CODEWITHSUNDEEP
mongodb
Mia
Mia

Reputation: 6531

How to .find() "url of the last array element"?

Here is my document structure:

{
        "name": "object name",
        "details": "object details",
        "images": [{
                "size": "large",
                "url": "http://domain.com/large.png"
        }, {
                "size": "medium",
                "url": "http://domain.com/medium.png"
        }, {
                "size": "small",
                "url": "http://domain.com/small.png"
        }]
}

Some documents only have the "large" images, some of them have all, some of them only have the "medium" image. What I want to do is, to get the url of the last element of the array.

Here is what I tried:

.find({},{"images.url":1})

Returns all url's. I tried using $slice as well, without any success (parameter errors).

Please note that I am trying to get the "url of the last element" not the "last element"

Looking for a solution.

Upvotes: 2

Views: 104

Answers (2)

Mia
Mia

Reputation: 6531

I realized that the reason $slice was not working for me was that my mongodb was version 2.6.6 I updated my mongodb to version 3.0.4 and it works perfectly as follows:

.find({},{"images.url":1,"images.url":{$slice:-1}})

Thank you for all the comments so far.

Upvotes: 1

Melvyn Marigny
Melvyn Marigny

Reputation: 131

You can use map() to filter the response : 

.find({}, { _id: 0, name: 0, details: 0, images: { $slice: -1 } }).map( function(u) { return u.images[0].url; } );

Upvotes: 1

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