在访问97中找到完整路径的目录部分(减去文件名)

问题描述

由于各种原因，我被困在Access 97中，只需要获取完整路径名的路径部分即可.

For various reasons, I'm stuck in Access 97 and need to get only the path part of a full pathname.

例如名称

c:\whatever dir\another dir\stuff.mdb

应该成为

c:\whatever dir\another dir\

此网站对如何操作有一些建议: http://www.ammara.com/access_image_faq/parse_path_filename.html

This site has some suggestions on how to do it:http://www.ammara.com/access_image_faq/parse_path_filename.html

但是它们看起来很丑陋.必须有更好的方法，对吧?

But they seem rather hideous. There must be a better way, right?

推荐答案

就是这样.没有魔术内置功能...

That's about it. There is no magic built-in function...

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