How can I run this simple node.js code synchronously using async.js?

Question

This is a simple function which will get a post url and return the post id of that url.

function findPostIdByUrl (url) {

  var id;

  Post.findOne({url}, '_id', function (err, post) {
    if (err) throw err;
    id = post.id;
  });

  return id;
}

but it doesn't return the actual id, because it run asynchronously. I want to first run Post.fin... code which will assign the post id to the id variable, and then run return id.

I've tried my best but I didn't figure out how can I do that. Is there any way to accomplish that?(whether by using async.js or not)

I don't know how, I've search on the internet but I didn't find anything. — Joseph
– Joseph, Commented Jul 17, 2017 at 12:31

kikiwie · Accepted Answer · 2017-07-17 12:37:00Z

2

What you can do here is get all the data from your request using async/await

so your code here will look like:

async function findPostIdByUrl (url) {
   var id;
   var post = await Post.findOne({url}, '_id')
   id = post.id
   return id;
}

answered Jul 17, 2017 at 12:37

kikiwie

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7 Comments

Joseph Over a year ago

It is not the valid code, when I run it I get: SyntaxError: Unexpected token function

kikiwie Over a year ago

can you post your function please?

kikiwie Over a year ago

and your node has to be v7.6 minimum or you cant use async/await (i m pretty sure its 7.6 but i m not 100% sure)

Joseph Over a year ago

async function findPostIdByUrl (url) { var id; var post = await Post.findOne({url}, '_id') id = post.id return id; } var id = findPostIdByUrl('i-am-gonna-create-a-post'); console.log(id);

Joseph Over a year ago

Oh I am using node 6.

|

Shivam · Accepted Answer · 2017-07-17 12:32:47Z

0

You can use Promises.

function findPostIdByUrl (url) {
  var id;
  return Post.findOne({url}, '_id').then((post) => {
      id = post.id
      return id;
  })
  .catch((err) => {/* Do something with err */})
}

You can actually skip setting the id.

return Post.findOne({url}, '_id').then((post) => {
   return post.id;
})

Also Post this, findPostIdByUrl should be used as

findPostIdByUrl(url).then((id) => {/* Whatever you need to do with id*/})

answered Jul 17, 2017 at 12:32

Shivam

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2 Comments

Joseph Over a year ago

It works but I want another thing. I want to be able to use the function like this: var id = findPostIdByUrl(url);

Shivam Over a year ago

Unfortunately, Thats not how async works in JS. You can very well use the async solution below by @kikiwie. That would to an extent make it look like var id = await findPostIdByUrl(url); But be advised. thats just a syntactic sugar

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