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MongoDB query to select one field if the other is null?
To select one field if the other is null, use $ifNull. Let us create a collection with documents −
> db.demo182.insertOne({"FirstName":"Chris","LastName":null});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e398ea19e4f06af55199802")
}
> db.demo182.insertOne({"FirstName":null,"LastName":"Miller"});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e398ead9e4f06af55199803")
}
>
> db.demo182.insertOne({"FirstName":"John","LastName":"Smith"});
{
"acknowledged" : true,
"insertedId" : ObjectId("5e398ebf9e4f06af55199804")
}
Display all documents from a collection with the help of find() method −
> db.demo182.find();
This will produce the following output −
{ "_id" : ObjectId("5e398ea19e4f06af55199802"), "FirstName" : "Chris", "LastName" : null }
{ "_id" : ObjectId("5e398ead9e4f06af55199803"), "FirstName" : null, "LastName" : "Miller" }
{ "_id" : ObjectId("5e398ebf9e4f06af55199804"), "FirstName" : "John", "LastName" : "Smith" }
Following is the query to select one field if the other is null −
> db.demo182.aggregate([
... {
... $project: {
... "item": 1,
... "Result": { "$ifNull": [ "$FirstName", "$LastName" ] }
... }
... }
...])
This will produce the following output −
{ "_id" : ObjectId("5e398ea19e4f06af55199802"), "Result" : "Chris" }
{ "_id" : ObjectId("5e398ead9e4f06af55199803"), "Result" : "Miller" }
{ "_id" : ObjectId("5e398ebf9e4f06af55199804"), "Result" : "John" }
Updated on: 2020-03-27T07:58:40+05:30
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