我有这份文件:
{
"$id": "1",
"DocType": "Unidade",
"Nome": "TONY",
"RG_InscricaoEstadual": "4347924938742",
"Setores": [
{
"$id": "9",
"Nome": "Child0",
"Setores": [
{
"$id": "10",
"Nome": "Child1",
"Setores": [
/* <n depth nested level> */
"$id": "11",
"Nome": "Child2",
"Id": "90228c56-eff2-46d2-a324-b04e3c69e15c",
"DocType": "Setor"
],
"Id": "60228c56-dff2-46d2-a324-b04e3c69e15b",
"DocType": "Setor"
}
],
"Id": "8457e1b7-39dc-462c-8f46-871882faea2c",
"DocType": "Setor"
}
]
}
如果我想检索Setor,如何查询此SubDocument,例如
"Id": "60228c56-dff2-46d2-a324-b04e3c69e15b"
我知道如果我现在知道它是嵌套的很多级别,我可以写一个查询来查找类似
的内容Unidade.Setor.Id=="8457e1b7-39dc-462c-8f46-871882faea2c"
但是如何针对未知数量的嵌套级别搜索它,例如1,2,3 n级?
如何找到具有 Id'90228c56-eff2-46d2-a324-b04e3c69e15c'的Setor?
关于如何解决这个问题的评论也将不胜感激。
答案 0 :(得分:3)
在c#中,我们可以创建一个递归方法来实现这种情况。
这是我创建的BsonDocument:
BsonDocument doc = new BsonDocument {
{ "id","1"},
{ "DocType", "Unidade"},
{ "Nome", "TONY"},
{ "RG_InscricaoEstadual", "4347924938742"},
{ "Setores",new BsonArray {
new BsonDocument {
{ "id","9" },
{ "Nome", "Child0"},
{ "Setores", new BsonArray { new BsonDocument {
{ "id","10" },
{ "Nome", "Child1"},
{ "Setores", new BsonArray { new BsonDocument {
{ "id","11" },
{ "Nome", "Child2"},
{ "Id","90228c56-eff2-46d2-a324-b04e3c69e15c" },
{ "DocType", "Setor"}
}
}
},
{ "Id","60228c56-dff2-46d2-a324-b04e3c69e15b" },
{ "DocType", "Setor"}
}
}
},
{ "Id","8457e1b7-39dc-462c-8f46-871882faea2c" },
{ "DocType", "Setor"}
}
}
}
};
您可以使用Mongo c# Query method从MongoDB获取此BsonDocument。
这是我用来通过“Id”查询文档的递归方法:
BsonDocument result = GetID(doc, "90228c56-eff2-46d2-a324-b04e3c69e15c");
public static BsonDocument GetID(BsonDocument doc, string queryId)
{
BsonDocument result = new BsonDocument();
if (doc.Elements.Where(c => c.Name == "Setores").Count() != 0)
{
foreach (var item in doc.GetElement("Setores").Value.AsBsonArray)
{
var id = item.AsBsonDocument.GetElement("Id").Value;
if (id == queryId)
{
result = item.AsBsonDocument;
break;
}
result = GetID(item.AsBsonDocument, queryId);
}
}
return result;
}
我希望这可以给你一些提示。
答案 1 :(得分:1)
你唯一能做的就是嵌套查询,即
find({"Unidade.Setor.Id": ObjectId("8457e1b7-39dc-462c-8f46-871882faea2c")
find({"Unidade.Setor.Setor.Id": ObjectId("8457e1b7-39dc-462c-8f46-871882faea2c")
find({"Unidade.Setor.Setor.Setor.Id": ObjectId("8457e1b7-39dc-462c-8f46-871882faea2c")
如果前一个失败,则依次运行。
但请勿!!!
您应该将这些Setor
记录存储为单独的文档。您可以使用对彼此的引用存储它们,然后使用lookup(如SQL中的连接)
https://docs.mongodb.com/manual/reference/operator/aggregation/lookup/