'Use a field's value and add it to a new field in an array - Mongodb
I am attempting to turn this
{
"fooList" : [
{
"bar" : {
"baz" : 100
}
},
{
"bar" : {
"baz" : 200
}
}
]
},
{
"fooList" : [
{
"bar" : {
"baz" : 300
}
},
{
"bar" : {
"baz" : 400
}
}
]
}
into this:
{
"fooList" : [
{
"baz" : 100,
"bar" : {
"baz" : 100
}
},
{
"baz" : 200,
"bar" : {
"baz" : 200
}
}
]
},
{
"fooList" : [
{
"baz" : 300,
"bar" : {
"baz" : 300
}
},
{
"baz" : 400,
"bar" : {
"baz" : 400
}
}
]
}
As you can see, I really just copy baz and its value out of bar, but my issue is that is occurs within an array.
db.getCollection(<collection_name>).updateMany(
{},
{ $set: { 'fooList.$[element].baz' : '$fooList.$[element].bar.baz' } },
{ upsert: true ,
arrayFilters: [{'element.bar' : { $exists : true }}]
}
)
But this will only set the string $fooList.$[element].bar.baz into baz, seen here through the results
[
{
"_id": ObjectId("5a934e000102030405000000"),
"fooList": [
{
"bar": {
"baz": 100
},
"baz": "$fooList.$[element].bar.baz"
}
]
}
]
Can anyone tell me what I might be doing wrong, or if this is even possible? Thanks
Solution 1:[1]
Here is a variation that demonstrates the v4.4 "merge onto self" capability that turns aggregate into an update. This is a useful approach when you want to process all the docs because it eliminates the "no filter" ({}) and {multi:true} bits required for update.
db.foo.aggregate([
// The $map/$mergeObjects is the SAME as the answer above, just with a little
// more compact format. It is import to $project here, NOT $addFields, because
// we seek to limit the pipeline to just _id (comes along automatically) and
// the fooList:
{$project: {
fooList: {$map: {
input: "$fooList",
in: {$mergeObjects: [
"$$this",
{$cond: [
{$ne: ["$$this.bar.baz", undefined]},
{baz: "$$this.bar.baz"},
{} // else empty object
]}
]}
}}
}}
// ...and now, using _id as the key (fast!), merge fooList back into doc:
,{$merge: {
into: "foo",
on: [ "_id" ],
whenMatched: "merge",
whenNotMatched: "fail"
}}
]);
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Buzz Moschetti |
