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Question

Save Button that keeps me on the same form.

  • April 19, 2017
  • 0 replies
  • 23 views
I'm having an issue trying to create a Save Button that returns to the form the Button is on. The Form ID I've placed the Button on is 17. When I click the Save Button it goes to the Main Form. I'm using this:

"Save"

Is there a place in the code where I can put the Form ID#? I've looked at several articles and it doesn't answer this exact question. Although I may be missing something as I'm new to code.
      This topic has been closed for replies.

      • GauravSharma3
      • Quickbase Alumni
      • April 19, 2017
      Hi,

      Create a formula URL button and insert this code:

      URLRoot() & "db/" & [_DBID_DOCUMENTS] & "?a=API_GenAddRecordForm" & URLEncode ([Record ID#])& "&z=" & Rurl() & "&dfid=17"

      Replace [_DBID_DOCUMENTS] with your table DBID which you can find from the table properties.

      Do not forget to check the option "Save parent record automatically when a child record is created." under the advanced settings from your table properties.

      Please let me know if you need any further help.

      Thanks,

      Gaurav Sharma
      Dear bgstatdex

      Can you post your current code? It did not come though in your post. I think the pre button at the bottom of the box helps retain the original code.
      • BarryGulino
      • Author
      • Quickbase Alumni
      • April 19, 2017
      Update
      • BarryGulino
      • Author
      • Quickbase Alumni
      • April 19, 2017
      This is the code I'm using. "Update
      • BarryGulino
      • Author
      • Quickbase Alumni
      • April 20, 2017
      I was using my iPad to post and it obviously didn't paste.  Try again:

      "<a class='Vibrant Success' onclick=$('#RedirectURL').val(document.location.href.replace(/&rl=[a-z0-9]*/,''));$('[name=rl]').val('');$('#saveButton').trigger('click'); href='#'>Update</a>
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