ataractic•12mo ago

Is there an idiomatic/common/modern way of creating a queue of jobs to be executed one after anot...

Is there an idiomatic/common/modern way of creating a queue of jobs to be executed one after another using Deno and typescript? Example: Multiple users wants to execute a resource heavy job by clicking a button on their dashboard, however to not lag the host, only one job should be executed at a time.

15 Replies

/|\ /|\•12mo ago

I would create an array with queued tasks, when system ends first task, it removes it from array and then start the next one

ataractic•12mo ago

so my idea is to create an array of objects that contain necessary information about a task but i would need to create a sub process that runs along the web server and checks for jobs? how can i do such a process? also how to create a global array

/|\ /|\•12mo ago

I would create tasks like: [function, callback] It executes the function and then calls the callback with the result of the function

ataractic•12mo ago

Since my task is always the same but for different conditions i can use the same function

/|\ /|\•12mo ago

Then maybe you could simply use: [args, callback]

ataractic•12mo ago

is the callback really needed?

/|\ /|\•12mo ago

Depends on what you want to do with the result

ataractic•12mo ago

i think i don't need a callback the real issue here is how to create a global variable and continuous running process that checks for the array

/|\ /|\•12mo ago

I think you have a globalThis object in deno No sorry it's just window So just use window[my array] Then create a function to execute the last queued task, and check it's running only once at a time. Each time you append something to the array, check if function is running Ex:

let running = false;
function startTask() {
    if (running) return;
    running = true; task(window["example"].shift());
running = false;
if (window["example"].length) startTask(); 
}

let running = false;
function startTask() {
    if (running) return;
    running = true; task(window["example"].shift());
running = false;
if (window["example"].length) startTask(); 
}

ataractic•12mo ago

i understand, will try this method your function has one problem we should check if there is more tasks in the queue out of the startTask() function, because executing the same function inside it can potentially never get out if there is always jobs to be done and i'm sure it would cause memory congestion or leaks at some point. but i see your point

ataractic•12mo ago

i think i found my answer with this deno workers https://deno.land/manual@v1.36.3/runtime/workers

Deno

Workers | Manual | Deno

Deno supports Web Worker API.

/|\ /|\•12mo ago

Yeah you should probably better use a while statement Looks perfect for your needs

Esente•12mo ago

Worker may not be available in Deno Deploy.

mxdvl•12mo ago

p-queue is a nice library that does exactly what you describe: https://deno.land/x/p_queue

ataractic•12mo ago

thank you everyone but i'm going with workers as a final choice