我是MongoDB的新手。而且我需要帮助。我有一个带有SQLite的小项目(7个表和1个视图)。我需要在MongoDB上创建该项目,我正在使用Studio 3T,已经将SQLite表迁移到MongoDB集合,但是现在我需要进行(VIEW / TEST)测试该项目。请帮助如何使用MongoDB编写此SQL脚本。SQLITE:
MongoDB:
我想用MongoDB创建的SQLIte脚本:
CREATE VIEW rezultatas AS
SELECT p.pavadinimas AS detales_pavadinimas,
SUM(d.pagamintas_kiekis) AS pagamintas_kiekis,
SUM(z.gamybos_islaidos) AS vidutine_kaina,
STRFTIME('%m', d.pagaminimo_data) AS menuo,
STRFTIME('%Y', d.pagaminimo_data) AS metai
FROM detales d,
zinynas z,
produktas p
WHERE (z.detale_id = p._id_) AND
(d.detale_id = z.detale_id) AND
(d.pagaminimo_data >= z.data_nuo) AND
NOT EXISTS (
SELECT *
FROM zinynas
WHERE (d.detale_id = detale_id) AND
(d.pagaminimo_data >= data_nuo) AND
(z.data_nuo < data_nuo)
)
GROUP BY p.pavadinimas,
STRFTIME('%m', d.pagaminimo_data),
STRFTIME('%Y', d.pagaminimo_data)
由于缺乏方案,我不得不猜测一些事情,但是基本布局应该可以工作。
我们将使用$createView,并将这些参数用作输入:
db.createView('rezultatas', 'produktas', pipeline)
意味着我们创建视图的管道从produktas
集合开始。
要使用的管道:
[
{ // match the documents from the zinyas collection.
$lookup:
{
from: "zinynas",
let: { produktas_id: "$_id" }, // i'm guessing its _id
pipeline: [
{ $match:
{ $expr: { $eq: [ "$detale_id", "$$produktas_id" ] }}
},
],
as: "z"
}
},
{
$unwind: "$z"
},
{ // match the documents from the detales collection. only keep the one with maximum data_nuo value.
$lookup:
{
from: "detales",
let: { z_detale_id: "$z.detale_id", z_data_nuo: "$z.data_nuo" },
pipeline: [
{ $match:
{
$and: [
{ $expr: { $eq: [ "$detale_id", "$$z_detale_id" ] }},
{ $expr: { $gte: [ "$pagaminimo_data", "$$z_data_nuo"]}}
]
}
},
{
$sort: {
data_nuo: -1
}
},
{
$limit: 1
}
],
as: "d"
}
},
{
$unwind: "$d"
},
{ // end up saving the fields we want.
$group: {
_id: { pavadinimas : "$pavadinimas", month: {$month: "$d.pagaminimo_data"}, year: {$year: "$d.pagaminimo_data"}},
pagamintas_kiekis: {$sum: "$d.pagamintas_kiekis"},
vidutine_kaina: {$sum: "$z.gamybos_islaidos"},
month: {$first: {$month: "$d.pagaminimo_data"}},
year: {$first: {$year: "$d.pagaminimo_data"}},
detales_pavadinimas: {$first: "$pavadinimas"}
}
}
]