Question

这是mongo db中的JSON对象：

{
   "_id": ObjectId("5a1e50efcb4ecdfe9b41d097"),
   "ProjectId": NumberInt(1),
   "TicketId": NumberInt(64),
   "Artifacts": [
     {
       "Slug": ObjectId("5a1e50e6a3d5592b583ec566"),
       "UploadedOn": ISODate("2017-11-29T06:17:10.0Z"),
       "UploadedBy": NumberInt(6),
       "Status": NumberInt(1),
       "ArtifactType": "image",
       "isThumbnailExist": NumberInt(0),
       "ThumbnailPath": "/files/story/thumbnails",
       "FileName": "7b179efffd03e92e495b802c24d3f357-B1",
       "OriginalFileName": "B1.png",
       "Extension": "png" 
    },
     {
       "Slug": ObjectId("5a1e515aa3d5592b583ec56a"),
       "UploadedOn": ISODate("2017-11-29T06:19:06.0Z"),
       "UploadedBy": NumberInt(6),
       "Status": NumberInt(0),
       "ArtifactType": "image",
       "isThumbnailExist": NumberInt(0),
       "ThumbnailPath": "/files/story/thumbnails",
       "FileName": "b83f3aa5d741fb18f1d527757cc002d2-B2",
       "OriginalFileName": "B2.png",
       "Extension": "png" 
    } 
  ] 
}

我只想检索 Artifacts.Status为1 的记录。！

我该怎么做。我不熟悉SQL。

这是我的mongo查询功能：

public static function getACtiveArtifacts($projectId, $ticketId) {
    try{
    $query = new Query();
$query->from('TicketArtifacts')
        ->select(array("Artifacts"))
        ->where(['TicketId' => (int) $ticketId, "ProjectId" => (int) $projectId]);
$ticketArtifactsDetails = $query->one();
return $ticketArtifactsDetails;   
} catch (\Throwable $ex) {
    Yii::error("TicketArtifactsCollection:getTicketArtifacts::" . $ex->getMessage() . "--" . $ex->getTraceAsString(), 'application');
    throw new ErrorException($ex->getMessage());
}
}

我的预期结果:(因为＆＃34;状态＆＃34;：NumberInt（1））

{
   "_id": ObjectId("5a1e50efcb4ecdfe9b41d097"),
   "ProjectId": NumberInt(1),
   "TicketId": NumberInt(64),
"Artifacts": [
     {
       "Slug": ObjectId("5a1e50e6a3d5592b583ec566"),
       "UploadedOn": ISODate("2017-11-29T06:17:10.0Z"),
       "UploadedBy": NumberInt(6),
       "Status": NumberInt(1),
       "ArtifactType": "image",
       "isThumbnailExist": NumberInt(0),
       "ThumbnailPath": "/files/story/thumbnails",
       "FileName": "7b179efffd03e92e495b802c24d3f357-B1",
       "OriginalFileName": "B1.png",
       "Extension": "png" 
    } 
  ] 
}

Answer 1

这可以使用聚合管道完成，使用$unwind和$match来实现结果。

Mongo Shell查询

db.collection.aggregate([
  {$unwind:"$Artifacts"},
  {$match:{"Artifacts.Status":1}}
])

当我们在我们的集合中应用$ unwind聚合管道来展开将导致文档的工件时，如下所示

{
        "_id" : ObjectId("5a1e50efcb4ecdfe9b41d097"),
        "ProjectId" : 1,
        "TicketId" : 64,
        "Artifacts" : {
                "Slug" : ObjectId("5a1e50e6a3d5592b583ec566"),
                "UploadedOn" : ISODate("2017-11-29T06:17:10Z"),
                "UploadedBy" : 6,
                "Status" : 1,
                "ArtifactType" : "image",
                "isThumbnailExist" : 0,
                "ThumbnailPath" : "/files/story/thumbnails",
                "FileName" : "7b179efffd03e92e495b802c24d3f357-B1",
                "OriginalFileName" : "B1.png",
                "Extension" : "png"
        }
}
{
        "_id" : ObjectId("5a1e50efcb4ecdfe9b41d097"),
        "ProjectId" : 1,
        "TicketId" : 64,
        "Artifacts" : {
                "Slug" : ObjectId("5a1e515aa3d5592b583ec56a"),
                "UploadedOn" : ISODate("2017-11-29T06:19:06Z"),
                "UploadedBy" : 6,
                "Status" : 0,
                "ArtifactType" : "image",
                "isThumbnailExist" : 0,
                "ThumbnailPath" : "/files/story/thumbnails",
                "FileName" : "b83f3aa5d741fb18f1d527757cc002d2-B2",
                "OriginalFileName" : "B2.png",
                "Extension" : "png"
        }
}

然后通过将条件Artifacts.Status:1应用$ match，我们将获得最终结果。

{
        "_id" : ObjectId("5a1e50efcb4ecdfe9b41d097"),
        "ProjectId" : 1,
        "TicketId" : 64,
        "Artifacts" : {
                "Slug" : ObjectId("5a1e50e6a3d5592b583ec566"),
                "UploadedOn" : ISODate("2017-11-29T06:17:10Z"),
                "UploadedBy" : 6,
                "Status" : 1,
                "ArtifactType" : "image",
                "isThumbnailExist" : 0,
                "ThumbnailPath" : "/files/story/thumbnails",
                "FileName" : "7b179efffd03e92e495b802c24d3f357-B1",
                "OriginalFileName" : "B1.png",
                "Extension" : "png"
        }
}

Answer 2

您可以使用$redact聚合阶段。

MongoDB聚合查询：

db.collection.aggregate([
    {
        $match: {"ProjectId":1}
    },
    {
        $redact: {
            $cond: {
                if: { $eq: ["$Status", 1] },
                then: "$$PRUNE",
                else: "$$DESCEND"
            }
        }
    }
])

P.S。抱歉，我没有使用Yii框架编写代码，因此我无法向您提供最终代码。

如何只返回mongo db中状态为1的对象参数

2 个答案: