command-line - 仅按名称打开文件，无扩展名

Question

如何通过仅提供文件名而不提供扩展名来打开 .bat 中的任何类型的文件？我想让 Windows 决定要使用的应用程序。

例子：

%SystemRoot%\explorer.exe E:\SomeFolder\

%SystemRoot%\explorer.exe E:\SomeFolder\file1

Answer 1

使用 START 命令：

start "Any title" E:\SomeFolder\
start "Any title" E:\SomeFolder\file1

取自开始帮助：

If Command Extensions are enabled, external command invocation
through the command line or the START command changes as follows:

.

non-executable files may be invoked through their file association just
    by typing the name of the file as a command.  (e.g.  WORD.DOC would
    launch the application associated with the .DOC file extension).
    See the ASSOC and FTYPE commands for how to create these
    associations from within a command script.

.

When searching for an executable, if there is no match on any extension,
then looks to see if the name matches a directory name.  If it does, the
START command launches the Explorer on that path.  If done from the
command line, it is the equivalent to doing a CD /D to that path.

请注意，前面的描述暗示纯文件名还必须执行正确的应用程序，没有 START 命令。要获取具有给定名称的第一个文件：

for %%f in (name.*) do set "filename=%%f" & goto continue
:continue

...并执行它：

%filename%

PS - 请注意，您希望“让 Windows 决定要使用的应用程序”，但在您的示例中，您明确选择%SystemRoot%\explorer.exe了要使用的应用程序。所以？