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¿Cómo excluir _id del resultado de mangosta?

Estoy teniendo problemas para obtener el resultado deseado como este

 [ { "status": "ONBOARD" "count": 1 }, { "status": "PENDING" "count": 1 }, ]

me sale asi:

 [ { "_id": { "status": "ONBOARD" }, "count": 1 }, { "_id": { "status": "CLIENT_ROUND" }, "count": 2 }, { "_id": { "status": "SCREENED" }, "count": 1 } ]

aquí está mi agregado:

 const result = await Lead.aggregate([ { $group: { _id: { status: '$status' }, count: { $sum: 1 } } $project: { _id: 0, status: 1, count: 1 } }

Ayúdame a obtener el resultado Soy nuevo, apliqué $ proyecto pero me da error

over 4 years ago · Santiago Trujillo
1 answers
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0

Ya casi estás ahí. Con status: "$_id.status" en la etapa de proyección.

 db.collection.aggregate([ { $group: { _id: { status: "$status" }, count: { $sum: 1 } } }, { $project: { _id: 0, status: "$_id.status", count: 1 } } ])

Ejemplo de parque infantil de Mongo


O simplemente $group con _id: "$status" .

 db.collection.aggregate([ { $group: { _id: "$status", count: { $sum: 1 } } }, { $project: { _id: 0, status: "$_id", count: 1 } } ])

Ejemplo de Mongo Playground 2

over 4 years ago · Santiago Trujillo Report
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