将部分屏幕截图复制到Pasteboard

时间:2012-09-20 02:07:55

标签: xcode uipasteboard

因此,将部分屏幕复制到粘贴板的代码可以正常工作,因为它已成功将其复制到我的相册中。但是,我希望能够将部分屏幕截图粘贴到新的SMS消息中。我知道它必须手动完成(长时间保持消息和粘贴),但它要么没有粘贴,要么没有粘贴选项(因为它将它保存为字符串)。代码的中间部分是我正在努力的部分。任何帮助都会很棒。我已经将forPasteboardType更改为“image”,但这也无效。

    //Capture part of Screen Shot
        UIGraphicsBeginImageContext(self.view.bounds.size);
        CGContextRef c = UIGraphicsGetCurrentContext();
        CGContextTranslateCTM(c, 0, 98);    //
        [self.view.layer renderInContext:c];
        UIImage *viewImage = UIGraphicsGetImageFromCurrentImageContext();
        UIGraphicsEndImageContext();

    //Send Screenshot to Pasteboard    
    UIPasteboard *pasteBoard = [UIPasteboard pasteboardWithName:UIPasteboardNameGeneral create:YES];
    pasteBoard.persistent = YES;
    NSData *data = UIImagePNGRepresentation(viewImage);
    [pasteBoard setData:data forPasteboardType:(NSString *)kUTTypePNG];     

    /////// Open SMS
    MFMessageComposeViewController *controller = [[[MFMessageComposeViewController alloc] init] autorelease];
    if([MFMessageComposeViewController canSendText])
    {
        controller.body = @"Hello from me, paste image here -->";
        controller.recipients = [NSArray arrayWithObjects:@"123456789", nil];
        controller.messageComposeDelegate = self;
        [self presentModalViewController:controller animated:YES];
    }
    ////// End SMS
}

1 个答案:

答案 0 :(得分:1)

//Capture part of Screen Shot
UIGraphicsBeginImageContext(self.view.bounds.size);
CGContextRef c = UIGraphicsGetCurrentContext();
CGContextTranslateCTM(c, 0, 98);    //
[self.view.layer renderInContext:c];
UIImage *viewImage = UIGraphicsGetImageFromCurrentImageContext();
UIGraphicsEndImageContext();

//Send Screenshot to Pasteboard    
UIPasteboard *pasteBoard = [UIPasteboard pasteboardWithName:UIPasteboardNameGeneral create:YES];
pasteBoard.persistent = YES;
NSData *data = UIImagePNGRepresentation(viewImage);
[pasteBoard setData:data forPasteboardType:(NSString *)kUTTypePNG];     

NSString *stringURL = @"sms:";
NSURL *url = [NSURL URLWithString:stringURL];
[[UIApplication sharedApplication] openURL:url];