MongoDB C＃从小组

时间：2016-02-14 00:05:55

标签： c# mongodb mongodb-query aggregation-framework mongodb-.net-driver

我有一组假装付款的状态，每个都有付款ID。

我想获取每个付款ID的最新状态。测试我创建了一些虚拟数据然后尝试查询它。我到目前为止：

[Test]
public void GetPaymentLatestStatuses()
{
    var client = new TestMongoClient();

    var database = client.GetDatabase("payments");

    var paymentRequestsCollection = database.GetCollection<BsonDocument>("paymentRequests");

    var statusesCollection = database.GetCollection<BsonDocument>("statuses");

    var payment = new BsonDocument { { "amount", RANDOM.Next(10) } };

    paymentRequestsCollection.InsertOne(payment);

    var paymentId = payment["_id"];

    var receivedStatus = new BsonDocument
                         {
                             { "payment", paymentId },
                             { "code", "received" },
                             { "date", DateTime.UtcNow }
                         };
    var acceptedStatus = new BsonDocument
                         {
                             { "payment", paymentId },
                             { "code", "accepted" },
                             { "date", DateTime.UtcNow.AddSeconds(-1) }
                         };
    var completedStatus = new BsonDocument
                          {
                              { "payment", paymentId },
                              { "code", "completed" },
                              { "date", DateTime.UtcNow.AddSeconds(-2) }
                          };

    statusesCollection.InsertMany(new [] { receivedStatus, acceptedStatus, completedStatus });

    var groupByPayments = new BsonDocument { {"_id", "$payment"} };

    var statuses = statusesCollection.Aggregate().Group(groupByPayments);

}

但现在我在砖墙上。

任何正确的方向戳都会有所帮助。我不确定我是不是在俯视望远镜的错误端。

更新

以下为我提供了正确文件的ID。

var groupByPayments = new BsonDocument
                      {
                          { "_id", "$payment" },
                          { "id", new BsonDocument { { "$first", "$_id" } } }
                      };

var sort = Builders<BsonDocument>.Sort.Descending(document => document["date"]);

var statuses = statusesCollection.Aggregate().Sort(sort).Group(groupByPayments).ToList();

我是否可以使用单个查询获取完整文档，或者我是否必须重新发出命令以获取该列表中的所有文档？

3 个答案:

答案 0 :(得分：10)

让我们从简单的方法开始，了解您想要实现的目标。在MongoDB的C＃Driver 2.X中，您可以找到AsQueryable扩展方法，让您从集合中创建LINQ查询。这个Linq提供程序是在MongoDB的Aggregation框架上构建的，所以最后你的链接查询将被转换为聚合管道。所以，如果你有这样一个类：

public class Status
{
  public ObjectId _id { get; set; }
  public ObjectId payment { get; set; }
  public string code { get; set; }
  public DateTime date { get; set; }
}

您可以创建如下所示的查询：

 var statusesCollection = database.GetCollection<Status>("statuses");
 var result= statusesCollection.AsQueryable()
                               .OrderByDescending(e=>e.date)
                               .GroupBy(e=>e.payment)
                               .Select(g=>new Status{_id =g.First()._id,
                                                     payment = g.Key,
                                                     code=g.First().code,
                                                     date=g.First().date
                                                    }
                                       )
                               .ToList();

现在您可能想知道为什么我必须将结果投影到Status类的新实例，如果我能从每个组中获得相同的结果调用First扩展方法？不幸的是，这还不支持。其中一个原因是因为Linq提供程序在构建聚合管道时使用$first操作，这就是$first操作的工作原理。另外，正如您在$first阶段中使用$group时在早期共享链接中看到的那样，$group阶段应遵循$sort阶段按照定义的顺序输入文档。

现在，假设您不想使用Linq，并且您希望自己创建聚合管道，则可以执行以下操作：

 var groupByPayments = new BsonDocument
                      {
                          { "_id", "$payment" },
                          { "statusId", new BsonDocument { { "$first", "$_id" } } },
                          { "code", new BsonDocument { { "$first", "$code" } } },
                          { "date", new BsonDocument { { "$first", "$date" } } }
                      };

var sort = Builders<BsonDocument>.Sort.Descending(document => document["date"]);

ProjectionDefinition<BsonDocument> projection = new BsonDocument
        {
            {"payment", "$_id"},
            {"id", "$statusId"},
            {"code", "$code"},
            {"date", "$date"},
        }; 
var statuses = statusesCollection.Aggregate().Sort(sort).Group(groupByPayments).Project(projection).ToList<BsonDocument>();

这个解决方案的优势在于您可以一次性获取数据，缺点是您必须预测所需的所有字段。我的结论是，如果文档没有很多字段，或者您不喜欢不需要你文档中的所有字段我会使用这个变体。

答案 1 :(得分：1)

这就是我实现它的方式。必须有更好的方法。

[Test]
public void GetPaymentLatestStatuses()
{
    var client = new TestMongoClient();

    var database = client.GetDatabase("payments");

    var paymentRequestsCollection = database.GetCollection<BsonDocument>("paymentRequests");

    var statusesCollection = database.GetCollection<BsonDocument>("statuses");

    var payment = new BsonDocument { { "amount", RANDOM.Next(10) } };

    paymentRequestsCollection.InsertOne(payment);

    var paymentId = payment["_id"];

    var receivedStatus = new BsonDocument
                         {
                             { "payment", paymentId },
                             { "code", "received" },
                             { "date", DateTime.UtcNow }
                         };
    var acceptedStatus = new BsonDocument
                         {
                             { "payment", paymentId },
                             { "code", "accepted" },
                             { "date", DateTime.UtcNow.AddSeconds(+1) }
                         };
    var completedStatus = new BsonDocument
                          {
                              { "payment", paymentId },
                              { "code", "completed" },
                              { "date", DateTime.UtcNow.AddSeconds(+2) }
                          };

    statusesCollection.InsertMany(new[] { receivedStatus, acceptedStatus, completedStatus });

    var groupByPayments = new BsonDocument
                          {
                              { "_id", "$payment" },
                              { "id", new BsonDocument { { "$first", "$_id" } } }
                          };

    var sort = Builders<BsonDocument>.Sort.Descending(document => document["date"]);

    var statuses = statusesCollection.Aggregate().Sort(sort).Group(groupByPayments).ToList();

    var statusIds = statuses.Select(x => x["id"]);

    var completedStatusDocumentsFilter =
        Builders<BsonDocument>.Filter.Where(document => statusIds.Contains(document["_id"]));

    var statusDocuments = statusesCollection.Find(completedStatusDocumentsFilter).ToList();

    foreach (var status in statusDocuments)
    {
        Assert.That(status["code"].AsString, Is.EqualTo("completed"));
    }
}

答案 2 :(得分：0)

  

必须有更好的方法。

从2.5.3开始，您可以访问聚合内的当前组。这使我们可以构建一个通用访问器，它将通过本机mongo查询从分组中检索第一个元素。

首先，反序列化的辅助类。 /// <summary> /// Mongo-ified version of <see cref="KeyValuePair{TKey, TValue}"/> /// </summary> class InternalKeyValuePair<T, TKey> { [BsonId] public TKey Key { get; set; } public T Value { get; set; } } //you may not need this method to be completely generic, //but have the sortkey be the same helps interface IDateModified { DateTime DateAdded { get; set; } } private List<T> GroupFromMongo<T,TKey>(string KeyName) where T : IDateModified { //mongo linq driver doesn't support this syntax, so we make our own bsondocument. With blackjack. And Hookers. BsonDocument groupDoc = MongoDB.Bson.BsonDocument.Parse(@" { _id: '$" + KeyName + @"', Value: { '$first': '$$CURRENT' } }"); //you could use the same bsondocument parsing trick to get a generic //sorting key as well as a generic grouping key, or you could use //expressions and lambdas and make it...perfect. SortDefinition<T> sort = Builders<T>.Sort.Descending(document => document.DateAdded); List<BsonDocument> intermediateResult = getCol<T>().Aggregate().Sort(sort).Group(groupDoc).ToList(); InternalResult<T, TKey>[] list = intermediateResult.Select(r => MongoDB.Bson.Serialization.BsonSerializer.Deserialize<InternalResult<T, TKey>>(r)).ToArray(); return list.Select(z => z.Value).ToList(); } 已被封存，因此我们自己动手。

