MongoDB 聚合 $group queryquery

Question

我有一个集合，里面的文档是这样的

{
    "_id" : "api",
    "titleAndTimeDetails" : [ 
        {
            "Title" : "api data",
            "timeTaken" : NumberLong(8091446063)
        }, 
        {
            "Title" : "class data",
            "timeTaken" : NumberLong(519163710)
        }, 
        {
            "Title" : "API or datasets for age world",
            "timeTaken" : NumberLong(34245103)
        }, 
        {
            "Title" : "What does a null result mean API?",
            "timeTaken" : NumberLong(171605137)
        }
 ]
}

/* 2 */
{
    "_id" : "government",
    "titleAndTimeDetails" : [ 
        {
            "Title" : "Is there an open standard for the CAFR (Comprehensive Annual Finance Report)?",
            "timeTaken" : NumberLong(574587563)
        }, 
        {
            "Title" : "College Scorecard full data base",
            "timeTaken" : NumberLong(9422714)
        }, 
        {
            "Title" : "List of German local politicians",
            "timeTaken" : NumberLong(691311396)
        }, 
        {
            "Title" : "Trying to extrapolate patient costs by physician from public Medicare pricing data",
            "timeTaken" : NumberLong(9590779130)
        }, 
        {
            "Title" : "Are there good examples of open read-write APIs in Federal government?",
            "timeTaken" : NumberLong(1784634763)
        } 
   ]
}

查询语句是显示Title每个“_id”花费的时间最少。请问有人可以帮我吗？

Answer 1

你应该先$unwind titleAndTimeDetails 然后是 $sort通过 titleAndTimeDetails.timeTaken 最后 $group通过 _id 并获取 Title 的 $first。

db.collection.aggregate([
    { $unwind: "$titleAndTimeDetails" },
    { $sort: { "titleAndTimeDetails.timeTaken": 1 } },
    { $group: { _id: "$_id", title: { $first: "$titleAndTimeDetails.Title" } } }
])