Question

在共享扩展程序中，我设法使用以下代码获取Safari页面的URL：

NSExtensionItem *item = self.extensionContext.inputItems.firstObject;
NSItemProvider *itemProvider = item.attachments.firstObject;
if ([itemProvider hasItemConformingToTypeIdentifier:(NSString *)kUTTypeURL]){
    [itemProvider loadItemForTypeIdentifier:(NSString *)kUTTypeURL
                                    options:nil
                          completionHandler:^(NSURL *url, NSError *error){
                              NSLog(@"%@", url.absoluteString);
                          }];
}

我还能获得该页面的HTML吗？

Answer 1

检查以下代码，

[itemProvider loadItemForTypeIdentifier: (NSString *) kUTTypePropertyList
                                options: 0
                      completionHandler: ^(id<NSSecureCoding> item, NSError *error) {
                          if (item != nil) {
                              NSDictionary *resultDict = (NSDictionary *) item;
                              NSString *jsString = resultDict[NSExtensionJavaScriptPreprocessingResultsKey][@"content"];
                          }
                      }];

Creating an iOS App Extension to perform custom actions with Safari content - swiftiostutorials.com

从Safari共享扩展获取HTML？

1 个答案: