我收集的名字是transactions
。 我分享交易徵收的對象如何在mongodb中查詢以獲得明顯的記錄數
{
"_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"
}
我想要得到的不同美食列表與總數
輸出
{
"name" : 'Mexican',
"count" : '2'
},
{
"name" : 'Australian',
"count" : '3'
},
我能有用mysql輕鬆完成,但我知道在mongo DB因爲我有新的mongodb
我曾試圖與榜樣,我什麼也沒找到:
db.transactions.aggregate(
{$group: {_id:'$cuisine'},count:{$sum:1}}
).result;
的可能的複製[MongoDB的SELECT COUNT(DISTINCT x)的上一個索引列 - 計算大數據集的唯一結果](http://stackoverflow.com/questions/11782566/mongodb-select-countdistinct-x -on-an-indexed-column-count-unique-results-for) – hassan
感謝@HassanAhmed的評論,但我沒有發現這篇文章非常有幫助 –