Error 53 when opening CSV file on mac - vba

Question

When I try and open a CSV file, I get:
Error 53: File not found
I get the error on the 4th line, Open FilePath For Input As #1
What am I doing wrong?
It is my first time opening a CSV, please be indulgent with my code.
Sub opentextfile()
Dim FilePath As String
FilePath = "/Users/christinekelly/Desktop/authors.csv"
file1 = FreeFile
Open FilePath For Input As #file1
row_number = 0
Do Until EOF(1)
Line Input #file1, LineFromFile
LineItems = Split(LineFromFile, ",")
ActiveCell.Offset(row_number, 0).Value = LineItems(2)
ActiveCell.Offset(row_number, 1).Value = LineItems(1)
ActiveCell.Offset(row_number, 2).Value = LineItems(0)
Number = row_number + 1
Loop
Close #file1
End Sub

Answer 1

So from looking at #Rebekare's answer to this question this is what worked for me.
I went to the file in question test.csv and opened the immediate window and typed ?ThisWorkbook.Path and got HDD:Users:USER:Desktop.
I then used the suggested concatenatation of this path with Application.PathSeparator & filename i.e.
FilePath = "HDD:Users:USER:Desktop" & Application.PathSeparator & "test.csv"
This is a useful approach as you get the actual file path syntax and then yield the path separator decision to the Application.
Following on from #Mat'sMug suggestion, I found this, which opens the file dialog, and you select the file you want the full path of and it is returned via the message box.
Sub PathofFile()
OpenFile = Application.GetOpenFilename()
MsgBox OpenFile
End Sub
Within that same link was a suggestion for using the Dir function to test if the filepath is valid. If valid you get the filename back, if not you get an error which you can use to determine your next action e.g. the following returns "test.csv" if it exists at that filepath.
MsgBox Dir("HDD:Users:USER:Desktop:test.csv")
If you are doing other operations with the file you might want to add a test to to see if the file is open first, using Microsoft's IsFileOpen function. An example as follows:
Sub Test
If Not IsFileOpen(FilePath) Then
Set wb = Workbooks.Open(FilePath)
End If
End Sub
Function IsFileOpen(filename As String)
Dim filenum As Integer, errnum As Integer
On Error Resume Next ' Turn error checking off.
filenum = FreeFile() ' Get a free file number.
' Attempt to open the file and lock it.
Open filename For Input Lock Read As #filenum
Close filenum ' Close the file.
errnum = Err ' Save the error number that occurred.
On Error GoTo 0 ' Turn error checking back on.
' Check to see which error occurred.
Select Case errnum
' No error occurred.
' File is NOT already open by another user.
Case 0
IsFileOpen = False
' Error number for "Permission Denied."
' File is already opened by another user.
Case 70
IsFileOpen = True
' Another error occurred.
Case Else
Error errnum
End Select
End Function