我们有一个具有以下形式的Mongo集合:
[
{
"_id" : "34210db0-6g91-83e8-ae8c-659f064f503f",
"dateReceived" : ISODate("2018-01-01T00:00:00.000Z"),
"group" : null,
"clientShortName" : "My Client Name",
"sourceShortName" : "Datasource Name",
"files" : [
{
"_id" : "807061f0-2d77-87e8-8610-9ff3cbc9c774"
"status" : 1,
"fileName" : "filename1.csv",
"numRows" : 15,
},
{
"_id" : "587036f0-2n65-55e8-8610-3ee3cbc9c814"
"status" : 8,
"fileName" : "filename2.csv",
"numRows" : 30,
}
]
}
]
我们有一个使用“find”和“map”的Mongo查询,以便生成转换后的输出。 find / map命令如下所示:
db.getCollection('batches')
.find({_id: "34210db0-6g91-83e8-ae8c-659f064f503f"}, {"__v": false, "files.diffHistory": false})
.map( doc =>
{
doc.id = doc._id;
doc.clientName = doc.clientShortName;
doc.dataSourceName = doc.sourceShortName;
delete doc._id;
delete doc.clientShortName;
delete doc.sourceShortName;
doc.numFiles = NumberInt(doc.files.length);
doc.files = doc.files.map( file =>
{
file.id = file._id;
delete file._id;
delete file.__v;
delete file.edits;
return file;
}
);
// broken....how should this be formatted?
doc.totalNumRows = {$sum: doc.files.numRows};
return doc;
}
)
此查找/映射有效并为“totalNumRows”求和产生预期输出EXCEPT。我们要完成的是添加所有“files.numRows”字段,以便在返回的数据集的顶层获得单个汇总条目。也就是说,我们会看到一个看起来像的结果集:
[
{
"id" : "34210db0-6g91-83e8-ae8c-659f064f503f",
"dateReceived" : ISODate("2018-01-01T00:00:00.000Z"),
"group" : null,
"clientName" : "My Client Name",
"dataSourceName" : "Datasource Name",
"files" : [
{
"id" : "807061f0-2d77-87e8-8610-9ff3cbc9c774"
"status" : 1,
"fileName" : "filename1.csv",
"numRows" : 15,
},
{
"id" : "587036f0-2n65-55e8-8610-3ee3cbc9c814"
"status" : 8,
"fileName" : "filename2.csv",
"numRows" : 30,
}
],
"totalNumRows": 45
}
]
到目前为止尝试的所有内容都会导致查询失败。有没有人知道用于总结“文件”子文档的“numRows”字段的正确命令/格式?
答案 0 :(得分:1)
那么你可以做得更好更快,那么......尝试$project
重命名你的实际字段名称,然后$sum
来计算总数numRows
db.collection.aggregate([
{ "$match": { _id: "34210db0-6g91-83e8-ae8c-659f064f503f" }},
{ "$project": {
"totalNumRows": {
"$sum": "$files.numRows"
},
"clientName": "$clientShortName",
"dataSourceName": "$sourceShortName",
"files": "$files"
}}
])
它将提供以下输出
[
{
"_id": 1111,
"clientName": "My Client Name",
"dataSourceName": "Datasource Name",
"files": [
{
"fileName": "filename1.csv",
"id": 2222,
"numRows": 15,
"status": 1
},
{
"fileName": "filename2.csv",
"id": 3333,
"numRows": 30,
"status": 8
}
],
"totalNumRows": 45
}
]