new line issue

[kausarkhanlp504-2]

on line 578.
you are converting -> \n to vbcrlf, which charactor code is 13 -> carriage return key. not \n key
the correct constant for -> \n is -> vblf

vbcr = vbcrlf = vbnewline -> charcode is 13, but vblf charcode is 10

old:

Private Function json_ParseString(json_String As String, ByRef json_Index As Long, ByRef json_Length As Long) As String
    Do While json_Index > 0 And json_Index <= json_Length
        json_Char = VBA.Mid$(json_String, json_Index, 1)

        Select Case json_Char
        Case "\"
            ' Escaped string, \\, or \/
            json_Index = json_Index + 1
            json_Char = VBA.Mid$(json_String, json_Index, 1)

            Select Case json_Char
            Case """", "\", "/", "'"
            Case "b"
            Case "f"
            Case "n"
                json_BufferAppend json_Buffer, vbCrLf, json_BufferPosition, json_BufferLength
                json_Index = json_Index + 1
            Case "r"
            Case "t"
            Case "u"
            End Select
        Case json_Quote
        Case Else
        End Select
    Loop
End Function

new:

```vba
Private Function json_ParseString(json_String As String, ByRef json_Index As Long, ByRef json_Length As Long) As String
    Do While json_Index > 0 And json_Index <= json_Length
        json_Char = VBA.Mid$(json_String, json_Index, 1)

        Select Case json_Char
        Case "\"
            ' Escaped string, \\, or \/
            json_Index = json_Index + 1
            json_Char = VBA.Mid$(json_String, json_Index, 1)

            Select Case json_Char
            Case """", "\", "/", "'"
            Case "b"
            Case "f"
            Case "n"
                json_BufferAppend json_Buffer, vbLf, json_BufferPosition, json_BufferLength
                json_Index = json_Index + 1
            Case "r"
            Case "t"
            Case "u"
            End Select
        Case json_Quote
        Case Else
        End Select
    Loop
End Function

[Nick-vanGemeren]

vbcr = vbcrlf = vbnewline

These are not the same - see here

The ECMA standard defines the escape sequence '\n' as representing a line feed (Chr(10)). So you are correct that parsing should convert '\n' to line feed (as the inverse of the conversion).

There may be earlier Issues which discuss why vbCrLf (which does contain a line feed) is used. My guess is that it relates to how the multi-line value will be used: displayed, stored in a database etc. I'm tempted to say it should be a settable option.

Note that this repository has been abandoned by @timhall. You are free to modify your local copy.