Need help with mongo db query
Mondo db query - search for parents with state good and children with state bad or missing. output should be an array of all the children with state bad or missing from parents with good state
Below is the JSON list
[
{
"name": "parent-a",
"status": {
"state": "good"
},
"children": [
"child-1",
"child-2"
]
},
{
"name": "child-1",
"state": "good",
"parent": "parent-a"
},
{
"name": "child-2",
"state": {},
"parent": "parent-a"
},
{
"name": "parent-b",
"status": {
"state": "good"
},
"children": [
"child-3",
"child-4"
]
},
{
"name": "child-3",
"state": "good",
"parent": "parent-b"
},
{
"name": "child-4",
"state": "bad",
"parent": "parent-b"
},
{
"name": "parent-c",
"status": {
"state": "bad"
},
"children": [
"child-5",
"child-6"
]
},
{
"name": "child-5",
"state": "good",
"parent": "parent-c"
},
{
"name": "child-6",
"state": "bad",
"parent": "parent-c"
}
]
Expected output
"children": [
{
"name": "child-2",
"state": {}
},
{
"name": "child-4",
"state": "bad"
}
]
Any inputs would be appreciated. Thanks in advance :)
One option is to use $lookup
* for this:
db.collection.aggregate([
{$match: {state: {$in: ["bad", {}]}}},
{$lookup: {
from: "collection",
localField: "parent",
foreignField: "name",
pipeline: [
{$match: {"status.state": "good"}}
],
as: "hasGoodParent"
}},
{$match: {"hasGoodParent.0": {$exists: true}}},
{$project: {name: 1, state: 1, _id: 0}}
])
See how it works on the playground example
*If your mongoDB version is lower than 5.0 you need to change the syntax a bit. Drop the localField
and foreignField
of the $lookup
and replace with let
and equality match on the pipeline