使用 MongoDB 检查字段是否存在

Question

So I'm attempting to find all records who have a field set and isn't null.

I try using $exists , however according to theMongoDB documentation, this query will return fields who equal null.

$exists does match documents that contain the field that stores the null value.

So I'm now assuming I'll have to do something like this:

 db.collection.find({ "fieldToCheck": { $exists: true, $not: null } })

Whenever I try this however, I get the error [invalid use of $not] Anyone have an idea of how to query for this?

Answer 1

Use $ne (for "not equal")

 db.collection.find({ "fieldToCheck": { $ne: null } })

Answer 2

Suppose we have a collection like below:

 { "_id":"1234" "open":"Yes" "things":{ "paper":1234 "bottle":"Available" "bottle_count":40 } }

We want to know if the bottle field is present or not?

Ans:

 db.products.find({"things.bottle":{"$exists":true}})

Answer 3

i find that this works for me

db.getCollection('collectionName').findOne({"fieldName": {$ne: null}})

Answer 4

db.<COLLECTION NAME>.find({ "<FIELD NAME>": { $exists: true, $ne: null } })

Answer 5

This comment is written in 2021 and applies for MongoDB 5.X and earlier versions.

If you value query performance never use $exists (or use it only when you have a sparse index over the field that is queried. the sparse index should match the criteria of the query, meaning, if searching for $exists:true , the sparse index should be over field:{$exist:true} , if you are querying where $exists:true the sparse index should be over field:{$exist:false}

Instead use:

 db.collection.find({ "fieldToCheck": { $ne: null } })

or

db.collection.find({ "fieldToCheck": { $eq: null } })

this will require that you include the fieldToCheck in every document of the collection, however - the performance will be vastly improved.

Answer 6

I Tried to convert it into boolean condition, where if document with table name already exist, then it will append in the same document, otherwise it will create one.

table_name is the variable using which i am trying to find the document

query = { table_name: {"$exists": "True"}} result = collection.find(query) flag = 0 for doc in result: collection.update_one({}, { "$push": { table_name: {'name':'hello'} } } ) flag = 1 if (flag == 0): collection.insert_one({ table_name: {'roll no': '20'}})

Answer 7

aggregate example

https://mongoplayground.net/p/edbKil4Zvwc

 db.collection.aggregate([ { "$match": { "finishedAt": { "$exists": true } } }, { "$unwind": "$tags" }, { "$match": { "$or": [ { "tags.name": "Singapore" }, { "tags.name": "ABC" } ] } }, { "$group": { "_id": null, "count": { "$sum": 1 } } } ])

Answer 8

You can use .Count()

count, err = collection.Find(bson.M{field: value}).Count()

but remember to set:

Session.SetSafe

db.SetSafe(&mgo.Safe{})

Otherwise it will return 0 on every call.