再次使用mongoDB。我真的很喜欢聚合，但仍然无法"获得它"。
这是我的数组：
{
"_id" : ObjectId("55951b2bf41edfc80b00002a"),
"orders" : [
{
"id" : "55929142f41edfdc0f00002f",
"name" : "XYZ",
"id_basket" : 1,
"card" : [
{
"id" : "250",
"serial" : "B",
"type" : "9cf4161002b9eda349bb9c5ae64b9f4a",
"name" : "Eco",
"ticket" : [
{
"id" : "55927d41f41edfd00f000030",
"name" : "ZZZ",
"price" : {
"name" : "Normal",
"price" : "10",
"price_disp" : "10 €",
}
},
{
"id" : "55927d41f41edfd00f000030",
"name" : "ZZZ",
"price" : {
"name" : "Normal",
"price" : "10",
"price_disp" : "10 €",
}
}
]
},
{
"id" : "250",
"serial" : "B",
"type" : "9cf4161002b9eda349bb9c5ae64b9f4a",
"name" : "Eco",
"ticket" : [
{
"id" : "55927d41f41edfd00f000030",
"name" : "ZZZ",
"price" : {
"name" : "Normal",
"price" : "10",
"price_disp" : "10 €",
}
},
{
"id" : "55927d41f41edfd00f000030",
"name" : "ZZZ",
"price" : {
"name" : "Normal",
"price" : "10",
"price_disp" : "10 €",
}
}
]
}
],
"full_amount" : "40",
},
{
"id" : "55929142f41edfdc0f00002f",
"name" : "XYZ",
"id_basket" : 1,
"card" : [
{
"id" : "250",
"serial" : "B",
"type" : "9cf4161002b9eda349bb9c5ae64b9f4a",
"name" : "Eco",
"ticket" : [
{
"id" : "55927d41f41edfd00f000030",
"name" : "ZZZ",
"price" : {
"name" : "Normal",
"price" : "10",
"price_disp" : "10 €",
}
},
{
"id" : "55927d41f41edfd00f000030",
"name" : "ZZZ",
"price" : {
"name" : "Normal",
"price" : "10",
"price_disp" : "10 €",
}
}
]
},
{
"id" : "250",
"serial" : "B",
"type" : "9cf4161002b9eda349bb9c5ae64b9f4a",
"name" : "Eco",
"ticket" : [
{
"id" : "55927d41f41edfd00f000030",
"name" : "ZZZ",
"price" : {
"name" : "Normal",
"price" : "10",
"price_disp" : "10 €",
}
},
{
"id" : "55927d41f41edfd00f000030",
"name" : "ZZZ",
"price" : {
"name" : "Normal",
"price" : "10",
"price_disp" : "10 €",
}
}
]
}
],
"full_amount" : "40",
},
],
"rate" : "0.23",
"date" : "2015-07-02 13:04:34",
"id_user" : 97,
}
我想输出这样的东西：
{
"_id" : ObjectId("55951b2bf41edfc80b00002a"),
"orders" : [
{
"id" : "55929142f41edfdc0f00002f",
"name" : "XYZ",
"card" : [
{
"id" : "250",
"serial" : "B",
"name" : "Eco",
"ticket" : [
{
"id" : "55927d41f41edfd00f000030",
"name" : "ZZZ",
"price" : "10 €"
},
{
"id" : "55927d41f41edfd00f000030",
"name" : "ZZZ",
"price" : "10 €"
}
]
},
{
"id" : "250",
"serial" : "B",
"name" : "Eco",
"ticket" : [
{
"id" : "55927d41f41edfd00f000030",
"name" : "ZZZ",
"price" : "10 €"
},
{
"id" : "55927d41f41edfd00f000030",
"name" : "ZZZ",
"price" : "10 €"
}
]
}
],
"full_amount" : "40",
},
{
"id" : "55929142f41edfdc0f00002f",
"name" : "XYZ",
"card" : [
{
"id" : "250",
"serial" : "B",
"name" : "Eco",
"ticket" : [
{
"id" : "55927d41f41edfd00f000030",
"name" : "ZZZ",
"price" : "10 €"
},
{
"id" : "55927d41f41edfd00f000030",
"name" : "ZZZ",
"price" : "10 €"
}
]
},
{
"id" : "250",
"serial" : "B",
"name" : "Eco",
"ticket" : [
{
"id" : "55927d41f41edfd00f000030",
"name" : "ZZZ",
"price" : "10 €"
},
{
"id" : "55927d41f41edfd00f000030",
"name" : "ZZZ",
"price" : "10 €"
}
]
}
],
"full_amount" : "40",
},
],
"rate" : "0.23",
"date" : "2015-07-02 13:04:34",
}
我尝试了许多展开、投影和分组的组合，但都没能得到我想要的。有人能帮我吗？
您可能不应该将聚合框架用于这样的任务，因为这些任务实际上不会在文档之间"聚合"任何内容。这实际上是一项"投影"任务，因为您所要求的只是"更改"文档的结构，而这项任务可能更适合在检索文档后在客户端中进行编码。
一个很好的原因是，像$unwind这样的操作在性能方面成本非常高。$unwind所做的是为出现的每个数组成员生成文档内容的"副本"，这将导致需要处理更多的文档。
把它想象成一个具有"一对多"关系的"SQL联接"，唯一的区别是数据本身包含在一个文档中。处理$unwind模拟"联接"结果，即为每个"子"（多）文档复制"主"（一个）文档内容。
为了应对人们进行的此类操作，MongoDB 2.6引入了$map运算符，该运算符处理文档中的数组元素。
因此，您不需要执行多个（或任何）$unwind操作，而是可以在$project阶段中使用$map来处理文档本身中的数组：
db.collection.aggregate([
{ "$project": {
"orders": { "$map": {
"input": "$orders",
"as": "o",
"in": {
"id": "$$o.id",
"name": "$$o.name",
"card": { "$map": {
"input": "$$o.card",
"as": "c",
"in": {
"id": "$$c.id",
"serial": "$$c.serial",
"name": "$$c.name",
"ticket": { "$map": {
"input": "$$c.ticket",
"as": "t",
"in": {
"id": "$$t.id",
"name": "$$t.name",
"price": "$$t.price.price_disp"
}
}}
}
}},
"full_amount": "$$o.full_amount"
}
}},
"rate": 1,
"date": 1
}}
])
操作相当简单，因为每个"数组"都有自己的变量名，对于这样的简单投影操作，真正剩下的就是选择您想要的字段。
在早期版本中，使用$unwind进行处理要困难得多：
db.collection.aggregate([
{ "$unwind": "$orders" },
{ "$unwind": "$orders.card" },
{ "$unwind": "$orders.card.ticket" },
{ "$group": {
"_id": {
"_id": "$_id",
"orders": {
"id": "$orders.id",
"name": "$orders.name",
"card": {
"id": "$orders.card.id",
"serial": "$orders.card.serial",
"name": "$orders.card.name"
},
"full_amount": "$orders.full_amount"
},
"rate": "$rate",
"date": "$date"
},
"ticket": {
"$push": {
"id": "$orders.card.ticket.id",
"name": "$orders.card.ticket.name",
"price": "$orders.card.ticket.price.price_disp"
}
}
}},
{ "$group": {
"_id": {
"_id": "$_id._id",
"orders": {
"id": "$_id.orders.id",
"name": "$_id.orders.name",
"full_amount": "$_id.orders.full_amount"
},
"rate": "$_id.rate",
"date": "$_id.date"
},
"card": {
"$push": {
"id": "$_id.orders.card.id",
"serial": "$_id.orders.card.serial",
