Question

我很难理解CouchDB的linked documents功能。

我有两个types数据存储在一个CouchDB数据库中：

{
  "id":"1",
  "type": "track",
  "title": "Bohemian Rhapsody"
}

{
  "id":"2",
  "type": "artist",
  "name": "Queen",
  "tracks": ["1"]
}

我的印象是我可以编写如下所示的视图，并获得以下文档：

{
  "id":"2",
  "type": "artist",
  "name": "Queen",
  "tracks": [
    {
      "id":"1",
      "type": "track",
      "title": "Bohemian Rhapsody"
    }
  ]
}

我一直在尝试这种观点，但它并没有按照我期望的方式运作：

function(doc) {
  if(doc.type == 'artist') {
    var tracks = [];
    for(var i = 0; i < doc.tracks.length; i++) {
      tracks.push({_id:doc.tracks[i]});
    }

    newdoc = eval(uneval(doc));
    newdoc.tracks = tracks;

    emit(doc._id,newdoc);
  }
}

示例：http://jphastings.iriscouch.com/_utils/database.html?music/_design/test/_view/linked

这不是我希望的回报 - 你有什么建议吗？感谢

Answer 1

好的，我终于明白你要做什么。这是可能的。这就是。

你有2个文件

{
"_id":"anyvalue",
"type": "track",
"title": "Bohemian Rhapsody"
}

{
"_id":"2",
"type": "artist",
"name": "Queen",
"tracks": ["anyvalue"]
}

你做错了的是没有关于曲目值（数组中的项目）的引号。

2）引用ID必须为_id才能生效。值得注意的是，因为你可以拥有id字段，但只有_id用于识别文档。

对于你想要这个视图的结果就足够了

function(doc) {
    if (doc.type === 'artist') {
        for (var i in doc.tracks) {
            var id = doc.tracks[i];
            emit(id, { _id: id });
        }
    }
}

你想要做的是在for循环中使用emit函数来发出每个艺术家的'track'的id字段。

然后，您想使用include_docs = true参数查询couch db视图。这是您在iris沙发上创建的数据库的最终结果。

http://jphastings.iriscouch.com/music/_design/test/_view/nested?reduce=false&include_docs=true

 {
"total_rows": 3,
"offset": 0,
"rows": [
 {
  "id": "0b86008d8490abf0b7e4f15f0c6a50a7",
  "key": "0b86008d8490abf0b7e4f15f0c6a463b",
  "value": {
    "_id": "0b86008d8490abf0b7e4f15f0c6a463b"
  },
  "doc": {
    "_id": "0b86008d8490abf0b7e4f15f0c6a463b",
    "_rev": "3-7e4ba3bfedd29a07898125c09dd7262e",
    "type": "track",
    "title": "Boheniam Rhapsody"
  }
},
{
  "id": "0b86008d8490abf0b7e4f15f0c6a50a7",
  "key": "0b86008d8490abf0b7e4f15f0c6a5ae2",
  "value": {
    "_id": "0b86008d8490abf0b7e4f15f0c6a5ae2"
  },
  "doc": {
    "_id": "0b86008d8490abf0b7e4f15f0c6a5ae2",
    "_rev": "2-b3989dd37ef4d8ed58516835900b549e",
    "type": "track",
    "title": "Another one bites the dust"
  }
},
{
  "id": "0b86008d8490abf0b7e4f15f0c6a695e",
  "key": "0b86008d8490abf0b7e4f15f0c6a6353",
  "value": {
    "_id": "0b86008d8490abf0b7e4f15f0c6a6353"
  },
  "doc": {
    "_id": "0b86008d8490abf0b7e4f15f0c6a6353",
    "_rev": "2-0383f18c198b813943615d2bf59c212a",
    "type": "track",
    "title": "Stripper Vicar"
  }
 }
]
}

杰森在这篇文章中解释得非常好

Best way to do one-to-many "JOIN" in CouchDB

此链接也有助于沙发数据库中的实体关系

http://wiki.apache.org/couchdb/EntityRelationship

CouchDB在视图中的链接文档

1 个答案: