SFDC to Marketo Sync

[katy1]

Hello,

I had a question about the SFDC and Marketo sync and how it handles duplicates. What would happen if person A is created in Marketo and is never synced with SFDC. Some time later,  person A is created in SFDC as the result of an import and is included in the normal SFDC to marketo sync. Would there be a match found between the two person A records? In this scenario they would be matching on email address. 

[Christiane_Rode · Accepted solution]

In this situation, provided you're using the out-of-the-box setup, you will have two records. Your CRM will push the person record over and will not merge a duplicate.

Now, with the reverse, if the lead is pushed from your CRM, and then someone fills out a form, Marketo will search for the contact and should not create a duplicate record.

Here's some additional information on working with duplicates.

[Darshil_Shah1]

Agreed, in the case there are non-connected records in the systems, those won't get synced/connected on their own - rather given that the sync user has access to the person record in the CRM, a new record will get created in Marketo from CRM. In this case, you'd have to merge the existing Marketo-only and the newly synced record from the CRM. FYR, Marketo only records don't get created in CRM automatically, it needs to be explicitly/implicitly synced to the CRM. However, records in the CRM that the sync user can see get automatically created using the native sync functionality.

[Katja_Keesom]

Entirely true. The basic issue to understand here is that there two different unique keys in play:

• When a record is created in Marketo, the email address is used as the unique key. A list upload or form fill will therefore update the existing record with the same email address.
• When a record is created in SFDC (or MSD for that matter) and then synced to Marketo, the unique key used is the CRM GUID of the record.

As a pre-existing Marketo-only record does not have a CRM GUID, the native sync will not recognize the first Person A as being the same record and indeed create a new one.