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Guillermo360i
Level: Powered On

Property toRecipients in payload has a value that does not match schema

Anyone else encountering this problem?!

1 ACCEPTED SOLUTION

Accepted Solutions
Community Support Team
Community Support Team

Re: Property toRecipients in payload has a value that does not match schema

Hi @Guillermo360i / @pmcopeland ,

 

Please try this method, using an array variable instead of a string variable.

50.PNG51.PNG

 

Best Regards,

Community Support Team _ Barry
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

View solution in original post

4 REPLIES 4
Community Support Team
Community Support Team

Re: Property toRecipients in payload has a value that does not match schema

Hi @Guillermo360i ,

 

Could you provide more details about this issue?

Please provide more information to help us analyze this issue.

 

Best Regards,

Community Support Team _ Barry
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
pmcopeland
Level: Powered On

Re: Property toRecipients in payload has a value that does not match schema

I'm also encountering this. 

 

For my flow, I look up a list of emails, and using For Each and the Append to String Variable action, I build a string of emails delimited by a semicolon.  

 

email@domain.com; email2@domain.com; email3@domain.com; etc

 

Recently whenever this string is used for the To field of Send an Email V2 I encounter this error.

pmcopeland
Level: Powered On

Re: Property toRecipients in payload has a value that does not match schema

Figured it out just as I posted this. I'm not sure what changed but Email V2 doesn't support spaces between emails anymore. 

 

The string needs to be: email@domain.com;email2@domain.com;email3@domain.com; 

Community Support Team
Community Support Team

Re: Property toRecipients in payload has a value that does not match schema

Hi @Guillermo360i / @pmcopeland ,

 

Please try this method, using an array variable instead of a string variable.

50.PNG51.PNG

 

Best Regards,

Community Support Team _ Barry
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

View solution in original post

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