forever | simple CLI tool for ensuring that a given script runs | Automation library

 by   foreversd JavaScript Version: 4.0.0 License: MIT

kandi X-RAY | forever Summary

kandi X-RAY | forever Summary

forever is a JavaScript library typically used in Automation applications. forever has no bugs, it has no vulnerabilities, it has a Permissive License and it has medium support. You can download it from GitHub.

A simple CLI tool for ensuring that a given script runs continuously (i.e. forever). Note that this project currently fully depends on the community for implementing fixes and new features. For new installations we encourage you to use pm2 or nodemon.

            kandi-support Support

              forever has a medium active ecosystem.
              It has 13769 star(s) with 987 fork(s). There are 278 watchers for this library.
              It had no major release in the last 12 months.
              There are 314 open issues and 606 have been closed. On average issues are closed in 369 days. There are 24 open pull requests and 0 closed requests.
              It has a neutral sentiment in the developer community.
              The latest version of forever is 4.0.0

            kandi-Quality Quality

              forever has 0 bugs and 0 code smells.

            kandi-Security Security

              forever has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
              forever code analysis shows 0 unresolved vulnerabilities.
              There are 0 security hotspots that need review.

            kandi-License License

              forever is licensed under the MIT License. This license is Permissive.
              Permissive licenses have the least restrictions, and you can use them in most projects.

            kandi-Reuse Reuse

              forever releases are available to install and integrate.
              Installation instructions, examples and code snippets are available.

            Top functions reviewed by kandi - BETA

            kandi has reviewed forever and discovered the below as its top functions. This is intended to give you an instant insight into forever implemented functionality, and help decide if they suit your requirements.
            • Stops the event and starts the process .
            • Detect worker protocol .
            • Get all connected processes in the process .
            • Send an action to the process .
            • Find the new worker .
            • tailing the process
            • get a new process
            • Called when a message is received
            • Checks if a given process has been removed
            • Generate random string
            Get all kandi verified functions for this library.

            forever Key Features

            No Key Features are available at this moment for forever.

            forever Examples and Code Snippets

            Sleep forever .
            pythondot img1Lines of Code : 14dot img1License : Non-SPDX (Apache License 2.0)
            copy iconCopy
            def SyncSleep(delay, name=None):
              """Pause for `delay` seconds (which need not be an integer).
              This is a synchronous (blocking) version of a sleep op. It's purpose is
              to be contrasted with Examples>AsyncSleep.
                delay: tf.Tensor w  
            Sleep forever .
            pythondot img2Lines of Code : 5dot img2License : Permissive (MIT License)
            copy iconCopy
            async def randsleep(a: int = 1, b: int = 5, caller=None) -> None:
                i = await randint(a, b)
                if caller:
                    print(f"{caller} sleeping for {i} seconds.")
                await asyncio.sleep(i)  

            Community Discussions


            Is there a way to skip await if it's too long? (Flutter)
            Asked 2022-Mar-14 at 04:39

            I use async await in main so the user has to wait in the splash screen before entering the app.



            Answered 2022-Mar-14 at 03:59

            You can use the timeout method on the Future class.

            You can pass a Duration, and an onTimeout callback which will be called if the duration of time has been exceeded.



            When does a mailbox processor stop running?
            Asked 2022-Mar-02 at 13:09

            I am having trouble understanding when a MailboxProcessor "finishes" in F#.

            I have collected some examples where the behavior is (perhaps) counter-intuitive.

            This mailbox processor prints nothing and halts the program:



            Answered 2022-Mar-02 at 13:09

            When started, MailboxProcessor will run the asynchronous computation specified as the body. It will continue running until the body finishes (either by reaching the end or by throwing an exception) or until the program itself is terminated (as the mailbox processor runs in the background).

            To comment on your examples:

            • This mailbox processor prints nothing and halts the program - I assume you run this in a console app that terminates after the mailbox processor is created. There is nothing blocking the program and so it ends (killing the mailbox processor in the background).

            • This mailbox processor counts up to 2207 then the program exits - I suspect this is for the same reason - your program creates the mailbox processor, which manages to run for a while, but then the program itself is terminated and the mailbox processor killed.

            • This mailbox processor prints 1 then the program exits - The body of the mailbox processor hangs and the next two messages are queued. The queue is never processed (because the body has hanged) and then your program terminates.

            You will get more useful insights if you add something like Console.ReadLine() to the end of your program, because this will prevent the program from terminating and killing the mailbox processor.

            For example, the following will process all 100000 items:



            Fill a 6x6 grid with 6 colors without same colors touching each other
            Asked 2022-Feb-09 at 11:39

            I'm trying to create a board game with p5.js (Javascript)

            To set up the game board which is a 6 by 6 grid, I have to fill the grid with 6 colors in a way that no horizontal or vertical touching cells have the same color. And all 6 colors have to be used in 6 cells.

            But now I'm struggling a bit creating an algorithm that places the colors randomly but keeping the rules.

            I tried to start at the top left corner, filling with a random color. Then I start to fill the cell to the left and the bottom with a different color.

            The problem is, that when the script wants to fill the last few cells, there are no colors left to use (either already 6 cells filled or a remaining color is a neighbor)

            Example: Still two cells need to be red, but only one place is left for red (under white):



            Answered 2022-Feb-06 at 18:57

            One way to look at this would be as searching for a path through a tree where each node has 6 possible children for the six colours which could come next. Ignoring all the constraints initially, you pick one of these at random 36 times, and have your order of placements.

            Using a recursive function (which will be useful in a moment), an unconstrained search would look like this:



            Why is `forever` in Haskell implemented this way?
            Asked 2022-Feb-05 at 20:34

            Haskell provides a convenient function forever that repeats a monadic effect indefinitely. It can be defined as follows:



            Answered 2022-Feb-05 at 20:34

            The execution engine starts off with a pointer to your loop, and lazily expands it as it needs to find out what IO action to execute next. With your definition of forever, here's what a few iterations of the loop like like in terms of "objects stored in memory":



            Angular / Ionic mobile app ios does not fetch from Firebase using angularfire
            Asked 2022-Feb-04 at 17:53

            I am trying to test a little Ionic/Angular sample app on an iOS Emulator.

            On the web, all the requests to firestore using angularfire work perfectly fine.

            Somehow if I try to execute the same app on the emulator, it keeps loading for the response of the request (if it was a empty response it would say that no results could be retrieved).

            What is going on? Do i need to set something specifically for the Emulator to work and perform requests to Firestore?



            Answered 2022-Jan-21 at 10:48
            import { initializeApp } from 'firebase/app';
            import { getFirestore } from 'firebase/firestore';
            import { Capacitor } from '@capacitor/core';
            import { initializeAuth, indexedDBLocalPersistence } from 'firebase/auth';
            import { getAuth } from 'firebase/auth';
            const firebaseApp = initializeApp({
             apiKey: process.env.VUE_APP_FIREBASE_API_KEY,
             authDomain: process.env.VUE_APP_FIREBASE_AUTH_DOMAIN,
             databaseURL: process.env.VUE_APP_FIREBASE_DATABASE_URL,
             projectId: process.env.VUE_APP_FIREBASE_PROJECT_ID,
             storageBucket: process.env.VUE_APP_FIREBASE_STORAGE_BUCKET,
             appId: process.env.VUE_APP_FIREBASE_APP_ID,
            function whichAuth() {
              let auth
              if (Capacitor.isNativePlatform()) {
                auth = initializeAuth(firebaseApp, {
                  persistence: indexedDBLocalPersistence
              } else {
                auth = getAuth()
              return auth
            export const auth = whichAuth()
            const db = getFirestore();
            export const auth = whichAuth();
            export { firebaseApp, db };



            AWS Checking StateMachines/StepFunctions concurrent runs
            Asked 2022-Feb-03 at 10:41

            I am having a lot of issues handling concurrent runs of a StateMachine (Step Function) that does have a GlueJob task in it.

            The state machine is initiated by a Lambda that gets trigger by a FIFO SQS queue.

            The lambda gets the message, checks how many of state machine instances are running and if this number is below the GlueJob concurrent runs threshold, it starts the State Machine.

            The problem I am having is that this check fails most of the time. The state machine starts although there is not enough concurrency available for my GlueJob. Obviously, the message the SQS queue passes to lambda gets processed, so if the state machine fails for this reason, that message is gone forever (unless I catch the exception and send back a new message to the queue).

            I believe this behavior is due to the speed messages gets processed by my lambda (although it's a FIFO queue, so 1 message at a time), and the fact that my checker cannot keep up.

            I have implemented some time.sleep() here and there to see if things get better, but no substantial improvement.

            I would like to ask you if you have ever had issues like this one and how you got them programmatically solved.

            Thanks in advance!

            This is my checker:



            Answered 2022-Jan-22 at 14:39

            You are going to run into problems with this approach because the call to start a new flow may not immediately cause the list_executions() to show a new number. There may be some seconds between requesting that a new workflow start, and the workflow actually starting. As far as I'm aware there are no strong consistency guarantees for the list_executions() API call.

            You need something that is strongly consistent, and DynamoDB atomic counters is a great solution for this problem. Amazon published a blog post detailing the use of DynamoDB for this exact scenario. The gist is that you would attempt to increment an atomic counter in DynamoDB, with a limit expression that causes the increment to fail if it would cause the counter to go above a certain value. Catching that failure/exception is how your Lambda function knows to send the message back to the queue. Then at the end of the workflow you call another Lambda function to decrement the counter.



            Scientific formats, subscripts and superscripts in RMarkdown table (docx output)
            Asked 2022-Jan-31 at 22:52

            Let's say I have the following rmd:



            Answered 2022-Jan-31 at 22:52

            Your code should look like this:



            Is this correct: std::views::reverse on infinite range?
            Asked 2022-Jan-30 at 17:17

            See this example code:



            Answered 2021-Dec-29 at 16:15

            According to [iterator.requirements.general-10]:

            A sentinel s is called reachable from an iterator i if and only if there is a finite sequence of applications of the expression ++i that makes i == s. If s is reachable from i, [i, s) denotes a valid range.

            And [iterator.requirements.general-12]:

            The result of the application of library functions to invalid ranges is undefined.

            Since ranges::iota_view(1) is not a valid range, applying views::reverse to it is undefined behavior.



            Express + Sequelize: hanging the app on connection
            Asked 2021-Dec-26 at 11:52

            I have an app with postgres as db, sequelize, and express, and whenever it receives a db query, it just stays there forever, no logging or anything I run postgres in a container which I can connect to through GUI normally When I swapped it for sqlite, it worked perfectly the application

            here is the relevant piece of code



            Answered 2021-Dec-12 at 05:49

            I think it is your "".

            If local, it should be just "localhost:5432". If deployed server is remote, it should be a certain IP address XXX.XXX.XXX.XXX:5432. If deployed server is home network, it should be "192.168.0.XXX:5432".

            Check your postgres network configuration



            pytest: full cleanup between tests
            Asked 2021-Dec-21 at 12:19

            In a module, I have two tests:



            Answered 2021-Dec-16 at 06:15

            The current structure of myfixture guarantee cleanup() is called between test_1 and test_2, unless prepare_stuff() is raising an unhandled exception. You will probably notice this, so the most likely issue is that cleanup() dosn't "clean" everything prepare_stuff() did, so prepare_stuff() can't setup something again.

            As for your question, there is nothing pytest related that can cause the hang between the tests. You can force cleanup() to be called (even if an exception is being raised) by adding finalizer, it will be called after the teardown part


            Community Discussions, Code Snippets contain sources that include Stack Exchange Network


            No vulnerabilities reported

            Install forever

            Note: If you are using forever programmatically you should install forever-monitor.


            For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .
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