mongoose concat and single element from array

Question

This is a two part mongoDB/mongoose question

I want to concat a First Name/Last Name into "name" AND I only want to show the single "current" item from an array.

So if my data looks like this

[{
    "fname":"bob",
    "lname":"jones",
    "role":"professional", 
    "active":true,
    "jobs":[{
        "job":"janitor",
        "current":true
    },{
        "job":"dog groomer"
        "current":false
    }]
},{
    "fname":"sally",
    "lname":"peterson",
    "role":"professional", 
    "active":true,
    "jobs":[{
        "job":"engineer",
        "current":false
    },{
        "job":"college admin"
        "current":true
    }]
},{
    "fname":"jackson",
    "lname":"smiley",
    "role":"professional", 
    "active":true,
    "jobs":[{
        "job":"car salesman",
        "current":false
    },{
        "job":"street sweeper"
        "current":false
    }{
        "job":"house painter"
        "current":true
    }]
},{
    "fname":"katie",
    "lname":"smiley",
    "role":"amature", 
    "active":true,
    "jobs":[{
        "job":"drone entheuast",
        "current":true
    }]
}]

And I want my return data to be

[{
    name:"bob jones", 
    job:"janitor"
},{
    name:"sally peterson", 
    job:"college admin"},
{
    name:"jackson smiley", 
    job:"house painter"
}]

Currently - I am using this mongoose syntax - but it's not enough...

module.exports.getActiveList = function( callback ) {
        const query = { "role":"professional", "active":true }
        People.find( query, 'name job', callback );
    }

How would I do that?

Ashh · Accepted Answer · 2018-09-09 18:15:06Z

1

You can try below aggregation

You can use $concat to combine both fname and lname as a name and $filter to obtain current active job from the jobs array

People.aggregate([
  { "$match": { "role": "professional", "active": true }},
  { "$project": {
    "name": { "$concat": ["$fname", " ", "$lname"] },
    "jobs": {
      "$filter": {
        "input": "$jobs",
        "as": "job",
        "cond": { "$eq": ["$$job.current", true] }
      }
    }
  }},
  { "$project": { "name": 1, "job": { "$arrayElemAt": ["$jobs.job", 0] }}}
])

Or using $let to complete in a single stage

People.aggregate([
  { "$match": { "role": "professional", "active": true }},
  { "$project": {
    "name": { "$concat": ["$fname", " ", "$lname"] },
    "job": {
      "$let": {
        "vars": {
          "jobs": {
            "$filter": {
              "input": "$jobs",
              "as": "job",
              "cond": { "$eq": ["$$job.current", true] }
            }
          }
        },
        "in": { "$arrayElemAt": ["$$jobs.job", 0] }
      }
    }
  }}
])

edited Sep 9, 2018 at 18:15

answered Sep 9, 2018 at 17:59

Ashh

46.6k16 gold badges111 silver badges137 bronze badges

Sign up to request clarification or add additional context in comments.

8 Comments

j-p Over a year ago

Anthony, thank you for your time. I've modified my question a bit, because I don't know how the aggregation syntax fits in with the query filter syntax..see above question modifications

j-p Over a year ago

I am trying your suggestions now

Ashh Over a year ago

ok great... I have updated the answer according to your question

j-p Over a year ago

hum - I get this error - MongoError: The 'cursor' option is required, except for aggregate with the explain argument -- also can you explain what "$arrayElemAt": ["$$jobs.job", 0] is doing...

Ashh Over a year ago

Here is your first answer stackoverflow.com/questions/50101104/…... and the $arrayElemAt takes the indexed element from the array (the second argument is for index)

|

Collectives™ on Stack Overflow

mongoose concat and single element from array

1 Answer 1

8 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

8 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related