cmd 命令转到打开批处理文件的子目录
所以这个标题很混乱。让我告诉你我的需求： 我想打开一个位于 c:\documents\test\ 的批处理文件，
在此批处理文件中我想收集 c:\documents\test\data 中的名称并转换它到一个文本文件。
现在，执行此操作的简单方法是：
CD c:\documents\test\data
dir/b/o:n > names.txt
但我的问题是我想移动文件夹“\test”，并且始终能够调用“\data”文件中的名称。
因此，这就是我真正需要的：
CD [variable that says current location] + "\data"
但我不知道该怎么做。请解释你的答案谢谢。
So that title was confusing. Let me tell you my needs:
I'm wanting to open a batch file that is located in c:\documents\test\
in this batch file i'm wanting to collect the names in c:\documents\test\data and convert it to a text file.
Now the easy way to do this would be:
CD c:\documents\test\data
dir/b/o:n > names.txt
but my issue is that i want to move the folder "\test" around and always be able to call on the names in the "\data" file.
Therefore this is what i really need:
CD [variable that says current location] + "\data"
but i dont know how to do this. Please explain your answers thanks.
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%~dp0是批处理文件的位置（运行FOR /?了解更多信息）%~dp0is the location of the batch file (RunFOR /?for more info)这 。字符代表当前工作目录...是否
满足您的需要？
编辑：
所以我看到没有人真正回答你的问题。根据您当前的编辑，您需要定义一个 Windows 环境变量来保存数据文件夹所在位置的路径，并且当您移动文件夹时，该过程的一部分需要更新环境变量，以便批处理脚本可以正确找到它。
因此，如果您将数据文件夹移动到 C:\archive\test\data 您将需要一些过程来调用命令行，
然后在批处理脚本中您将执行以下操作：
the . character represents the current working directory... does
do what you need?
edit:
So I see no one really answered your issue. Based on your current edit you would want to define a windows environment variable that held the path to wherever your data folder is, and when you move the folder part of that process would need to update the environment variable so your batch script can find it appropriately.
so if you moved your data folder to C:\archive\test\data you would need some process to call the command line with
then in your batch script you would do something like: