HubSpot Ideas

mleblanc

Social Media Reporting: Filter By Brand

A company marketing team works across multiple social media brands, each of which has a different audience, for example, one is an employer brand where the audience is employees and prospective recruits and the other is a sales/customer brand. 

 

Currently, you can filter social media reporting by channel (Twitter vs. LinkedIn) but this team needs to report across channels and filter by brand. Please add functionality to select by social media account to filter and then add the filtered report to the custom dashboard. 

4 Replies
rwong
HubSpot Employee

Upvoting this thread on behalf of a customer who's managing multi-language content in his portal.

 

Context: he has multiple language accounts for each social network (e.g. FB pages for English, Norwegian, Finnish) connected to HubSpot. It'd be great if the social Reports tab has more nuanced options in the dropdown filter. It could look like:

 

  • All networks
    • Facebook accounts
      • [Brand name] EN
      • [Brand name] NO
      • [Brand name] FI
    • LinkedIn accounts
      • [Brand name] EN
      • [Brand name] FI

This would be useful not just on the Reports tab but also in the Publishing and Monitoring tabs 🙂

CHolzschuh
Member

Thumbs up, I have the same issue, that would be really helpful and probably is not too difficult to implement. So my case is like: 

 

Brand 1:  Social media report for brand 1 on Linkedin, Twitter, Facebook, Instagram 

Brand 2:  Social media report for brand 2 on Linkedin, Twitter

danmoyle
Key Advisor | Diamond Partner

Yes, please. A client I'm working with has 4 brands across the multiple social platforms, and cannot report on posts, audience, etc. in their dashboards. It's only all together. We can measure traffic through custom analytics views, but we can't through the social tools. Help! 🙂 

Tri-Nat_Tree
Member

Also chiming in. We need to be able to segment and share social reports by brand. I'm baffled why this isn't a default feature in Hubspot.