有没有人尝试将 VBA(或 VB6)函数(通过 AddressOf ?)传递给 SQLite 创建 UDF 函数(http://www.sqlite.org/c3ref/create_function.html).

VBA 将如何处理生成的回调参数?

要调用的函数将具有以下签名......

空白 (xFunc)(sqlite3_context,int,sqlite3_value**)

有帮助吗?

解决方案

不幸的是,您不能直接使用 VB6/VBA 函数作为回调,因为 VB6 只生成 stdcall 函数而不是 cdecl SQLite 期望的功能。

您将需要编写一个 C dll 来代理来回调用或重新编译 SQLite 以支持您自己的自定义扩展。

重新编译 dll 将函数导出为 stdcall, ,您可以使用以下代码注册一个函数:

'Create Function
Public Declare Function sqlite3_create_function Lib "SQLiteVB.dll" (ByVal db As Long, ByVal zFunctionName As String, ByVal nArg As Long, ByVal eTextRep As Long, ByVal pApp As Long, ByVal xFunc As Long, ByVal xStep As Long, ByVal xFinal As Long) As Long

'Gets a value
Public Declare Function sqlite3_value_type Lib "SQLiteVB.dll" (ByVal arg As Long) As SQLiteDataTypes    'Gets the type
Public Declare Function sqlite3_value_text_bstr Lib "SQLiteVB.dll" (ByVal arg As Long) As String        'Gets as String
Public Declare Function sqlite3_value_int Lib "SQLiteVB.dll" (ByVal arg As Long) As Long                'Gets as Long

'Sets the Function Result
Public Declare Sub sqlite3_result_int Lib "SQLiteVB.dll" (ByVal context As Long, ByVal value As Long)
Public Declare Sub sqlite3_result_error_code Lib "SQLiteVB.dll" (ByVal context As Long, ByVal value As Long)

Public Declare Sub CopyMemory Lib "kernel32" Alias "RtlMoveMemory" (dest As Any, source As Any, ByVal bytes As Long)

Public Property Get ArgValue(ByVal argv As Long, ByVal index As Long) As Long
    CopyMemory ArgValue, ByVal (argv + index * 4), 4
End Property

Public Sub FirstCharCallback(ByVal context As Long, ByVal argc As Long, ByVal argv As Long)
    Dim arg1 As String
    If argc >= 1 Then
        arg1 = sqlite3_value_text_bstr(ArgValue(argv, 0))
        sqlite3_result_int context, AscW(arg1)
    Else
        sqlite3_result_error_code context, 666
    End If
End Sub

Public Sub RegisterFirstChar(ByVal db As Long)
    sqlite3_create_function db, "FirstChar", 1, 0, 0, AddressOf FirstCharCallback, 0, 0
    'Example query: SELECT FirstChar(field) FROM Table
End Sub
许可以下: CC-BY-SA归因
不隶属于 StackOverflow
scroll top