Open PDF using cell value in VBA-CodePudding

I'd like to open a specific PDF based on the file path in another sheet "Paths" cell C2, when I click a button (in another sheet) with a macro assigned.

I've just started learning vba last month so far I have this but I encounter an error -

Sub Sample()

    ActiveWorkbook.FollowHyperlink Sheets("Paths").Range("C2")

End Sub

The error is:

Run-time error '-214722104 (800401ea)': Cannot open the specified file.

Please help.

CodePudding user response：

I think there is no error on the script. please check below points.

Checking address of "sheets", "cells"
Checking value of "C2" (as expected, the most likely) "C2" value like this C:\Users\aaa\Desktop\KR01.pdf

please check it.

CodePudding user response：

I just made this roundabout way to open the pdf ...

''' Sub OpenCIS()

Sheets("Paths").Select
Range("C3").Select
Selection.Copy

Dim x As Variant
Dim Path As String

' Set the Path variable equal to the path of your program's installation
Path = Sheets("Paths").Range("C2")

x = Shell(Path, vbNormalFocus)

Application.Wait Now   0.00002
Application.SendKeys "^{o}", True

Application.Wait Now   0.00002
Application.SendKeys "^{v}", True
Application.SendKeys "{HOME}", True
Application.SendKeys "{DEL}", True 'deletes the ? that appears before the path
Application.SendKeys "{ENTER}", True

End Sub ''' I don't know why the path appears as ?C:\Users\Renee Arianne\Desktop\Desktop - Filled CIS\CIS_2016 v5 filled.pdf.