    /// <summary>
    /// Mongo-ified version of <see cref="KeyValuePair{TKey, TValue}"/>
    /// </summary>
    class MongoKeyValuePair<T, TKey>
    {
        [BsonId]
        public TKey Key { get; set; }
        public T Value { get; set; }
    }
    private MongoKeyValuePair<T, TKey>[] GroupFromMongo<T, TKey>(Expression<Func<T, TKey>> KeySelector, Expression<Func<T, object>> SortSelector)
    {
        //mongo linq driver doesn't support this syntax, so we make our own bsondocument. With blackjack. And Hookers. 
        BsonDocument groupDoc = MongoDB.Bson.BsonDocument.Parse(@"
         {
                _id: '$" + GetPropertyName(KeySelector) + @"',
                Value: { '$first': '$$CURRENT' }
        }");
        SortDefinition<T> sort = Builders<T>.Sort.Descending(SortSelector);
        List<BsonDocument> groupedResult = getCol<T>().Aggregate().Sort(sort).Group(groupDoc).ToList();
        MongoKeyValuePair<T, TKey>[] deserializedGroupedResult = groupedResult.Select(r => MongoDB.Bson.Serialization.BsonSerializer.Deserialize<MongoKeyValuePair<T, TKey>>(r)).ToArray();
        return deserializedGroupedResult;
    }

    /* This was my original non-generic method with hardcoded strings, PhonesDocument is an abstract class with many implementations */
    public List<T> ListPhoneDocNames<T>() where T : PhonesDocument
    {
        return GroupFromMongo<T,String>(z=>z.FileName,z=>z.DateAdded).Select(z=>z.Value).ToList();
    }


    public string GetPropertyName<TSource, TProperty>(Expression<Func<TSource, TProperty>> propertyLambda)
    {
        Type type = typeof(TSource);

        MemberExpression member = propertyLambda.Body as MemberExpression;
        if (member == null)
            throw new ArgumentException(string.Format(
                "Expression '{0}' refers to a method, not a property.",
                propertyLambda.ToString()));

        PropertyInfo propInfo = member.Member as PropertyInfo;
        if (propInfo == null)
            throw new ArgumentException(string.Format(
                "Expression '{0}' refers to a field, not a property.",
                propertyLambda.ToString()));

        if (type != propInfo.ReflectedType &&
            !type.IsSubclassOf(propInfo.ReflectedType))
            throw new ArgumentException(string.Format(
                "Expresion '{0}' refers to a property that is not from type {1}.",
                propertyLambda.ToString(),
                type));

        return propInfo.Name;
    }

好的..我在https://stackoverflow.com/a/672212/346272

的帮助下对其进行了通用化 
    class MongoKeyValuePair<T, TKey>
    {
        [BsonId]
        public TKey Key { get; set; }
        public T Value { get; set; }
        public long Count { get; set; }
    }

        BsonDocument groupDoc = MongoDB.Bson.BsonDocument.Parse(@"
         {
                _id: '$" + GetPropertyName(KeySelector) + @"',
                Value: { '$first': '$$CURRENT' },
                Count: { $sum: 1 }
        }");

对于奖励积分，您现在可以轻松地执行任何mongos其他分组操作，而无需对抗linq助手。有关所有可用的分组操作，请参阅https://docs.mongodb.com/manual/reference/operator/aggregation/group/。我们加一个计数。

import { Injectable } from '@angular/core';

@Injectable()
export class ErrorService {
  errorInfo: string;
}

运行与之前完全相同的聚合，您的count属性将填入与您的groupkey匹配的文档数量。整齐！

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