Question

我想在Safari中通过我的应用程序在Cocoa中打开一个URL。我正在使用：

[[NSWorkspace sharedWorkspace] openURL:[NSURL URLWithString: @"my url"]];

但问题是，如果我的默认浏览器不是Safari，那么URL会在其他浏览器中打开。但我希望我的URL只能在Safari中打开。请告诉解决方案。

谢谢：）

Answer 1

let url = URL(string:"https://twitter.com/intent/tweet")!
NSWorkspace.shared.open([url], 
                        withAppBundleIdentifier:"com.apple.Safari", 
                        options: [],
                        additionalEventParamDescriptor: nil,
                        launchIdentifiers: nil)

Answer 2

使用NSWorkspace's openURLs(_:withAppBundleIdentifier:options:additionalEventParamDescriptor:launchIdentifiers:)：

let url = NSURL(string:"http://example.com")!
let browserBundleIdentifier = "com.apple.Safari"

NSWorkspace.sharedWorkspace().openURLs([url],
               withAppBundleIdentifier:browserBundleIdentifier,
                               options:nil,
        additionalEventParamDescriptor:nil,
                     launchIdentifiers:nil)

Answer 3

使用safari中的scripting bridge在Safari中打开网址，您会找到一种在文件Safari.h中打开网址的方法。要了解有关使用Scripting bridge的更多信息，请参阅link并使用带有safari的脚本桥并生成Safari.h，请在此处参阅我的answer。

在Safari中打开URL的方法是：

NSDictionary *theProperties = [NSDictionary dictionaryWithObject:@"https://www.google.co.in/" forKey:@"URL"];
SafariDocument *doc = [[[sfApp classForScriptingClass:@"document"] alloc] initWithProperties:theProperties];
[[sfApp documents] addObject:doc];
[doc release];

Answer 4

你不能使用URL，你需要一个NSString

if(![[NSWorkspace sharedWorkspace] openFile:fullPath
                            withApplication:@"Safari.app"])
    [self postStatusMessage:@"unable to open file"];

Answer 5

要使用任何应用程序打开URL，您可以使用启动服务。您要查看的功能是LSOpenURLsWithRole;

修改
您必须将SystemConfiguration框架链接到项目才能使此方法可用。

Apple doc reference here

例如，如果您想使用safari打开http://www.google.com：

//the url CFURLRef url = (__bridge CFURLRef)[NSURL URLWithString:@"http://www.google.com"]; //the application NSString *fileString = @"/Applications/Safari.app/"; //create an FSRef of the application FSRef appFSURL; OSStatus stat2=FSPathMakeRef((const UInt8 *)[fileString UTF8String], &appFSURL, NULL); if (stat2<0) { NSLog(@"Something wrong: %d",stat2); } //create the application parameters structure LSApplicationParameters appParam; appParam.version = 0; //should always be zero appParam.flags = kLSLaunchDefaults; //use the default launch options appParam.application = &appFSURL; //pass in the reference of applications FSRef //More info on params below can be found in Launch Services reference appParam.argv = NULL; appParam.environment = NULL; appParam.asyncLaunchRefCon = NULL; appParam.initialEvent = NULL; //array of urls to be opened - in this case a single object array CFArrayRef array = (__bridge CFArrayRef)[NSArray arrayWithObject:(__bridge id)url]; //open the url with the application OSStatus stat = LSOpenURLsWithRole(array, kLSRolesAll, NULL, &appParam, NULL, 0); //kLSRolesAll - the role with which the applicaiton is to be opened (kLSRolesAll accepts any) if (stat<0) { NSLog(@"Something wrong: %d",stat); }

Answer 6

产生一个过程并执行打开-a＆＃34; Safari＆＃34; http://someurl.foo 也可以解决问题

通过cocoa app在safari中打开网址

6 个答案: