对于mongodb是完全新的,我试图以我需要的格式获取某些报告的查询。如何查询mongo以我想要的格式显示结果?
这里是我的资产保全的结果:
db.getCollection('Asset').find({})
结果:
/* 1 */
{
"_id" : ObjectId("58176238f23ee4040a48846c"),
"AssetTypeId" : "58176225f23ee4040a48846b",
"AssetTypeName" : "Chairs",
"Fields" : [
{
"_id" : "5770ae6cf500fd7675c234c5",
"Title" : "Asset Name",
"Value" : "Fancy Chair"
},
{
"_id" : "576b7593f500fd7675c234b9",
"Title" : "Status",
"Value" : "Missing"
},
{
"_id" : "57b2b52b31f7c7d80db34a23",
"Title" : "Location",
"Value" : "Office")
}
]
}
/* 2 */
{
"_id" : ObjectId("5817638ff23ee4040a48846e"),
"AssetTypeId" : "58176225f23ee4040a48846b",
"AssetTypeName" : "Chairs",
"Fields" : [
{
"_id" : "5770ae6cf500fd7675c234c5",
"Title" : "Asset Name",
"Value" : "Basic Chair"
},
{
"_id" : "576b7593f500fd7675c234b9",
"Title" : "Status",
"Value" : "Stores"
},
{
"_id" : "57b2b52b31f7c7d80db34a23",
"Title" : "Location",
"Value" : "Home")
}
]
}
/* 3 */
{
"_id" : ObjectId("581767f3f23ee4040a488471"),
"AssetTypeId" : "5817678ff23ee4040a488470",
"AssetTypeName" : "Table",
"Fields" : [
{
"_id" : "5770ae6cf500fd7675c234c5",
"Title" : "Asset Name",
"Value" : "Fancy Table"
},
{
"_id" : "576b7593f500fd7675c234b9",
"Title" : "Status",
"Value" : "Active"
},
{
"_id" : "57b2b52b31f7c7d80db34a23",
"Title" : "Location",
"Value" : "Office")
}
]
}
这是最终的结果我想从查询得到:
[
{"Asset Name":"Fancy Chair", "Status":"Missing","Location":"Office","AssetTypeName":"Chairs"},
{"Asset Name":"Basic Chair", "Status":"Stores","Location":"Office","AssetTypeName":"Chairs"},
{"Asset Name":"Fancy Table", "Status":"Active","Location":"Office","AssetTypeName":"Table"}
]
请添加任何你试过到目前为止并通过迭代的结果,并创建对象的一个新的数组d格式有人会和你一起合作,这样做会吸引更多的人。 – Veeram
这是我得到的: – bobbinson
db.getCollection('Asset')。find({},{'Fields.Title':1,'Fields.Value':1,“AssetTypeName”:1}); – bobbinson