Hello,
I have a form in which i'm searching for UserID and storing that into frm_user_id for a view which will filter based on that parameter.
I would like when i run the search to present in the view the value of the display_name of the userID. i've tried to run a wordpress shortcode which display the name when you present the userid but is not working in the if_get_param statement.
Example :
[if_get_param frm_user_id="userid" type="not_equal" value=""] Filter : [get param=frm_user_id] [su_user user_id="[get param=frm_user_id]"] [/if_get_param] [su_user user_id="[get param=frm_user_id]"]
The su_user is a plugin shortcode which is showing the Display name of the user. The shortcode is executed normaly after the statement but in the statement is not executed.
A hit please?
Thanks,
Adrian
You don't need to use su_user. Formidable shows the display name by adding a "show" parameter to the field's shortcode. The shortcode you use is [frm_user_id show="display_name"]. See this: https://formidableforms.com/knowledgebase/user-id/#kb-display-user-id-fields.
Thanks for the suggestion, I'll try.
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