Question

我正在使用MongoDB。这是我的文件。集合名称为test

{"group_id":"111","image":"x1.jpg",create_time:129}
{"group_id":"111","image":"x2.jpg",create_time:94}
{"group_id":"222","image":"x3.jpg",create_time:129}
{"group_id":"222","image":"x4.jpg",create_time:101}

预期输出的示例

{"group_id":"111","count":2,"image":["x1.jpg","x2.jpg"]}
{"group_id":"222","count":2,"image":["x3.jpg","x4.jpg"]}

然后我尝试

db.test.aggregate([{"$group":{"_id":{"group_id":"$group_id"},"count":{"$sum":1},"image":{"$push":"$image"}}},{"$sort":{"count":1}}],{allowDiskUse:true})

但是我想得到图像按create_time排序的结果。像这样：

{"group_id":"111","count":2,"image":["x2.jpg","x1.jpg"]}
{"group_id":"222","count":2,"image":["x4.jpg","x3.jpg"]}

任何人都可以指导我进行分组依据和排序的正确组合吗？

Answer 1

再使用一种

{'$ sort'：{“ group_id”：1，“ image”：1}}

在聚合管道的开头

db.test.aggregate([
    {'$sort' : {"group_id" : 1,"image" : 1}},
    {"$group":{"_id":{"group_id":"$group_id"},"count":{"$sum":1},"image":{"$push":"$image"}}},
    {"$sort":{"count":1}}],
{allowDiskUse:true})

希望您会得到预期的结果

如何通过一个字段对匹配，分组和排序进行汇总

1 个答案: