如何将来自不同数组的两个匹配对象合并为一个对象?
How to merge two matching objects from different array into one object?
我有一种情况,我从聚合中得到了一个结果,我正在获取这种格式的数据。
{
"_id" : ObjectId("5a42432d69cbfed9a410e8ad"),
"bacId" : "BAC0023444",
"cardId" : "2",
"defaultCardOrder" : "2",
"alias" : "Finance",
"label" : "Finance",
"for" : "",
"cardTooltip" : {
"enable" : true,
"text" : ""
},
"dataBlocks" : [
{
"defaultBlockOrder" : "1",
"blockId" : "1",
"data" : "0"
},
{
"defaultBlockOrder" : "2",
"blockId" : "2",
"data" : "0"
},
{
"defaultBlockOrder" : "3",
"blockId" : "3",
"data" : "0"
}
],
"templateBlocks" : [
{
"blockId" : "1",
"label" : "Gross Profit",
"quarter" : "",
"data" : "",
"dataType" : {
"typeId" : "2"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
},
{
"blockId" : "2",
"label" : "Profit Forecast",
"quarter" : "",
"data" : "",
"dataType" : {
"typeId" : "2"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
},
{
"blockId" : "3",
"label" : "Resource Billing",
"quarter" : "",
"data" : "",
"dataType" : {
"typeId" : "2"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
}
]
},
{
"_id" : ObjectId("5a42432d69cbfed9a410e8ad"),
"bacId" : "BAC0023444",
"cardId" : "3",
"defaultCardOrder" : "3",
"alias" : "Staffing",
"label" : "Staffing",
"for" : "",
"cardTooltip" : {
"enable" : true,
"text" : ""
},
"dataBlocks" : [
{
"defaultBlockOrder" : "1",
"blockId" : "1",
"data" : "1212"
},
{
"defaultBlockOrder" : "2",
"blockId" : "2",
"data" : "1120"
},
{
"defaultBlockOrder" : "3",
"blockId" : "3",
"data" : "1200"
}
],
"templateBlocks" : [
{
"blockId" : "1",
"label" : "Staffing Planner",
"quarter" : "",
"data" : "",
"dataType" : {
"typeId" : "1"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
},
{
"blockId" : "2",
"label" : "Baseline",
"quarter" : "",
"data" : "",
"dataType" : {
"typeId" : "1"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
},
{
"blockId" : "3",
"label" : "Projected",
"quarter" : "",
"data" : "",
"dataType" : {
"typeId" : "1"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
}
]
}
现在我想比较每一行的两个对象数组,在这种情况下,它的 "dataBlocks" 和 "templateBlocks" 基于 "blockId" s,我想得到结果采用以下格式。
{
"_id" : ObjectId("5a42432d69cbfed9a410e8ad"),
"bacId" : "BAC0023444",
"cardId" : "2",
"defaultCardOrder" : "2",
"alias" : "Finance",
"label" : "Finance",
"for" : "",
"cardTooltip" : {
"enable" : true,
"text" : ""
},
"blocks" : [
{
"defaultBlockOrder" : "1",
"blockId" : "1",
"data" : "0",
"label" : "Gross Profit",
"quarter" : "",
"dataType" : {
"typeId" : "2"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
},
{
"defaultBlockOrder" : "2",
"blockId" : "2",
"data" : "0",
"label" : "Profit Forecast",
"quarter" : "",
"dataType" : {
"typeId" : "2"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
},
{
"defaultBlockOrder" : "3",
"blockId" : "3",
"data" : "0",
"label" : "Resource Billing",
"quarter" : "",
"dataType" : {
"typeId" : "2"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
}
]
},
{
"_id" : ObjectId("5a42432d69cbfed9a410e8ad"),
"bacId" : "BAC0023444",
"cardId" : "3",
"defaultCardOrder" : "3",
"alias" : "Staffing",
"label" : "Staffing",
"for" : "",
"cardTooltip" : {
"enable" : true,
"text" : ""
},
"dataBlocks" : [
{
"defaultBlockOrder" : "1",
"blockId" : "1",
"data" : "1212",
"label" : "Staffing Planner",
"quarter" : "",
"dataType" : {
"typeId" : "1"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
},
{
"defaultBlockOrder" : "2",
"blockId" : "2",
"data" : "1120",
"label" : "Baseline",
"quarter" : "",
"dataType" : {
"typeId" : "1"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
},
{
"defaultBlockOrder" : "3",
"blockId" : "3",
"data" : "1200",
"label" : "Projected",
"quarter" : "",
"dataType" : {
"typeId" : "1"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
}
]
}
是否可以用 mongodb 完成它?我正在使用 3.4 并尝试使用聚合来实现这一点。
提前致谢。
执行以下查询:
db.merge.aggregate([
// unwind twice
{$unwind: "$templateBlocks"},
{$unwind: "$dataBlocks"},
// get rid of documents where dataBlocks.blockId and
// templateBlocks.blockId are not equal
{$redact: {$cond: [{
$eq: [
"$dataBlocks.blockId",
"$templateBlocks.blockId"
]
},
"$$KEEP",
"$$PRUNE"
]
}
},
// merge dataBlocks and templateBlocks into a single document
{$project: {
bacId: 1,
cardId: 1,
defaultCardOrder: 1,
alias: 1,
label: 1,
for: 1,
cardTooltip: 1,
dataBlocks: {
defaultBlockOrder: "$dataBlocks.defaultBlockOrder",
blockId: "$dataBlocks.blockId",
data: "$dataBlocks.data",
label: "$templateBlocks.label",
quarter: "$templateBlocks.quarter",
data: "$templateBlocks.data",
dataType: "$templateBlocks.dataType",
tooltip: "$templateBlocks.tooltip"
}
}
},
// group to put correspondent dataBlocks to an array
{$group: {
_id: {
_id: "$_id",
bacId: "$bacId",
cardId: "$cardId",
defaultCardOrder: "$defaultCardOrder",
alias: "$alias",
label: "$label",
for: "$for",
cardTooltip: "$cardTooltip"
},
dataBlocks: {$push: "$dataBlocks" }
}
},
// remove the unnecessary _id object
{$project: {
_id: "$_id._id",
bacId: "$_id.bacId",
cardId: "$_id.cardId",
defaultCardOrder: "$_id.defaultCardOrder",
alias: "$_id.alias",
label: "$_id.label",
for: "$_id.for",
cardTooltip: "$_id.cardTooltip",
dataBlocks: "$dataBlocks"
}
}
])
考虑到性能取决于数据集的大小,因为查询会展开两次,并且可能会产生大量中间文档。
您可以在 3.6 中尝试以下聚合。
下面的查询迭代数据块数组并将数据块元素与模板块元素合并。使用 $indexofArray
查找模板块,它定位具有匹配块 ID 的数组索引和 $arrayElemAt
以访问找到的索引处的元素。
db.collection_name.aggregate([{"$addFields":{
"blocks":{
"$map":{
"input":"$dataBlocks",
"in":{
"$mergeObjects":[
"$$this",
{"$arrayElemAt":[
"$templateBlocks",
{"$indexOfArray":["$templateBlocks.blockId","$$this.blockId"]}
]
}
]
}
}
}
}}])
对于 3.4,将 $mergeObjects
替换为 $arrayToObject
、$objectToArray
和 $concatArrays
的组合以合并两个数组中的每个数组元素。
db.collection_name.aggregate([{"$addFields":{
"blocks":{
"$map":{
"input":"$dataBlocks",
"in":{
"$arrayToObject":{
"$concatArrays":[
{"$objectToArray":"$$this"},
{"$objectToArray":{
"$arrayElemAt":[
"$templateBlocks",
{"$indexOfArray":["$templateBlocks.blockId","$$this.blockId"]
}
]
}}
]
}
}
}
}
}}])
您可以使用排除项目作为最后阶段从输出中删除数组字段。
{"$project":{"templateBlocks":0,"dataBlocks":0}}
我有一种情况,我从聚合中得到了一个结果,我正在获取这种格式的数据。
{
"_id" : ObjectId("5a42432d69cbfed9a410e8ad"),
"bacId" : "BAC0023444",
"cardId" : "2",
"defaultCardOrder" : "2",
"alias" : "Finance",
"label" : "Finance",
"for" : "",
"cardTooltip" : {
"enable" : true,
"text" : ""
},
"dataBlocks" : [
{
"defaultBlockOrder" : "1",
"blockId" : "1",
"data" : "0"
},
{
"defaultBlockOrder" : "2",
"blockId" : "2",
"data" : "0"
},
{
"defaultBlockOrder" : "3",
"blockId" : "3",
"data" : "0"
}
],
"templateBlocks" : [
{
"blockId" : "1",
"label" : "Gross Profit",
"quarter" : "",
"data" : "",
"dataType" : {
"typeId" : "2"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
},
{
"blockId" : "2",
"label" : "Profit Forecast",
"quarter" : "",
"data" : "",
"dataType" : {
"typeId" : "2"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
},
{
"blockId" : "3",
"label" : "Resource Billing",
"quarter" : "",
"data" : "",
"dataType" : {
"typeId" : "2"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
}
]
},
{
"_id" : ObjectId("5a42432d69cbfed9a410e8ad"),
"bacId" : "BAC0023444",
"cardId" : "3",
"defaultCardOrder" : "3",
"alias" : "Staffing",
"label" : "Staffing",
"for" : "",
"cardTooltip" : {
"enable" : true,
"text" : ""
},
"dataBlocks" : [
{
"defaultBlockOrder" : "1",
"blockId" : "1",
"data" : "1212"
},
{
"defaultBlockOrder" : "2",
"blockId" : "2",
"data" : "1120"
},
{
"defaultBlockOrder" : "3",
"blockId" : "3",
"data" : "1200"
}
],
"templateBlocks" : [
{
"blockId" : "1",
"label" : "Staffing Planner",
"quarter" : "",
"data" : "",
"dataType" : {
"typeId" : "1"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
},
{
"blockId" : "2",
"label" : "Baseline",
"quarter" : "",
"data" : "",
"dataType" : {
"typeId" : "1"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
},
{
"blockId" : "3",
"label" : "Projected",
"quarter" : "",
"data" : "",
"dataType" : {
"typeId" : "1"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
}
]
}
现在我想比较每一行的两个对象数组,在这种情况下,它的 "dataBlocks" 和 "templateBlocks" 基于 "blockId" s,我想得到结果采用以下格式。
{
"_id" : ObjectId("5a42432d69cbfed9a410e8ad"),
"bacId" : "BAC0023444",
"cardId" : "2",
"defaultCardOrder" : "2",
"alias" : "Finance",
"label" : "Finance",
"for" : "",
"cardTooltip" : {
"enable" : true,
"text" : ""
},
"blocks" : [
{
"defaultBlockOrder" : "1",
"blockId" : "1",
"data" : "0",
"label" : "Gross Profit",
"quarter" : "",
"dataType" : {
"typeId" : "2"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
},
{
"defaultBlockOrder" : "2",
"blockId" : "2",
"data" : "0",
"label" : "Profit Forecast",
"quarter" : "",
"dataType" : {
"typeId" : "2"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
},
{
"defaultBlockOrder" : "3",
"blockId" : "3",
"data" : "0",
"label" : "Resource Billing",
"quarter" : "",
"dataType" : {
"typeId" : "2"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
}
]
},
{
"_id" : ObjectId("5a42432d69cbfed9a410e8ad"),
"bacId" : "BAC0023444",
"cardId" : "3",
"defaultCardOrder" : "3",
"alias" : "Staffing",
"label" : "Staffing",
"for" : "",
"cardTooltip" : {
"enable" : true,
"text" : ""
},
"dataBlocks" : [
{
"defaultBlockOrder" : "1",
"blockId" : "1",
"data" : "1212",
"label" : "Staffing Planner",
"quarter" : "",
"dataType" : {
"typeId" : "1"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
},
{
"defaultBlockOrder" : "2",
"blockId" : "2",
"data" : "1120",
"label" : "Baseline",
"quarter" : "",
"dataType" : {
"typeId" : "1"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
},
{
"defaultBlockOrder" : "3",
"blockId" : "3",
"data" : "1200",
"label" : "Projected",
"quarter" : "",
"dataType" : {
"typeId" : "1"
},
"tooltip" : {
"enable" : true,
"text" : ""
}
}
]
}
是否可以用 mongodb 完成它?我正在使用 3.4 并尝试使用聚合来实现这一点。
提前致谢。
执行以下查询:
db.merge.aggregate([
// unwind twice
{$unwind: "$templateBlocks"},
{$unwind: "$dataBlocks"},
// get rid of documents where dataBlocks.blockId and
// templateBlocks.blockId are not equal
{$redact: {$cond: [{
$eq: [
"$dataBlocks.blockId",
"$templateBlocks.blockId"
]
},
"$$KEEP",
"$$PRUNE"
]
}
},
// merge dataBlocks and templateBlocks into a single document
{$project: {
bacId: 1,
cardId: 1,
defaultCardOrder: 1,
alias: 1,
label: 1,
for: 1,
cardTooltip: 1,
dataBlocks: {
defaultBlockOrder: "$dataBlocks.defaultBlockOrder",
blockId: "$dataBlocks.blockId",
data: "$dataBlocks.data",
label: "$templateBlocks.label",
quarter: "$templateBlocks.quarter",
data: "$templateBlocks.data",
dataType: "$templateBlocks.dataType",
tooltip: "$templateBlocks.tooltip"
}
}
},
// group to put correspondent dataBlocks to an array
{$group: {
_id: {
_id: "$_id",
bacId: "$bacId",
cardId: "$cardId",
defaultCardOrder: "$defaultCardOrder",
alias: "$alias",
label: "$label",
for: "$for",
cardTooltip: "$cardTooltip"
},
dataBlocks: {$push: "$dataBlocks" }
}
},
// remove the unnecessary _id object
{$project: {
_id: "$_id._id",
bacId: "$_id.bacId",
cardId: "$_id.cardId",
defaultCardOrder: "$_id.defaultCardOrder",
alias: "$_id.alias",
label: "$_id.label",
for: "$_id.for",
cardTooltip: "$_id.cardTooltip",
dataBlocks: "$dataBlocks"
}
}
])
考虑到性能取决于数据集的大小,因为查询会展开两次,并且可能会产生大量中间文档。
您可以在 3.6 中尝试以下聚合。
下面的查询迭代数据块数组并将数据块元素与模板块元素合并。使用 $indexofArray
查找模板块,它定位具有匹配块 ID 的数组索引和 $arrayElemAt
以访问找到的索引处的元素。
db.collection_name.aggregate([{"$addFields":{
"blocks":{
"$map":{
"input":"$dataBlocks",
"in":{
"$mergeObjects":[
"$$this",
{"$arrayElemAt":[
"$templateBlocks",
{"$indexOfArray":["$templateBlocks.blockId","$$this.blockId"]}
]
}
]
}
}
}
}}])
对于 3.4,将 $mergeObjects
替换为 $arrayToObject
、$objectToArray
和 $concatArrays
的组合以合并两个数组中的每个数组元素。
db.collection_name.aggregate([{"$addFields":{
"blocks":{
"$map":{
"input":"$dataBlocks",
"in":{
"$arrayToObject":{
"$concatArrays":[
{"$objectToArray":"$$this"},
{"$objectToArray":{
"$arrayElemAt":[
"$templateBlocks",
{"$indexOfArray":["$templateBlocks.blockId","$$this.blockId"]
}
]
}}
]
}
}
}
}
}}])
您可以使用排除项目作为最后阶段从输出中删除数组字段。
{"$project":{"templateBlocks":0,"dataBlocks":0}}