在VBA中打开方法 [英] Workbooks.Open Method in VBA

问题描述

我的vba脚本在 myMacro.xls Workbooks.Open 方法工作正常如下，

My vba script in myMacro.xls Workbooks.Open Method work well as below,

Workbooks.Open Filename:="D:\ExcelMacroProj\myTest.xls", ReadOnly:=True

但是，当我尝试将文件名值更改为新路径时，如下所示，工作显示运行时错误1004 。

But when I try to change the Filename value to a new path as below, but all my practices didn't work. Show Run time error 1004.

Workbooks.Open Filename:="myTest.xls", ReadOnly:=True
or
Workbooks.Open Filename:="./myTest.xls", ReadOnly:=True
or
Workbooks.Open Filename:=".\myTest.xls", ReadOnly:=True

实际上 myMacro.xls 和 myTest.xls 被放置在同一个文件夹中。这就是为什么我要更改为灵活的文件夹目录。

Actually myMacro.xls and myTest.xls was placed in the same folder. That's why I want to change to a flexible folder directory.

我该如何解决这个问题？感谢您的阅读和回复。

how could I fix this issue? Appreciated for your read and reply.

推荐答案

您可以尝试使用 ThisWorkbook.Path 制作绝对路径。它返回运行宏的工作簿的文件夹路径（不包括文件名）。这样的东西应该可以工作：

You might try using ThisWorkbook.Path to make an absolute path. It returns the folder path of the workbook running the macro (excluding the filename). Something like this should work:

Workbooks.Open Filename:=ThisWorkbook.Path & "\myTest.xls", ReadOnly:=True

确保包含反斜杠，因为工作簿路径不以一个结尾。

Make sure to include a backslash, since the workbook path doesn't end with one.

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