如何显示“打开”未注册文件扩展名的对话框？ [英] How can I display the "open with" dialog for an unregistered file extension?

问题描述

我想让用户为当前未注册的文件扩展名选择一个关联（打开方式）。

I want to let the user select an association (open with) for an currently unregistered file extension.

目前我正在告诉API将文件打开使用 ShellExecute 并返回 ERROR_NO_ASSOCIATION 错误代码。

Currently I'm telling the API to open the file by using ShellExecute and it returns an ERROR_NO_ASSOCIATION error code.

有没有办法告诉API它应该让用户选择一个新的关联？

Is there a way to tell the API that it should let the user select a new association?

推荐答案

我使用

procedure ShellOpenAs(const AFileName: string; AHandle: HWND);
begin
  ShellExecute(AHandle, 'open', PChar('rundll32.exe'), PChar('shell32.dll,OpenAs_RunDLL ' + AFileName), nil, SW_SHOWNORMAL);
end;

修改（灵感来自于David的评论， http://stackoverflow.com/a/13229516/1431618 ）：
一个可以省略 ShellExecute 和 RunDll32 通过调用 OpenAs_RunDLL 直接：

Edit (inspired by David's comment and http://stackoverflow.com/a/13229516/1431618): One can omit ShellExecute and RunDll32 by calling OpenAs_RunDLL directly:

procedure OpenAs_RunDLL(hwnd: HWND; hinst: HINST; lpszCmdLine: LPSTR; nCmdShow: Integer); stdcall; external shell32;

procedure ShellOpenAs(AHandle: HWND; const AFileName: string);
begin
  OpenAs_RunDLL(AHandle, HInstance, PChar(AFileName), SW_SHOWNORMAL);
end;

还有一个 SHOpenWithDialog 。 （我觉得有趣的是，Microsoft写了一个RunDLL兼容的入口点，但直到Vista没有打扰提供一个常规的API函数。）

There is also SHOpenWithDialog on Windows Vista and later. (I find it interesting that Microsoft wrote a RunDLL compatible entry point but until Vista didn't bother to provide a regular API function.)

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