Mongodb mongo中如何按任意列表对对象数组进行排序
我在寻找一种按任意列表对对象数组进行排序的方法。假设我有这个对象数组Mongodb mongo中如何按任意列表对对象数组进行排序,mongodb,mongoose,Mongodb,Mongoose,我在寻找一种按任意列表对对象数组进行排序的方法。假设我有这个对象数组 [ { "_id": "4JEEuhNIae", "category": "grocery" }, { "_id": "4JW7miNITl", "category": "food" }, { "_id": "4Je4kmrrbZ", "category": "coffee" }, { "_id": "4JgAh3N86x", "category":
[
{
"_id": "4JEEuhNIae",
"category": "grocery"
},
{
"_id": "4JW7miNITl",
"category": "food"
},
{
"_id": "4Je4kmrrbZ",
"category": "coffee"
},
{
"_id": "4JgAh3N86x",
"category": "coffee"
}
]
这是我想用作排序标准的数组。食品的记录应该首先出现,然后是咖啡和杂货
['food','coffee','grocery']
结果应该是:
[
{
"_id": "4JW7miNITl",
"category": "food"
},
{
"_id": "4Je4kmrrbZ",
"category": "coffee"
},
{
"_id": "4JgAh3N86x",
"category": "coffee"
},
{
"_id": "4JEEuhNIae",
"category": "grocery"
},
]
如何使用mongoose在mongodb上进行此类排序?我真的不想在获取数据后对代码进行任何操作。您可以尝试在从方法返回的数组上使用本机JavaScript方法运行自定义比较器函数:
var order = ["food", "coffee", "grocery"];
var docs = db.collection.find().toArray().sort(function(a, b) {
return order.indexOf(a.category) - order.indexOf(b.category);
});
printjson(docs);
样本输出
[
{
"_id" : "4JW7miNITl",
"category" : "food"
},
{
"_id" : "4Je4kmrrbZ",
"category" : "coffee"
},
{
"_id" : "4JgAh3N86x",
"category" : "coffee"
},
{
"_id" : "4JEEuhNIae",
"category" : "grocery"
}
]
使用新的MongoDB 3.4版本,您应该能够在聚合框架中利用本机MongoDB操作符和
- 管道步骤允许您在不知道所有其他现有字段的情况下,将新字段添加到现有文档中李>
- 表达式返回给定数组中特定元素的位置
您的代码是文档中的列表字段吗?或者它是一个文档列表?@SergiuZaharie它是一个文档列表似乎我不能直接为mongoose编写自定义排序函数,我需要使用
db.eval
好的,谢谢
var order = ["food", "coffee", "grocery"],
projection = {
"$addFields" : {
"__order" : { "$indexOfArray" : [ order, "$category" ] }
}
},
sort = { "$sort" : { "__order" : 1 } };
db.collection.aggregate([ projection, sort]);