我一直试图让一个窗口出现,要求该人选择一个文件,我最终做到了。问题是,Xcode 抱怨我使用的方法已被弃用。我查看了类参考,但从 Mac OS 10.6 开始,“运行面板”部分下的所有内容都已被弃用。我现在应该使用不同的课程吗?
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11026 次
3 回答
30
在 10.6 中,对此类进行了一些更改。好处之一是现在有一个基于块的 API。
这是有关如何使用它的代码片段:
NSOpenPanel *panel = [[NSOpenPanel openPanel] retain];
// Configure your panel the way you want it
[panel setCanChooseFiles:YES];
[panel setCanChooseDirectories:NO];
[panel setAllowsMultipleSelection:YES];
[panel setAllowedFileTypes:[NSArray arrayWithObject:@"txt"]];
[panel beginWithCompletionHandler:^(NSInteger result){
if (result == NSFileHandlingPanelOKButton) {
for (NSURL *fileURL in [panel URLs]) {
// Do what you want with fileURL
// ...
}
}
[panel release];
}];
于 2011-12-27T18:07:34.877 回答
27
据我所知,您可以使用runModal
如下所示的方法:
NSOpenPanel *openPanel = [[NSOpenPanel alloc] init];
if ([openPanel runModal] == NSOKButton)
{
NSString *selectedFileName = [openPanel filename];
}
于 2011-10-08T00:01:25.587 回答
4
看到六年后我如何发现这个问题很有用,并且由于没有快速的答案,这里有一个快速的解决方案。
您会找到两个样本,一个作为独立窗口,另一个作为工作表。
斯威夫特 3.0
func selectIcon() {
// create panel
let panel = NSOpenPanel()
// configure as desired
panel.canChooseFiles = true
panel.canChooseDirectories = false
panel.allowsMultipleSelection = false
panel.allowedFileTypes = ["png"]
// *** ONLY USE ONE OF THE FOLLOWING OPTIONS, NOT BOTH ***
// ********************** OPTION 1 ***********************
// use this if you want a selection window to display that is
// displayed as a separate stand alone window
panel.begin { [weak self] (result) in
guard result == NSFileHandlingPanelOKButton, panel.urls.isEmpty == false, let url = panel.urls.first else {
return
}
let image = NSImage.init(contentsOf: url)
DispatchQueue.main.async {
self?.iconImageView.image = image
}
}
// ********************** OPTION 2 ***********************
// use this if you want a sheet style view that displays sliding
// down from your apps window
panel.beginSheetModal(for: self.view.window!) { [weak self] (result) in
guard result == NSFileHandlingPanelOKButton, panel.urls.isEmpty == false, let url = panel.urls.first else {
return
}
let image = NSImage.init(contentsOf: url)
DispatchQueue.main.async {
self?.iconImageView.image = image
}
}
}
于 2017-06-30T21:02:17.857 回答