Ok step 1 first: I removed your line of code "Call Open_Folder"
So My code looks the following for the hyperlink button:
I tested it and it works, Column B gets the hyperlink button with the text to display using the column letter. (See attachment: Excel-VBA button hyperlink Test.xlsm)
Step 2: What I need it to do is also create a button for column C and column D using the same referal from column A.
But the hyperlink output and TextToDisplay needs to be different for each of the columns.
Something like this:
Column B: https://google.com=100" & x, TextToDisplay:=B
Column C: https://yahoo.com=100" & x, TextToDisplay:=X
Column D: https://excelforum.com=100" & x, TextToDisplay:=E
Once this step is set and working I will move on to my next point is implementing your recommendation of adding the "Call Open_Folder" to this Worksheet_SelectionChange procedure
Which eventually has to implement the following code which makes a button out of the column A data and searches and opens a folder on the V: disk.
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