I don't want the message to count as "read" but I'd like to know what's in the queue. The documentation:
http://boto.s3.amazonaws.com/ref/sqs.html#module-boto.sqs
Isn't very straight forward about what absorbs a message and what doesn't. The dump message seems close, but I'd rather do this in memory rather than to a file.
The faq:
http://aws.amazon.com/articles/1343#12
Has some sketchy solution:
How do I peek at a message?
With version 2008-01-01, the PeekMessage action has been removed from
Amazon SQS. This functionality was used mainly to debug small systems
— specifically to confirm a message was successfully sent to the queue
or deleted from the queue. To do this with version 2008-01-01, you can
log the message ID and the receipt handle for your messages and
correlate them to confirm when a message has been received and
deleted.
Has anyone had any luck with this? It seems like very basic queue functionality and I'd be shocked if there wasn't a clean way to do this.
Right click no longer works in the new SQS console.
To view queue messages in the SQS console you now need to click into a queue > Send and receive messages > Poll for messages
There is no longer a true peek function available in SQS but you can probably accomplish what you want by simply using get_messages and setting the visibility_timeout quite low. As long as you don't delete the messages you have read, they will reappear on the queue after the visibility_timeout has expired and will be available for reading. The only tricky part is trying to figure out how long the timeout should be. If you have lots and lots of messages in the queue, you will have to make multiple calls to get_messages to retrieve them all and you probably don't want previously read messages reappearing while you are still peeking at the messages.
Update 11/11/2020
Right-clicking no longer works on new SQS console.
See #marmor's answer
Original Answer (old dashboard)
If you have access to Amazon's AWS Console, on the queue list page, you can right-click on a queue.
Then select View/Delete Messages from the pop-up menu.
This will pop-up a window where you can start polling for messages in the queue.
Refer to images below:
I have created a desktop application named SQSCLI
A Graphical Desktop Application available for Windows & Mac
It has a Long Polling Support and it uses AWS profiles available in your local.
Video Demo of this application
https://www.youtube.com/watch?v=ALNHHvts9oo
More details on this link
https://www.middlewareinventory.com/blog/sqscli-app/
In the new AWS Console:
Visit the main queue page at https://console.aws.amazon.com/sqs
Click on the name of your queue - this will take you to its details page
Click on the "Send and Receive Messages" button (top right)
Click on the "Poll for Messages" button
Click on the message id to view message details
Related
There is a receive function at https://boto3.amazonaws.com/v1/documentation/api/latest/reference/services/sqs.html#SQS.Client.receive_message to get SQS message,
Is there a function that I can just take a Peek at the SQS messages, without actually receiving it. Because If I receive the messages, it will be deleted from the queue. But I want the messages to be stay in the queue after peeking.
you can check
https://docs.aws.amazon.com/AWSSimpleQueueService/latest/SQSDeveloperGuide/sqs-visibility-timeout.html
aws sqs sdk's (and client libraries written on top of them) by default they dont delete messages. but they have 'Visibility Timeout' which is 30 seconds by default. That means after you read the message , it wont be visible to other consumers for 30 seconds. It is up to a client to delete it within that time frame so that no one else will get that message ever.
So You can reduce that visility time out to something really small like 1 second. So you can download the message and within 1 second it will be available to other consumers . you can even set it to 0 so everyone can read the message at any point.
But this still means you will receieve the message. SQS is pretty simple queue system. you might want to check other queue system like Kafka or different way to design your system like using Notification services such as SNS
I'm looking for a good solution to implement a handshake between a python backend server and a react frontend connected through a websocket.
The frontend allows the user to upload a file and then send it to the backend for processing. Now it might be possible that the processing encounters some issues and likes the user's confirmation to proceed or cancel - and that's where I'm stuck.
My current implementation has different "endpoints" in the backend which call then different function implementations and a queue which is continuously processed and content (messages) is sent to the frontend. But these are always complete actions, they either succeed or fail and the returned message is accordingly. I have no system in place to interupt a running task (e.g. file processing), send a request to the frontend and then wait for response before I continue the function.
Is there a design pattern or common approach for this kind of problem?
How long it takes to process? Maybe a good solution is set up a message broker like RabbitMQ and create a queue for this process. In the front-end you have to create a panel to see the state of the process, which is running in an async task, and if it has found some issues, let the user know and ask what to do.
I need to execute a command on a simple button press event in my Django project (for which I'm using "subprocess.Popen()" in my views.py ).
After I execute this script it may take anywhere from 2 minutes to 5 minutes to complete. So while the script executes I need to disable the html button but I want the users to continue using other web pages while the script finishes in the background. Now the real problem is that I want to enable the html button back, when the process finishes!
I'm stuck at this from many days. Any help or suggestion is really really appreciated.
I think you have to use some "realtime" libraries for django. I personally know django-realtime (simple one) and swampdragon (less simple, but more functional). With both of this libraries you can create web-socket connection and send messages to clients from server that way. It may be command for enabling html button or javascript alert or whatever you want.
In your case I advice you first option, because you can send message to client directly from any view. And swampdragon needs model to track changes as far I know.
Like valentjedi suggested, you should be using swampdragon for real time with django.
You should take the first tutorial here: http://swampdragon.net/tutorial/part-1-here-be-dragons-and-thats-a-good-thing/
Then read this as it holds knowledge required to accomplish what you want:
http://swampdragon.net/tutorial/building-a-real-time-server-monitor-app-with-swampdragon-and-django/
However there is a difference between your situation and the example given above, in your situation:
Use Celery or any other task queue, since the action you wait for takes long time to finish, you will need to pass it to the background. (You can also make these tasks occur one after another if you don't want to freeze your system with enormous memory usage).
Move the part of code that runs the script to your celery task, in this case, Popen should be called in your Celery task and not in your view (router in swampdragon).
You then create a channel with the user's unique identifier, and add relevant swampdragon javascript code in your html file for the button to subscribe to that user's channel (also consider disabling the feature on your view (router) since front-end code can be tempered with.
The channel's role will be to pull the celery task state, you
then disable or enable the button according to the state of
the task.
overview:
Create celery task for your script.
Create a user unique channel that pulls the task state.
Disable or enable the button on the front-end according to the state of the taks, consider displaying failure message in case the script fails so that the user restart again.
Hope this helps!
I would like to create a Twilio app in python that runs in the background on my mac osx desktop and performs an action whenever a new text message arrives. I'm only interested in processing the most recently received text message.
My current solution is a polling mechanism where I have an infinite loop and in the loop I call TwilioRestClient(...).messages.list() every single time and process the first message in the returned array. I sleep for 1 second in each iteration in order to avoid taxing the processor too much.
The current implementation usually works but sometimes lags at the TwilioRestClient(...).messages.list() call for multiple seconds. I'm assuming this is because I'm using the trial version of Twilio. However, any advice on how to fix this would be greatly appreciated. The other main issue is that the implementation is extremely wasteful. The app should really be sleeping for the vast majority of the time it is running and there should be brief spurts of activity when new messages arrive.
Here's the current implementation:
def run(self):
last_message = self.get_most_recent_message_body().lower()
self.is_running = True
while self.is_running:
message = self.get_most_recent_message_body().lower()
if (message != last_message):
last_message = message
self.commander.execute(message)
sleep(1);
def get_most_recent_message_body(self):
messages = TwilioRestClient(self.TWILIO_ACCOUNT_SID, self.TWILIO_AUTH_TOKEN).messages.list()
most_recent_message = messages[0]
return most_recent_message.body
I want the app to be as responsive as possible. The optimal solution would be some kind of Observer pattern or some kind of signal use where Twilio alerts my app to "wake up" whenever a new message arrives. Is there a way to implement this?
Thanks!
Twilio developer evangelist here.
You can absolutely do that without polling. Let me explain.
When you buy a number with Twilio you can set two URLs that point to an application that you build, one for voice and one for messaging. Then, when Twilio receives a call or a text message it will make an HTTP request (a webhook) to the url you set with the details of the call or the message. Your application can respond with some XML (called TwiML) to tell Twilio what to do with the message or call.
I'm guessing you're writing a python application as you tagged this question with Python. Here's a quick guide to getting started receiving SMS messages with Python.
I have a some EC2 servers pulling work off of a SQS queue. Occasionally, they encounter a situation where the can't finish the job. I have the process email me of the condition. As it stands now, the message stays "in flight" until it times out. I would like for the process to immediately release it back to the queue after the email is sent. But, I'm not sure how to accomplish this. Is there a way? If so, can you please point me to the call or post a code snippet.
I'm using Python 2.7.3 and Boto 2.5.2.
If you have read a message and decide, for whatever reason, that you do not want to process it and would rather make it immediately available to other readers of the queue, you can simply set that message's visibility timeout to zero using the change_visibility method of the Message object in boto. See The SQS Developer's Guide for details.