Question

我正在努力寻找一种方法来使用ExecuteMultipleRequest来创建记录并同时将其与其他记录相关联。到目前为止，我只是成功地首先创建了记录，然后在另一个请求中建立了我的关联。

使用此代码会抛出一个异常，它需要一个目标（需要先提交以创建记录并获取ID）

private AssociateRequest GetConfigurationRequest(VariableDataRequestModel receivedRequest)
{
    var configurationRelationship = new Relationship(ConfigurationRelationshipName);            
    var configurationEntities = new EntityReferenceCollection(new EntityReference[]{new EntityReference(ConfigurationEntityLogicalName,new Guid(receivedRequest.ConfigurationId))});                        
    var rtn = new AssociateRequest() {  RelatedEntities = configurationEntities, Relationship = configurationRelationship };
    return rtn;
}

Answer 1

使用ExecuteMultipleRequest实现的唯一方法是在创建记录之前指定新记录的Id：

Guid accountId = new Guid("c85dca5d-6520-4436-a5f6-178b633af819");

Entity myAccount = new Entity("account");
myAccount["name"] = "TEST ACCOUNT";
entity.Id = accountId;
// ...
EntityReferenceCollection relatedEntities = new EntityReferenceCollection();
relatedEntities.Add(new EntityReference("account", accountId));

指定允许Id，但不建议。

Answer 2

您可以做的另一件事是在ExecuteMultipleRequest中提交记录，然后在ExecuteMultipleResponse中获取GUID，然后您可以将该GUID与您创建的Record一起使用来执行AssociateRequest。

示例：

ExecuteMultipleResponse response = (ExecuteMultipleResponse)service.Execute(updateCRM);

List<Entity> addedRecordList = new List<Entity>();

foreach (ExecuteMultipleResponseItem item in response.Responses)
{
    Guid responseReturnedGuid = (Guid)item.Response.Results.Values.FirstOrDefault(x => x is Guid);
    if (responseReturnedGuid != Guid.Empty || responseReturnedGuid != null)
    {
        Entity addedRecord = (Entity)updateCRM.Requests[item.RequestIndex].Parameters.Values.FirstOrDefault(x => x is Entity);
        addedRecord.Id = responseReturnedGuid;
        addedRecordList.Add(addedRecord);
    }
}

在CRM 2011中如何创建记录并将其关联到同一个ExecuteMultipleRequest

2 个答案: