Multiple choice in file import

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Posted on by Giuseppe Caradonna

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Hi all,

I need to implement a dialog that provides, a user's, the possibility to choose more than one file, in order to run an unique import procedure.

How can I get what I want?

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Anton Venter 20,346 Super User 2025 Season 2 on at

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Hi Giuseppe,

How do you determine from the file which import to run?

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Check out this link, it should be what you are looking for

http://dynamicsuser.net/forums/p/31783/171187.aspx

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Try:

WinAPI::getOpenFileName(0, ['CSV', '*.CSV'], @'C:\', 'Select a file', '', '', 512);

512 flag allows you to select more than one file.

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Douglas Noel 3,905 on at

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Hy,

I don't know if this will really work in all cases. The File names are #0 delimied each and the last hat two #0 entries. (historical c/c++string arrays). I'm not sure what the kernel will transfer from the bindata buffer to the string? But you can try. Maybe it works.

But why not use .NET class

see the following entry (sorry for the 'google')

groups.google.com/forum

regards

Douglas

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Giuseppe Caradonna 1,421 on at

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Sorry M Z, but the result of the getOpenFileName() method is a string that contains the parent folder of files that I have selected. I need a list of the selected files...

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Douglas Noel 3,905 on at

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Hi Giuseppe,

I thought as much, therefore my apprehension. So give the .NET class a chance (the link is a full link, don't know why it is abbreviated)

think that can be the better way. But haven't tested.

Please give a short statement if the multislect using the .NET class can be used.

regards

Douglas

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Anton Venter 20,346 Super User 2025 Season 2 on at

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Hi Giuseppe, here is a job with code that I believe is what you are looking for.

static void AV_MultiFileSelectDlgTest(Args _args) { int idx; int cnt; boolean result; System.String[] files; Microsoft.Win32.OpenFileDialog dlg = new Microsoft.Win32.OpenFileDialog(); dlg.set_Multiselect(true); dlg.set_DefaultExt(".txt"); dlg.set_Filter("Text documents|*.txt|All files|*.*"); result = dlg.ShowDialog(); if (result) { files = dlg.get_FileNames(); cnt = files.get_Count(); for (idx = 0; idx <= cnt; idx++) { info(files.get_Item(idx)); } } }

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