Hi,
I have very little knowledge of how to write code for excel and typically use google for help. Here's my question: using a calendar I created in Excel, how can I only display two rows about 14 days of a calendar month based on the current date? So the reader would see the the row (week) containing today's date plus 14 days forward. Everything else would be hidden. I have the calendar months vertically stacked. Any advice you can offer is greatly appreciated.
Regards,
Saabra
Depending on how your calendar is formatted a filter could do the job or you'd need to add some code.
Can you upload this calendar?
Cheers!
Tsjallie
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If you think design is an expensive waste of time, try doing without ...
Hi Jsjallie,
I've attached the census with calendar. Census Report.xlsm Your help is greatly appreciated.
Regards,
Saabra
Hi Saabra,
here's the workbook I added filtering of the calendars to.
I made some notes on the Census sheet of the changes I made.
And left some work for you to do of course
Wow, thank you. I'm not sure I understand it yet and will take some time today to see what you did. Just an FYI that the following screens popped up upon opening: Two_Screens.docx. I'll be in touch.
Saabra
That has most probably to do with Office executing the Workbook_Open procedure while still in blocked mode for security.
Once you saved the workbook or made a trusted document this error will no longer occur.
Okay, I'll have more time to look this over tomorrow and get back to you.
Thanks again,
Barb
Tsjallie,
It appears that the ISFORMULA function used in column A is not compatible with Excel 2010. Is there another function we can use compatible with Excel 2010?
Saabra
Oops, overlooked that you're using Office 2010.
Yes, the ISFORMULA-function is available since Office 2013.
Here's the workbook again.
Added a user defined function ISFORMULA and adjusted the formula to accommodate the new function.
Wow, this looks great. I'll get back to you after I make the adjustments. Thanks a million.
Saabra
Tsjallie,
Can you resend the adjusted attachment in unprotected mode? I can't enable editing on the spreadsheet.
Saabra
Tsjallie,
See attached screen shot upon opening the revised census report with filter. Screen shot upon opening of revised calendar.docx. Please advise.
Thanks,
Saabra
As I see in the screenshot that the workbook is still in Protected View. That's done by Windows for security reasons, because the workbook comes from the Internet.
The workbook or sheet itself is not protected.
When the error messages start to pop up click End to stop processing the user defined function.
Then quit protected view by clicking Enable Editing.
After that save the workbook.
If you're asked if you want to make it a trusted document, choose Yes.
BTW, the IsFormula-function is called when the workbook is recalculated.
And I think recalculating is triggered by saving the workbook in the Workbook_Open() procedure. Why do you want to save it on opening?
I'm uploading a version of the workbook with this line disabled for you to see if it then opens without errors.
Thanks for the insight. I wasn't trying to save it upon opening. I was trying enable editing.
Barb
I was referring to this procedure in ThisWorkbook:
In the latest version I uploaded I disabled that line, so it should open without errors now.
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Tsjallie,
I've attached the first 5 screen shots upon opening your spreadsheet. Five Screens.docx It's not working, and I'm sorry I don't understand why. Do the screen shots shed anymore light on the problem?
Thanks,
Saabra
Can you check the references in the Extra menu of the VB Editor.
Do any of the checked references show 'Missing'?
SaabraRefs.JPG
Tsjallie,
I just want to thank you for your persistent help on this project. Your spreadsheet works like a charm. Your efforts will help expedite the care of hundreds of people suffering with mental illness. Thank you again.
Saabra
Good to hear that it's working now.
Was it something with a reference?
Glad I could help. Keep up the good work!Your efforts will help expedite the care of hundreds of people suffering with mental illness.
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