使用MongoTemplate聚合过滤MongoDB中的嵌套元素

我正在尝试筛选标题与给定输入标题数组匹配的字段数组，并显示完整的文档，不包括不匹配的字段。下面是我的文档。实际上，我想使用MongoTemplate实现这一点，但首先我尝试通过mongo查询来获得它。以下是我的文件：

{
    "version": 2,
    "pageName": "Content_2",
    "domain": "bingo.com",
    "locale": "en-us",
    "contents": [
        {
            "contentName": "Template_2",
            "fields": [
                {
                    "fieldType": "Plain Text",
                    "id": "companyName456",
                    "title": "Company Name",
                    "alternateText": "Company Name",
                    "value": "Microsoft",
                    "placeholder": "Enter your Company name"
                },
                {
                    "fieldType": "Plain Text",
                    "id": "designation789",
                    "title": "Designation",
                    "alternateText": "Designation",
                    "value": "Software Engineer",
                    "placeholder": "Enter your designation name"
                }
            ]
        }
    ]
}

我尝试了以下查询，但返回的结果为空：

db.contents.aggregate(
  [
    { $match: { locale: "en-us" } },
    {
      $redact: {
        $cond: {
          if: { $in: [ "$title", ["Designation"] ] },
          then: "$$DESCEND",
          else: "$$PRUNE"
        }
      }
    }
  ]
);

我期待以下结果：

{
    "pageName": "Home",
    "link": "hello.com",
    "locale": "en-us",
    "contents": [
        {
            "contentName": "Template_2",
            "fields": [
                {
                    "fieldType": "Plain Text",
                    "id": "designation789",
                    "title": "Designation",
                    "alternateText": "Designation",
                    "value": "Software Engineer",
                    "placeholder": "Enter your designation name"
                }
            ]
        }
    ]
}

---

请导游。我对MongoDB很陌生

这样的方法应该行得通

db.contents.aggregate([
# Match that local
     {$match: { locale: "en-us" } },
# Unwind by contents and contents.fields
     {$unwind: "$contents"},
     {$unwind: "$contents.fields"},
# Match them
     {$match: { "contents.fields.title": "Designation" } },
# group back by _id of the document
     {$group: {
         "_id": "$_id",
         "contents": { "$push": "$contents" }
     }}
 ])

如果您需要所有其他字段，也可以按如下方式分组

db.contents.aggregate([
     {$match: { locale: "en-us" } },
     {$unwind: "$contents"},
     {$unwind: "$contents.fields"},
     {$match: { "contents.fields.title": "Designation" } },
     {$group: {
         "_id": {
            "_id": "$_id",
            "version": "$version",
            "pageName": "$pageName",
            "domain": "$domain",
            "locale": "$locale",

        },
         "contents": { "$push": "$contents" }
     }},
     {$project: {"version": "$_id.version", 
                 "_id": "$_id._id",
                 "pageName": "$_id.pageName",
                 "locale": "$_id.locale",
                 "domain": "$_id.domain",
                 "contents": "$contents",}}
 ])

---

谢谢它帮助了我。在$project内部，是否有较短的方式来投影所有外部字段？我已经尝试过这种方法，但我正在寻找一个较短的解决方案。据我所知，但我也是mongoI的新手，我能够通过以下查询实现我的结果：

db.contents.aggregate（[{$match:{pageName:{Content_2}}，{$unwind:$contents}，{$unwind:$contents.fields}，{$match:{“contents.fields.title”：{$in:[“公司名称”]}}}，{$group:{“{U id”：“$\U id”，“内容”：“{$push”：“$contents”}，{$first:$$root”}}，{$replaceRoot:{newRoot:{$mergeObjects:[“$root”，“{contents:'$contents''.$contents'}}}}}}}}，{$code>