I have three functions
async function inventories() {
print("starting inventories")
sleepFor(5000)
// Here I am querying a mongodb database
const inventoriesRes = await db.inventories.find({})
return inventoriesRes
}
async function costs() {
print("starting costs")
const costsRes = await db.costs.find({})
return costsRes
}
async function users() {
print("starting users")
const users = await db.users.find({})
return usersRes
}
I want :
Here's my solution
const dbqueries = async () => {
const [result1, result2, result3] = await Promise.all([inventories(), costs(), users()])
return {result1, result2, result3}
}
dbqueries.then(results => {
print(JSON.stringify(results))
}).catch(error => {
throw error
});
The problem is that I am not this solution works as I want. To verify, I've put a sleep (see code below) in the inventories() function.
What's happening is
printing starting inventories
waiting for 5 seconds
printing costs
printing users
This let me think that costs() didn't start in parallel. If it had started earlier, I would have
printing starting inventories
printing costs
printing users
...
function sleepFor(sleepDuration){
var now = new Date().getTime();
while(new Date().getTime() < now + sleepDuration){
/* Do nothing */
}
}
Either my solution is wrong, or the sleepFor function is doing something more than just putting to sleep only the inventories() function. Any help?
PS. print works exactly like console.log() in mongoshell which is a fully-functional JS interpreter
…or the
sleepForfunction is doing something more than just putting to sleep only theinventories()function
Precisely that. You've wrote a synchronous, blocking sleep function, which stops the entire program. You've called it before any await in the function, so it's not even sleeping later while the concurrent work is already under way, you're sleeping before inventories() returns its promise and the other two functions can get called.
Your original program did work as expected. For verification, you should have added another print statement to log when each database call is finished:
async function inventories() {
print("starting inventories")
// Here I am querying a mongodb database
const inventoriesRes = await db.inventories.find({})
print("finished inventories")
return inventoriesRes
}
Remember that JS code does not run in parallel, on multiple threads or anything. All you can do is to start multiple asynchronous tasks (in your case the database queries), without immediately waiting for each to finish. Rather, you start them all, then wait for the results while they do their work concurrently.