根据您自己变量值的条件添加聚合

Question

我有两个函数可以使用聚合从monga获取数组。功能完全相同，除了一个管道"；match(｛startTime:｛$gte:start｝｝)"；。我怎样才能离开一个函数并添加一个"；数学"；仅通过"；"开始"；变量，哪个是日期过滤器？

let groupedByUserSessions
if (lastDay) {
const start = getDateFromString(lastDay);
groupedByUserSessions = await getValuesByDate(start)
} else {
groupedByUserSessions = await getAllValues();
}

功能完全相同，

function getValuesByDate(start) {
return Sessions.aggregate()
.match({ startTime: { $gte: start } })
.group({
_id: { departament: "$departament", userAdName: "$userAdName" },
cleanTime: { $sum: { $subtract: ["$commonTime", "$idlingTime"] } }
})
.group({
_id: { departament: "$_id.departament"},
users: { $push: {value: '$cleanTime', name: '$_id.userAdName'} },
commonCleanTime: { $sum: "$cleanTime" }
})
.project({
departament: '$_id.departament',
users: '$users',
commonCleanTime: '$commonCleanTime',
performance: { $divide: [ "$commonCleanTime",  { $size: "$users" }] }
});

}

Answer 1

您可以将管道分成一个"头部；和一个"；尾部；并在给定start自变量时以不同的方式构造头：

function getValuesByDate(start) {
let agg = Sessions.aggregate();
if (start) {
agg = agg.match({ startTime: { $gte: start } });
}
return agg
.group({
_id: { departament: '$departament', userAdName: '$userAdName' },
cleanTime: { $sum: { $subtract: ['$commonTime', '$idlingTime'] } },
})
.group({
_id: { departament: '$_id.departament' },
users: { $push: { value: '$cleanTime', name: '$_id.userAdName' } },
commonCleanTime: { $sum: '$cleanTime' },
})
.project({
departament: '$_id.departament',
users: '$users',
commonCleanTime: '$commonCleanTime',
performance: { $divide: ['$commonCleanTime', { $size: '$users' }] },
});
}