"name": "$_id.orders.card.name",
"ticket": "$ticket"
}
}
}},
{ "$group": {
"_id": "$_id._id",
"orders": {
"$push": {
"id": "$_id.orders.id",
"name": "$_id.orders.name",
"card": "$card",
"full_amount": "$_id.orders.full_amount"
}
},
"rate": { "$first": "$_id.rate" },
"date": { "$first": "$_id.date" }
}}
])
因此，仔细阅读，您应该看到，由于您$unwind三次，因此也有必要$group"三次"，同时仔细分组每个"级别"的所有不同值，并通过$push重新构建数组。
这真的不建议，如前所述：
您"不是分组/聚合任何东西"，并且每个子文档"必须"包含的"唯一"itentifier，因为重新构造数组需要"分组"操作。（请参阅：注意）
此处的
$unwind操作成本非常高。所有文档信息都是由"n"数组X"n"个数组元素等因子重新生成的。因此，聚合管道中的数据比集合或查询选择本身实际包含的数据多得多。
因此，总之，对于"重新格式化数据"的一般处理，您应该处理代码中的每个文档，而不是将其"扔"到聚合管道中。
如果您的文档数据需要"足够"的操作，这会对返回的结果大小产生"实质性的差异"，并且您认为这比提取整个文档并在客户端中进行操作更有效，那么您应该使用$project表单，如$map操作所示。
边栏
你原来的"标签"在这里提到了"PHP"。
包括聚合在内的所有MongoDB查询都没有特定的语言，只是"数据结构"，对于这些语言（PHP、JavaScript、python等），它们大多以"原生形式"表示，对于那些没有"原生"自由结构表达格式的语言（C、C#、Java），它们使用"构建器方法"表示。
在所有情况下，都有可用于JSON的简单解析器，这在这里是一个常见的"linqua-franca"，因为MongoBShell本身是基于JavaScript的，并在本地理解JSON结构（作为实际的JavaScript对象）。
因此，在处理此类示例时，请使用以下工具：
json_decode：深入了解如何构建本机数据结构。
json_encode：以便根据任何json表示的样本检查您的本地数据结构。
这里的所有内容都只是简单的"键/值"array()表示法，尽管是嵌套的。但了解这些工具并定期使用它们可能是一种很好的做法。
注意：
您提供的数据示例看起来非常像您为了创建多个项目而"剪切并粘贴"数据，因为各种"子项目"都共享相同的"id"值。
您的"真实"数据不应该这样做！所以我希望它不会，但如果是这样，那么就修复它
为了使第二个示例可行（第一个示例非常好），需要将数据更改为包含每个子元素的"唯一"id"值。
正如我在这里使用的：
{
"_id" : ObjectId("55951b2bf41edfc80b00002a"),
"orders" : [
{
"id" : "55929142f41edfdc0f00002a",
"name" : "XYZ",
"card" : [
{
"id" : "250",
"serial" : "B",
"name" : "Eco",
"ticket" : [
{
"id" : "55927d41f41edfd00f000031",
"name" : "ZZZ",
"price" : "10 €"
},
{
"id" : "55927d41f41edfd00f000032",
"name" : "ZZZ",
"price" : "10 €"
}
]
},
{
"id" : "251",
"serial" : "B",
"name" : "Eco",
"ticket" : [
{
"id" : "55927d41f41edfd00f000033",
"name" : "ZZZ",
"price" : "10 €"
},
{
"id" : "55927d41f41edfd00f000034",
"name" : "ZZZ",
"price" : "10 €"
}
]
}
],
"full_amount" : "40",
},
{
"id" : "55929142f41edfdc0f00002b",
"name" : "XYZ",
"card" : [
{
"id" : "252",
"serial" : "B",
"name" : "Eco",
"ticket" : [
{
"id" : "55927d41f41edfd00f000035",
"name" : "ZZZ",
"price" : "10 €"
},
{
"id" : "55927d41f41edfd00f000036",
"name" : "ZZZ",
"price" : "10 €"
}
]
},
{
"id" : "253",
"serial" : "B",
"name" : "Eco",
"ticket" : [
{
"id" : "55927d41f41edfd00f000037",
"name" : "ZZZ",
"price" : "10 €"
},
{
"id" : "55927d41f41edfd00f000038",
"name" : "ZZZ",
"price" : "10 €"
}
]
}
],
"full_amount" : "40",
}
],
"rate" : "0.23",
"date" : "2015-07-02 13:04:34",
}