Question

如何获取与当前实体相关的所有注释的guid列表（或例如集合）？

我知道它在某种程度上与service.Retreive（...）有关，但无法生成有效的代码。

附加1：

public void Execute(IServiceProvider serviceProvider)
{
    Entity entity = null;

    var context = (IPluginExecutionContext)serviceProvider.GetService(typeof(IPluginExecutionContext));

    if (entity.Contains("new_parentopportunity"))
    {
        try
        {
            IOrganizationServiceFactory factory = (IOrganizationServiceFactory)serviceProvider.GetService(typeof(IOrganizationServiceFactory));
            IOrganizationService service = factory.CreateOrganizationService(context.UserId);

            Guid projectGuid = entity.Id;
            Guid oppGuid = ((EntityReference)entity["new_parentopportunity"]).Id;

            for (int i = 0; i < 100; i++) //Instead of 100, I'll place obtained collection.Length
            {
                Entity opp = service.Retrieve("", oppGuid, new Microsoft.Xrm.Sdk.Query.ColumnSet(true)); //RETREIVE ENTITY NOTE HERE
                var temop = CreateNoteAttachment(opp);
                service.Create(temp);
            }

        }
        catch (Exception ex)
        {
            throw new InvalidPluginExecutionException("An error occurred in the plug-in.", ex);
        }
    }
}

private Entity CreateNoteAttachment(Entity oppNote)
{
    //some code here
}

更新1：

我也是在查询中写的，你能看一下吗，那里的一切都很好吗？ PasteBin

Answer 1

查询相关注释的两种方法：

早期约束：

 var  notes =
                service.AnnotationSet.Where(annotation => annotation.Opportunity_Annotation.Id == entity.Id)
                    .Select(annotation => annotation.Id)
                    .ToList();

FetchXml：

  var fetchXml = string.Format("<fetch version='1.0' output-format='xml-platform' mapping='logical' distinct='false'>" +
                                    "<entity name='annotation'>" +
                                    "<attribute name='annotationid' />" +
                                    "<filter type='and'>"+
                                    "<condition attribute='objectid' operator='eq' value='{0}'/>"+
                                    "</filter>"+
                                    "</entity>" +
                                    "</fetch>", entity.Id);

            var response = service.RetrieveMultiple(new FetchExpression(fetchXml));
            var notes = response.Entities;

获取笔记列表

1 个答案: