从cmd中打开“打开”对话框 [英] Open the "open"-dialogbox from cmd

问题描述

我想知道是否可以从cmd打开打开-dialogbox，所以我可以使用它来选择正确的文件打开。
（我已经知道如何直接从cmd打开文件，但这不是我感兴趣的）

I want to know if I can open the "open"-dialogbox from the cmd, so I can use that to choose the right file to open. (I already know how to open the file directly from the cmd, but thats not what I interested in)

现在我正在做一个项目其中我使用一个程序（例如program.exe）。这个程序需要一个modelfile（例如modelfile.mod）。
我在CMD中执行程序，编写：program.exe modelfile.mod，它的工作。
我有很多modelfiles有不同的名字，但程序文件总是有相同的名称。而不是写在CMD每次我需要执行程序，我想创建一个批处理，我可以选择我喜欢执行的模块，然后它执行程序与选择的modfile作为输入。

At the moment I'm doing a project where i use a program (e.g. program.exe). This program needs a modelfile (e.g. modelfile.mod). I execute the program in the CMD by writing: program.exe modelfile.mod, and it work. I have many modelfiles with different names, but the program file always has the same name. Instead of writing in the CMD every time I need to execute the program I would like to create a batch where I can choose the modfile I like to execute whereafter it executes the program with the chosen modfile as input.

因此，我需要知道如何从cmd打开open-dialogbox，以及如何获取所选文件的名称以用于程序的执行。

Therefore I need to know how to open the "open"-dialogbox from the cmd, and also how to get the name of the chosen file for use in execution of the program.

推荐答案

如果你安装了PowerShell，你可以这样做：

If you've got PowerShell installed, you can do something like this:

@echo off
setlocal
set ps_cmd=powershell "Add-Type -AssemblyName System.windows.forms|Out-Null;$f=New-Object System.Windows.Forms.OpenFileDialog;$f.Filter='Model Files (*.mod)|*.mod|All files (*.*)|*.*';$f.showHelp=$true;$f.ShowDialog()|Out-Null;$f.FileName"

for /f "delims=" %%I in ('%ps_cmd%') do set "filename=%%I"

if defined filename (
    echo You chose %filename%
) else (
    echo You didn't choose squat!
)

goto :EOF

或者如果您想要分解powershell cmdlet以更容易维护：

Or if you want to break down the powershell cmdlets for easier maintenance:

@echo off
setlocal
set "ps=Add-Type -AssemblyName System.windows.forms | Out-Null;"
set "ps=%ps% $f=New-Object System.Windows.Forms.OpenFileDialog;"
set "ps=%ps% $f.Filter='Model Files (*.mod)|*.mod|All files (*.*)|*.*';"
set "ps=%ps% $f.showHelp=$true;"
set "ps=%ps% $f.ShowDialog() | Out-Null;"
set "ps=%ps% $f.FileName"

for /f "delims=" %%I in ('powershell "%ps%"') do set "filename=%%I"

if defined filename (
    echo You chose %filename%
) else (
    echo You didn't choose squat!
)

goto :EOF

（PowerShell命令无情地从只是修补博客。 ）有关其他属性，请参见 OpenFileDialog类文档您可以设置，例如标题和 InitialDirectory 。

(PowerShell command mercilessly leeched from the Just Tinkering Blog.) See the OpenFileDialog Class documentation for other properties you can set, such as Title and InitialDirectory.

这篇关于从cmd中打开“打开”对话框的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！