Update list item at Any Site


#1

Hello , Thank you for the product!
I am trying to use some actions in my workflows, but your examples in documentation are not really helpful sometimes :frowning: You did a great job explaining how to use them here, could you please do the same extended explanation on how to use Update List Item at Any Site?
Thank you!


Update a list item in any site - finding a common ID
#2

Hello katy,
Thank you for your message.

To use the “Update List Item at Any Site” you need to first define a dictionary with the values you want to use to update the list item. You can learn more about working with dictionaries here.
Remember to use the internal name of each column and the correct data type also for this.

Then you can use the “Update List Item at Any Site” action along with the dictionary you’ve just created.

Please notice you’ll specify the item’s ID of the one you want to update. You can use a simple lookup to do it too.

You also have to specify the list name or URL but that’s quite simple.

And that’s basically it. If you need anything just let me know.

Best regards,
Andre Lima
Plumsail Team.


#3

Thank you, Andre for the detailed instructions. Sorry i haven’t replied before. It looks fairly simple, but when i am creating a workflow with the action (it’s actually Create new item in Any List, but i guess they have the same logic, right?) i am getting the error and the workflow terminated: Exception: 0x80070003
I’ve put a check and it shows that dictionary is created successfully, but this Create item action is not executed. Any ideas why?


#4

Hi Katy,

Do you create items on a different site? In this case, you need to specify SiteUrl and List Url parameters as shown in the screenshot below:


If it doesn’t help, please, send us a screenshot of your workflow.

Best Regards,
Anna Dorokhova
Plumsail team


#5

Thank you, Anna!
It works perfectly!!! That is now the complete instructions for such dummies like me :-))


#6

:slight_smile:

Please, feel free to contact us again if you need any other help.