我的收藏名称是transactions
。
我正在分享交易集合的对象
{
"_id" : ObjectId("58aaec83f1dc6914082afe31"),
"amount" : "33.00",
"coordinates" : {
"lat" : "4.8168",
"lon" : "36.4909"
},
"cuisine" : "Mexican",
"date" : ISODate("0062-02-22T11:46:52.738+05:30"),
"location" : {
"address" : "2414 Trudie Rue",
"city" : "West Alisa",
"state" : "New York",
"zip" : "10000"
},
"place_name" : "Outdoors",
"place_type" : "Wooden"
},
{
"_id" : ObjectId("58aaec83f1dc6914082afe32"),
"amount" : "557.00",
"coordinates" : {
"lat" : "-36.6784",
"lon" : "131.3698"
},
"cuisine" : "Australian",
"date" : ISODate("1294-10-04T19:53:15.562+05:30"),
"location" : {
"address" : "5084 Buckridge Cove",
"city" : "Sylviaview",
"state" : "Hawaii",
"zip" : "51416-6918"
},
"place_name" : "Toys",
"place_type" : "Cotton"
},
{
"_id" : ObjectId("58aaec83f1dc6914082afe33"),
"amount" : "339.00",
"coordinates" : {
"lat" : "45.1468",
"lon" : "91.4097"
},
"cuisine" : "Mexican",
"date" : ISODate("1568-11-25T02:54:53.046+05:30"),
"location" : {
"address" : "94614 Harry Island",
"city" : "Cartwrightside",
"state" : "Louisiana",
"zip" : "18825"
},
"place_name" : "Clothing",
"place_type" : "Frozen"
},
{
"_id" : ObjectId("58aaec83f1dc6914082afe34"),
"amount" : "173.00",
"coordinates" : {
"lat" : "-57.2738",
"lon" : "19.6381"
},
"cuisine" : "Australian",
"date" : ISODate("0804-05-07T03:00:07.724+05:30"),
"location" : {
"address" : "1933 Lewis Street",
"city" : "Aufderharville",
"state" : "Louisiana",
"zip" : "23416"
},
"place_name" : "Beauty",
"place_type" : "Fresh"
},
{
"_id" : ObjectId("58aaec83f1dc6914082afe34"),
"amount" : "173.00",
"coordinates" : {
"lat" : "-57.2738",
"lon" : "19.6381"
},
"cuisine" : "Australian",
"date" : ISODate("0804-05-07T03:00:07.724+05:30"),
"location" : {
"address" : "1933 Lewis Street",
"city" : "Aufderharville",
"state" : "Louisiana",
"zip" : "23416"
},
"place_name" : "Beauty",
"place_type" : "Fresh"
}
我希望总数
获取不同美食的列表输出
{
"name" : 'Mexican',
"count" : '2'
},
{
"name" : 'Australian',
"count" : '3'
},
我本可以轻松完成 mysql 但我知道mongodb因为我是mongodb的新手
我试过这个例子但我什么都没找到:
db.transactions.aggregate(
{$group: {_id:'$cuisine'},count:{$sum:1}}
).result;
答案 0 :(得分:1)
请尝试以下代码。你应该按照烹饪记录进行分组并获得它们的数量。稍后在项目管道中,您可以定义最终外观。
db.transactions.aggregate([
{ $group: { _id: "$cuisine", count: { $sum: 1 } } },
{ $project:{ _id: 0, name: "$_id", count:"$count" } }
]);