can anyone help me for how can run mongoose query in forEach loop in nodejs and suggest for inner join result need of both collections
like below details
userSchema.find({}, function(err, users) {
if (err) throw err;
users.forEach(function(u,i){
var users = [];
jobSchema.find({u_sno:s.u.sno}, function(err, j) {
if (err) throw err;
if (!u) {
res.end(JSON.stringify({
status: 'failed:Auction not found.',
error_code: '404'
}));
console.log("User not found.");
return
}
users.push(j);
})
})
res.send(JSON.stringify({status:"success",message:"successfully done",data:{jobs:j,users:u}}));
})
A nice elegant solution is to use the cursor.eachAsync()
function. Credit to https://thecodebarbarian.com/getting-started-with-async-iterators-in-node-js.
The eachAsync() function executes a (potentially async) function for each document that the cursor returns. If that function returns a promise, it will wait for that promise to resolve before getting the next document. This is the easiest way to exhaust a cursor in mongoose.
// A cursor has a `.next()` function that returns a promise. The promise
// will resolve to the next doc if there is one, or null if they are no
// more results.
const cursor = MyModel.find().sort({name: 1 }).cursor();
let count = 0;
console.log(new Date());
await cursor.eachAsync(async function(doc) {
// Wait 1 second before printing first doc, and 0.5 before printing 2nd
await new Promise(resolve => setTimeout(() => resolve(), 1000 - 500 * (count++)));
console.log(new Date(), doc);
});
Schema.find() is an async function. So your last line of code will execute while you wait for the first job search is executed in your loop. I suggest change it to Promises and use Promise.all(array).
To do so, first you have to change to use Promise with mongoose. you can do this with bluebird like this:
var mongoose = require('mongoose');
mongoose.Promise = require('bluebird');
Then you can use Promises instead of callbacks like this:
userSchema.find({}).then(function(users) {
var jobQueries = [];
users.forEach(function(u) {
jobQueries.push(jobSchema.find({u_sno:s.u.sno}));
});
return Promise.all(jobQueries );
}).then(function(listOfJobs) {
res.send(listOfJobs);
}).catch(function(error) {
res.status(500).send('one of the queries failed', error);
});
EDIT How to list both jobs and users
If you want to have a structure like:
[{
user: { /* user object */,
jobs: [ /* jobs */ ]
}]
you could merge the lists together. listOfJobs is in the same order as the jobQueries list, so they are in the same order as the users. Save users to a shared scope to get access to the list in the 'then function' and then merge.
..
}).then(function(listOfJobs) {
var results = [];
for (var i = 0; i < listOfJobs.length; i++) {
results.push({
user: users[i],
jobs: listOfJobs[i]
});
}
res.send(results);
}).catch(function(error) {
res.status(500).send('one of the queries failed', error);
});
No need to use forEach()
which is synchronous and being called in an asynchronous fashion, that will give you wrong results.
You can use the aggregation framework and use $lookup
which performs a left outer join to another collection in the same database to filter in documents from the "joined" collection for processing.
So the same query can be done using a single aggregation pipeline as:
userSchema.aggregate([
{
"$lookup": {
"from": "jobs", /* underlying collection for jobSchema */
"localField": "sno",
"foreignField": "u_sno",
"as": "jobs"
}
}
]).exec(function(err, docs){
if (err) throw err;
res.send(
JSON.stringify({
status: "success",
message: "successfully done",
data: docs
})
);
})
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