MongoDB query with array of IDs

Question

I have this collection:

{
        "_id": "5f1bd6c7d90cb03e845adbbf",
        "name": "John Doe",
        "likes": [
            {
                "_id": "5f378b105d337347e8958a50",
                "user": "5f357ac9c0d2019cc82b5a2f",
                "post": "5f3032894cd4a000175d333c",
            },
            {
                "_id": "5f354e351a98dd7c7081d72d",
                "user": "5f357ac9c0d2019cc82b5a2f",
                "post": "5f3bf7a44ed4f031bc575c2b"
            },
            {
                "_id": "5f354cc2879f8a81b0ab28fe",
                "user": "5f30380288a63e001755401e",
                "post": "5f310ce5e18bad001792fb60"
            }
        ]

Now I want to find all posts that match the IDs of likes.post.

This works but obviously only gives me one post:

const posts = await Post.find({ _id: [user.likes[0].post] });
res.json(posts);

I tried a for-loop but it only gives back one post. When I console log this I get all posts that match:

for (let i = 0; i < user.likes.length; i++) {
  const posts = await Post.find({ _id: [user.likes[i].post] });
  res.json(posts);
}

Lee Jeonghyun · Accepted Answer · 2020-08-19 09:21:54Z

1

You can do that by using $in operator.

const posts = await Post.find({ _id: { $in: user.likes.map(l => l.post) } });
res.json(posts);

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answered Aug 19, 2020 at 9:21

Lee Jeonghyun

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Comments

SamAko · Accepted Answer · 2020-08-19 10:57:05Z

I tried a for-loop but it only gives back one post. When I console log this I get all posts that match:

The loop construct fails because you're dispatching a response after the first iteration and the connection is closed. Here's how to fix that:

const posts = [];
for (let i = 0; i < user.likes.length; i++) {
  const post = await Post.find({ _id: [user.likes[i].post] });
  posts.push(post)
}

res.json(posts)

Or use a map like so:

const postPromises = user.likes.map(({ post }) => {
  return Post.find({ _id: [user.likes[i].post] });
})

const posts = Promise.all(postPromises);
res.json(post)

But then, instead of running two separate queries, you can modify whatever query you run to obtain that document. Assuming you are using mongoose:

Author
  .findOne({ name: 'John Doe' })
  .populate({
    path: 'likes',
    populate: { path: 'post' }
  })
;

I'm on the move and can't test this now, but if that fails, try this instead:

Author
  .findOne({ name: 'John Doe' })
  .populate('likes.post')
;

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