Question

我有一个FilterDefinition构建，它将根据非空属性查找地址。

public static FilterDefinition<TU> FindPointByAddress<TU>(Address address)
{
    var filterBuilder = Builders<TU>.Filter;
    var filterItems = new List<FilterDefinition<TU>>();

    if (!String.IsNullOrWhiteSpace(address.Street))
    {
        filterItems.Add(filterBuilder.Eq("Address.Street", address.Street));
    }

    if (!String.IsNullOrWhiteSpace(address.City))
    {
        filterItems.Add(filterBuilder.Eq("Address.City", address.City));
    }

    if (!String.IsNullOrWhiteSpace(address.StateProvince))
    {
        filterItems.Add(filterBuilder.Eq("Address.StateProvince", address.StateProvince));
    }

    if (!String.IsNullOrWhiteSpace(address.PostCode))
    {
        filterItems.Add(filterBuilder.Eq("Address.PostCode", address.PostCode));
    }

    return filterBuilder.And(filterItems);
}

IMO这个查询感觉很脏，有没有更好的方法来构建这种类型的查询或者这是正确的方法吗？

Answer 1

前几天我遇到了类似的情况。我写了一个简单的方法，它将字段名称和字段值作为字符串。

public void AddEqualCompareFilter(string fieldName, string fieldValue)
    {
        if (String.IsNullOrEmpty(fieldValue) == false) {
            if (Filter != null) {
                Filter = Filter & Builders<TranslationsDocument>.Filter.Eq(fieldName, fieldValue);
            }
            else {
                FilterCount++;
                Filter = Builders<TranslationsDocument>.Filter.Eq(fieldName, fieldValue);
            }

        }
    }

然后我使用此代码段来决定基于FilterCount：

if (FilterCount > 0) {
            Result = collection.Find(Filter).ToListAsync().GetAwaiter().GetResult();
            return true;
        }
        else {

            Result = collection.Find(new BsonDocument()).ToListAsync().GetAwaiter().GetResult();
            return true;
        }

C＃Mongo查询效率

1 个答案: