我有以下MongoDB查询,该查询按日期和结果分组并计数.我想让查询还返回特定日期和结果的计数为0(如果当天没有数据).
I have the following MongoDB query that groups by date and result and gives a count. I'd like to have the query also return a count of 0 for a particular date and result if data doesn't exist for that day.
例如,我有以下结果状态:SUCCESS和FAILED.如果在21日没有失败的结果,我希望返回的计数为0:
For example I have the following result statuses: SUCCESS and FAILED. If on the 21st there were no results of FAILED I would want a count returned of 0:
{ "_id" : { "month" : 1, "day" : 21, "year" : 2014, "buildResult" : "FAILURE" }, "count" : 0 }我已经对关系数据库和日历表做了类似的事情,但是我不确定如何使用MongoDB来解决这个问题.这可能吗?还是应该在运行查询后以编程方式执行某些操作?
I've done something similar with a relational database and a calendar table, but I'm not sure how to approach this with MongoDB. Is this possible or should I do something programatically after running the query?
以下是数据库中文档的示例(简体):
Here is an example of a document (simplified) in the database:
{ "_id" : ObjectId("52deab2fe4b0a491abb54108"), "type" : "build", "time" : ISODate("2014-01-21T17:15:27.471Z"), "data" : { "buildNumber" : 43, "buildDuration" : 997308, "buildResult" : "SUCCESS" } }这是我当前的查询:
db.builds.aggregate([ { $match: { "data.buildResult" : { $ne : null} }}, { $group: { _id: { month: { $month: "$time" }, day: { $dayOfMonth: "$time" }, year: { $year: "$time" }, buildResult: "$data.buildResult", }, count: { $sum: 1} } }, { $sort: { "_id.year": 1, "_id.month": 1, "_id.day": 1} } ])推荐答案
如果我正确理解了您想要什么,则可以尝试以下方法:
If I correctly understand what do you want, you could try this:
db.builds.aggregate([ { $project: { time: 1, projectedData: { $ifNull: ['$data.buildResult', 'none'] } } }, { $group: { _id: { month: { $month: "$time" }, day: { $dayOfMonth: "$time" }, year: { $year: "$time" }, buildResult: "$projectedData" }, count: { $sum: { $cond: [ { $eq: [ "$projectedData", "none" ] }, 0, 1 ] } } } }, { $sort: { "_id.year": 1, "_id.month": 1, "_id.day": 1 } } ])更新: 您要从输出中获取输入中的更多文档,只能使用与数组一起使用的unwind运算符,但是您没有任何数组,因此据我所知,您无法获取更多文档.因此,您应该在查询结果之后添加一些逻辑,以便为另一种buildResult类型的现有日期创建新数据(计数为0)
Update: You want to get from output more documents that been in input, it is possible only with unwind operator that works with arrays, but you haven't any arrays, so as I know it is impossible to get more documents in your case. So you should add some logic after query result to create new data for existing dates with 0 count for another type of buildResult...
更多推荐
如果没有结果,则MongoDB的总返回计数为0
发布评论