如何在“打开”对话框WinAPI C ++中设置文件的路径

Question

I wonder how can I set a file path in the Open Dialog, that was started by another process. I mean this dialog

I need to set path to file.

What is the best way to do this ?

Of course I can always send keys using SendInput but is there any more elegant way to do this ? For example to send WM_COMMAND or something else ?

Thanks

Answer 1

What you are asking can only be solved with hacks.

The best approach is to use MSAA / UI Automation as much as possible.

Assuming you know the HWND of the Open dialog you can walk all the child items looking for something with the role of ROLE_SYSTEM_COMBOBOX. Inside this item you need to find something with the role of ROLE_SYSTEM_TEXT that is editable (enabled and visible). Just looking for a ROLE_SYSTEM_TEXT item might find the search box in the top right corner.

If you can't find this object then it might be the older style dialog, restart the search and only look for ROLE_SYSTEM_TEXT.

It might be a good idea to verify this detection by comparing the HWNDs id (I will discuss these later).

When you find the item you can change the text with put_accValue() .

There is also a CDM_SETCONTROLTEXT message but you need to be inside the process to use it.

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If you don't know the HWND of the dialog or want to do this to all open dialogs on the system then you need a hook so you know when there is a new HWND candidate. Detecting these can be tricky.

The only option I know is to look for certain controls with certain ids. In particular there will be a ListBox with the id of lst1 and either a Edit box or a Combo box with a specific id.

You need to do a lot of testing to get this right. Remember that on Vista and later there are two dialog implementations ( IFileOpenDialog and GetOpenFileName ) and you need to test both. GetOpenFileName even has a old and a new mode (compatibility mode when old applications use templates and/or hooks).