Question

我正在尝试从Contact实体检索数据，将其连接到Function实体，然后将Function联接到Account实体。

这个想法是使用一个查询来检索联系人数据以及另外两个查询中的一些额外字段。

我能够从Function链接的实体中进行联接和检索数据，但是在理解如何使用嵌套的链接实体时遇到了麻烦。

QueryExpression queryExpression = new QueryExpression(Contact.EntityLogicalName);

queryExpression.ColumnSet = new ColumnSet(Contact.Fields.pix_GZGId, Contact.Fields.Id, Contact.Fields.FirstName, Contact.Fields.LastName, 
    Contact.Fields.MiddleName, Contact.Fields.GenderCode, Contact.Fields.EMailAddress1, Contact.Fields.EMailAddress2, 
    Contact.Fields.Telephone2, Contact.Fields.MobilePhone, Contact.Fields.pix_Gastlessenoptin);
FilterExpression emailFilter = new FilterExpression(LogicalOperator.Or);

emailFilter.AddCondition(Contact.Fields.EMailAddress1, ConditionOperator.Equal, email);
emailFilter.AddCondition(Contact.Fields.EMailAddress2, ConditionOperator.Equal, email);

FilterExpression nameFilterAll = new FilterExpression(LogicalOperator.And);
nameFilterAll.AddCondition(Contact.Fields.FirstName, ConditionOperator.Equal, firstname);
nameFilterAll.AddCondition(Contact.Fields.LastName, ConditionOperator.Equal, lastname);

if (!String.IsNullOrEmpty(middlename))
{
    nameFilterAll.AddCondition(Contact.Fields.MiddleName, ConditionOperator.Equal, middlename);
}

FilterExpression nameFilterPartial = new FilterExpression(LogicalOperator.And);
nameFilterPartial.AddCondition(Contact.Fields.FirstName, ConditionOperator.BeginsWith, firstname.Substring(0,1));
nameFilterPartial.AddCondition(Contact.Fields.LastName, ConditionOperator.Equal, lastname);

FilterExpression nameFilterLast = new FilterExpression();
nameFilterLast.AddCondition(Contact.Fields.LastName, ConditionOperator.Equal, lastname);

FilterExpression completeNameFilter = new FilterExpression(LogicalOperator.Or);
completeNameFilter.Filters.Add(nameFilterAll);
completeNameFilter.Filters.Add(nameFilterPartial);
completeNameFilter.Filters.Add(nameFilterLast);

queryExpression.Criteria = new FilterExpression(LogicalOperator.And);
queryExpression.Criteria.Filters.Add(emailFilter);
queryExpression.Criteria.Filters.Add(completeNameFilter);

queryExpression.AddOrder(Contact.Fields.LastName, OrderType.Descending);

//this is the problem part - it doesn't seem to do anything.
LinkEntity linkedFunctions = new LinkEntity(Contact.EntityLogicalName, pix_functie.EntityLogicalName, 
    Contact.Fields.Id, pix_functie.Fields.pix_Persoon, JoinOperator.LeftOuter);
linkedFunctions.Columns = new ColumnSet(pix_functie.Fields.Id, pix_functie.Fields.pix_Organisatie, 
    pix_functie.Fields.pix_isgzgfunction, pix_functie.Fields.pix_omschrijving, pix_functie.Fields.CreatedOn);
linkedFunctions.EntityAlias = ConfigKeys.PixFunctieAlias;

FilterExpression accountFilterExpression = new FilterExpression();
//accountFilterExpression.Conditions.Add(new ConditionExpression(pix_functie.Fields.pix_isgzgfunction, ConditionOperator.Equal, true));

linkedFunctions.AddLink(Account.EntityLogicalName, pix_functie.Fields.pix_Organisatie, Account.Fields.Id, JoinOperator.LeftOuter);
linkedFunctions.LinkEntities[0].LinkCriteria = accountFilterExpression;
linkedFunctions.LinkEntities[0].EntityAlias = "linkedAccountAlias";

queryExpression.LinkEntities.Add(linkedFunctions);

request.Query = queryExpression;

RetrieveMultipleResponse response = (RetrieveMultipleResponse)_context.Execute(request);

这使我从Function链接实体获得的数据毫无问题，然后我对其进行分组，排序和处理：

    IEnumerable<Microsoft.Xrm.Sdk.Entity> entities = response.EntityCollection.Entities   
            .OrderByDescending(x => (x.GetAttributeValue<AliasedValue>(ConfigKeys.PixFunctieAlias + "." + pix_functie.Fields.pix_isgzgfunction) != null ? (bool?)
                x.GetAttributeValue<AliasedValue>(ConfigKeys.PixFunctieAlias + "." + pix_functie.Fields.pix_isgzgfunction).Value : new Nullable<bool>()).HasValue)
            .ThenByDescending(x => (x.GetAttributeValue<AliasedValue>(ConfigKeys.PixFunctieAlias + "." + pix_functie.Fields.pix_isgzgfunction) != null ? (bool?)
                x.GetAttributeValue<AliasedValue>(ConfigKeys.PixFunctieAlias + "." + pix_functie.Fields.pix_isgzgfunction).Value : null))
            .ThenByDescending(x => (x.GetAttributeValue<AliasedValue>(ConfigKeys.PixFunctieAlias + "." + pix_functie.Fields.CreatedOn) != null ? (DateTime?)
                x.GetAttributeValue<AliasedValue>(ConfigKeys.PixFunctieAlias + "." + pix_functie.Fields.CreatedOn).Value : new Nullable<DateTime>()).HasValue)
            .ThenByDescending(x => (x.GetAttributeValue<AliasedValue>(ConfigKeys.PixFunctieAlias + "." + pix_functie.Fields.CreatedOn) != null ? (DateTime?)
                x.GetAttributeValue<AliasedValue>(ConfigKeys.PixFunctieAlias + "." + pix_functie.Fields.CreatedOn).Value : null));


    List<TtNugGzgDataContract.Entities.Contact> lvl1Contacts = new List<TtNugGzgDataContract.Entities.Contact>();
    List<TtNugGzgDataContract.Entities.Contact> lvl2Contacts = new List<TtNugGzgDataContract.Entities.Contact>();

    foreach(var entity in entities)
    {
        Contact contact = entity.ToEntity<Contact>();
        TtNugGzgDataContract.Entities.Contact contactToAdd = new TtNugGzgDataContract.Entities.Contact()
        {
            externalID = contact.Id,
            firstname = contact.FirstName,
            lastname = contact.LastName,
            insertion = contact.MiddleName,
            gender = Converter.GetApiGender(contact.GenderCodeEnum),
            phone = !String.IsNullOrEmpty(contact.Telephone2) ? contact.Telephone2 : contact.MobilePhone,
            newsletter = contact.pix_GastlessenoptinEnum == pix_nieuwsbriefoptin.Toestaan
        };


        contactToAdd.entityID = entity.GetAttributeValue<AliasedValue>("linkedAccountAlias" + "." + Account.Fields.pix_GZGId) != null ?
            (string)entity.GetAttributeValue<AliasedValue>("linkedAccountAlias" + "." + Account.Fields.pix_GZGId).Value : null;

        contactToAdd.function = entity.GetAttributeValue<AliasedValue>(ConfigKeys.PixFunctieAlias + "." + pix_functie.Fields.pix_omschrijving) != null ?
            (string)entity.GetAttributeValue<AliasedValue>(ConfigKeys.PixFunctieAlias + "." + pix_functie.Fields.pix_omschrijving).Value : null;

        contactToAdd.isGzg = entity.GetAttributeValue<AliasedValue>(ConfigKeys.PixFunctieAlias + "." + pix_functie.Fields.pix_isgzgfunction) != null ?
            (bool?)entity.GetAttributeValue<AliasedValue>(ConfigKeys.PixFunctieAlias + "." + pix_functie.Fields.pix_isgzgfunction).Value : null;

        contactToAdd.createdOn = entity.GetAttributeValue<AliasedValue>(ConfigKeys.PixFunctieAlias + "." + pix_functie.Fields.CreatedOn) != null ?
            (DateTime?)entity.GetAttributeValue<AliasedValue>(ConfigKeys.PixFunctieAlias + "." + pix_functie.Fields.CreatedOn).Value : null;

        EntityReference reference = entity.GetAttributeValue<AliasedValue>(ConfigKeys.PixFunctieAlias + "." + pix_functie.Fields.pix_Organisatie) != null ?
            (EntityReference)entity.GetAttributeValue<AliasedValue>(ConfigKeys.PixFunctieAlias + "." + pix_functie.Fields.pix_Organisatie).Value : null;



        contactToAdd.accId = (entity.GetAttributeValue<AliasedValue>(ConfigKeys.PixFunctieAlias + "." + pix_functie.Fields.pix_Organisatie) != null ?
            (EntityReference)entity.GetAttributeValue<AliasedValue>(ConfigKeys.PixFunctieAlias + "." + pix_functie.Fields.pix_Organisatie).Value : new EntityReference()).Id;


        if (!String.IsNullOrEmpty(contact.EMailAddress1) && contact.EMailAddress1.Equals(email, StringComparison.InvariantCultureIgnoreCase))
        {
            contactToAdd.email = contact.EMailAddress1;
            contactToAdd.isSecondaryEmail = false;
            lvl1Contacts.Add(contactToAdd);
        } else
        {
            contactToAdd.email = contact.EMailAddress2;
            contactToAdd.isSecondaryEmail = true;
            lvl2Contacts.Add(contactToAdd);
        }
    }

但是我无法从嵌套的链接实体中获取entityID。 AFAIK别名是正确的，我什至尝试将其与第一级链接实体的别名一起放在前缀之前。

有人能指出我正确的方向吗？

Answer 1

我建议您不要使用手动滚动QueryExpression的方法来为您完成此操作。

下载并安装XrmToolBox（http://www.xrmtoolbox.com）。
从插件商店（在工具菜单上）下载FetchXml Builder。
在图形界面中使用FetchXml构建查询。
在“查看”菜单上，选择“ QueryExpression”。
右侧将弹出一个窗口，向您显示为FetchXml查询生成的QueryExpression。

希望有帮助！

Dynamics CRM QueryExpression嵌套的LinkedEntity

1 个答案: