$group and sum add all greater than-CodePudding

Let's say my documents look like this:

[
  {
    name: "Example 1",
    year: "2012"
  },
  {
    name: "Example 2",
    year: "2012"
  },
  {
    name: "Example 3",
    year: "2013"
  },
  {
    name: "Example 4",
    year: "2014"
  }
]

Using an aggregation, is there a way to group by year and sum the document count, but additionally add the sum of all later years?

The result I want is this:

[
    {
        _id: "2012",
        count: 4 // years 2012-2014
    },
    {
        _id: "2013",
        count: 2 // years 2013-2014
    },
    {
        _id: "2014",
        count: 1 // only year 2014
    }
]

Right now, I'm using a normal $group $sum, which gives me the counts for each year individually and then I sort them in JavaScript. I was hoping that there was a simpler way that gets rid of the additional JS code:

yearCounts: [           
    { $group: { _id: "$year", count: { $sum: 1 } } }
]

const yearCounts: { _id: string, count: number }[] = aggregationResult[0].yearCounts || [];
const yearCountsSummed = yearCounts.map((yearCount: { _id: string, count: number }) => {
    const yearsUntil = yearCounts.filter(year => year._id >= yearCount._id);
    const countSummed = yearsUntil.map(yearCount => yearCount.count).reduce((a, b) => a   b) || 0;
    return countSummed;
});

CodePudding user response：

Yes, this was recently made quite easy using the

Hope, this solution might help or lead to solve your problem.