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  • Jan Højriis Dragsbæk 12 posts 63 karma points
    Jan 04, 2013 @ 11:19
    Jan Højriis Dragsbæk
    0

    Using dictionary items on the submit button

    I am trying to set up a contour formula and have successfully gotten everything except the submit button attached to the dictionary. My colleague told me I had to use either #submit or #Submit, but none of these seem to work.

    Google tells me that it works in the earlier versions of contour, but what about 3.0.5? How do i do this?

  • Comment author was deleted

    Jan 04, 2013 @ 12:00

    Are you using the razor macro?

  • Jan Højriis Dragsbæk 12 posts 63 karma points
    Jan 04, 2013 @ 12:01
    Jan Højriis Dragsbæk
    0

    Yes, does the razor macro disable that functionality?

  • Comment author was deleted

    Jan 04, 2013 @ 12:02

    Might have forgot to add that, will check

  • Comment author was deleted

    Jan 04, 2013 @ 12:35

    Oops look like the button value is hardcoded, will update in next maintenance release

    For a quick solution simply open the file \Umbraco\plugins\umbracoContour\Views\Form.cshtml and update to what you want it to be

  • Wessel Spoelder 3 posts 26 karma points
    Jan 09, 2013 @ 22:05
    Wessel Spoelder
    0

    Wow, I was pulling my hair out over this... Thanks for your reply Tim.

  • Comment author was deleted

    Jan 10, 2013 @ 08:41

    Updated in the upcoming 3.0.6 release http://issues.umbraco.org/issue/CON-199

  • Rodske 74 posts 104 karma points
    Feb 05, 2013 @ 10:53
    Rodske
    0

    Hey Tim,

    Any chance of a quick code snippet of how to reference the Umbraco (not Umbraco.Form) currentPage model from within the \Umbraco\plugins\umbracoContour\Views\Form.cshtml ? Any use of @Umbraco.Field("pageName") errors. 

    Just FYI, I'm using MVC 4, Umbraco 6.0.0 & Contour 3.0.6.

    Cheers!

  • David Peck 690 posts 1896 karma points c-trib
    Jan 06, 2014 @ 23:34
    David Peck
    0

    I reckon this should do it for anyone in the future:

    @umbraco.library.GetDictionaryItem("YourDictionaryAlias")

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