更新时间:2023-12-05 22:53:40
您可以map()
.将 Array.map()
与猫鼬一起使用因为它返回一个数组,所以***使用 $group
_id
比使用 $push
You can map()
. Use Array.map()
with mongoose as it returns an array, and you are better off simply using the $group
_id
than using $push
const subCategory = (await SubCategory.aggregate([
{ '$match': { category: "dining" } },
{ '$group': { '_id': "$name" } }
])).map(({ _id }) => _id);
或使用 Cursor.map()
如果使用核心驱动程序中的基础Collection
:
const subCategory = await SubCategory.collection.aggregate([
{ '$match': { category: "dining" } },
{ '$group': { '_id': "$name" } }
]).map(({ _id }) => _id).toArray();
如果您不希望获得与众不同"的结果,则与find()
大致相同:
Much the same with find()
if you don't want the "distinct" results:
const subCategory = (await Subcategory.find({ category: "dining" }))
.map(({ name }) => name);
或使用 Cursor.map()
const subCategory = await Subcategory.collection.find({ category: "dining" })
.map(({ name }) => name).toArray();
您还可以使用 distinct()
,它基本上是对聚集过程和map()
幕后"(仅返回字段部分"而不是不同的聚集方法):
You can also use distinct()
, which basically does a variation of the aggregation process and the map()
"under the hood" ( the "return just the field part" and not the distinct aggregation method ):
const subCategory = await SubCategory.distinct("name",{ category: "dining" });
MongoDB本身不会返回BSON文档以外的任何内容,并且简单的字符串不是BSON文档.
MongoDB itself won't return anything other than a BSON Document, and a simple string is NOT a BSON Document.