deleteting file using file url.
bucket = storage.bucket('<BUCKET_NAME>')
def deleteFile(imageName):
try:
bucket.delete_blob('profile_images/' + imageName)
bucket.delete()
return True
except Exception as e:
print(e)
return False
The Storage client in the Firebase Admin SDK is a thin wrapper around the Node.js client for Google Cloud Storage. And as far as I know the latter doesn't have a way to map the download URL back to a File.
This means that you'll have to find the path from the client using FirebaseStorage.getReferenceFromUrl(), and then pass that path to your web server. That way the JavaScript code can use the path to create a reference to the File.
I have deployed an ASP.NET CORE web API project on Azure app services. I have copied a file using an FTP client to /site/wwwroot. Now let suppose file name is xyz.jpg, so it should be accessible with link somename.azurewebsites.net/xyz.jpg but ITS NOT. I have tried pasting the file in other folders to but nothing works.
I also have a controller for uploading pictures. It's also working fine. It uploads the picture in desired folder, i can see the picture via FTP client but still the picture is not accessible via any link. What am I doing wrong here ?
For a Web API application, you have to define the request and response yourself in the controller, or your link can't be recognized by the application.
For example, you can add the method to your controller. It works on my side.
[Route("myroute/{pic}")]
public IActionResult Get(string pic)
{
Byte[] b = System.IO.File.ReadAllBytes("image/"+pic);
return File(b, "image/jpeg");
}
In my code, pictures are stored in the folder called image in the root directory, and I define a route called myroute.
Here's my link to access the picture.https://myappname.azurewebsites.net/myroute/mypicname.jpg
Hope it helps.
I have created a small web application with NodeJS Express. Basically a webserver that has a 'webserver.properties' file. With a very basic app.yaml file.
After deploying it to Google Cloud by use of 'gcloud app deploy' I get the everything up and running.
However...when I open the following URL in the browser: https://webserverurl.com/webserver.properties , the webserver.properties file can be approached and is in turn downloaded immediately.
How can I prevent this from happening and make sure that such properties files are inaccessible from outside?
The problem is that when you use this line:
app.use('/', express.static(__dirname + '/'));
you are giving access to your root directory. See this for a definition of __dirname. If you want to give access to a specific folder you can do this:
Lets say your root directory is src and you fave a dir with static files called src/myfiles. In order to give acces to files in myfiles you can use this line:
app.use('/mypathname', express.static('myfiles'));
where:
'/mypathname' is the part pertaining your URL. In your case it would be https://webserverurl.com/mypathname/any-file-name.jpg
express.static('myfiles') is the name of your local dir.
See this guide.
Hope this helps