Question

在我的应用程序中，我有一个webview，并希望打开一个单独的视图控制器中的网页中加载的图像，这很好，我需要做的就是获取图像来源的URL＆amp;将它加载到不同的视图控制器中，我可以这样做。

以下是我用来获取图片来源的网址的代码。

-(BOOL)gestureRecognizer:(UIGestureRecognizer *)gestureRecognizer shouldReceiveTouch:(UITouch *)touch {
    if ([gestureRecognizer isKindOfClass:[UITapGestureRecognizer class]]) {
        CGPoint touchPoint = [touch locationInView:self.view];

        NSString *imageSRC = [NSString stringWithFormat:@"document.elementFromPoint(%f, %f).src", touchPoint.x, touchPoint.y];
        NSString *srcOfImage = [webView stringByEvaluatingJavaScriptFromString:imageSRC];
        NSLog(@"src:%@",srcOfImage);
    }
    return YES;
}

现在，我的问题是，有时，当图像可能有链接（即标记）时，webview将在我的单独视图控制器中打开图像时加载链接。我想做的是，停止webview打开链接（只有图像中的链接）（如果存在）。有关我如何实现这一目标的任何指示？

Answer 1

最后想出来了，答案在UIWebViewDelegate中撒谎。对于任何有兴趣的人，这是我如何解决它..

bool isImage;

-(BOOL)gestureRecognizer:(UIGestureRecognizer *)gestureRecognizer shouldReceiveTouch:(UITouch *)touch {
    if ([gestureRecognizer isKindOfClass:[UITapGestureRecognizer class]]) {
        CGPoint touchPoint = [touch locationInView:self.view];

        NSString *imageSRC = [NSString stringWithFormat:@"document.elementFromPoint(%f, %f).src", touchPoint.x, touchPoint.y];
        NSString *srcOfImage = [webView stringByEvaluatingJavaScriptFromString:imageSRC];
        NSLog(@"src:%@",srcOfImage);
        NSURL *imgsrcURL = [NSURL URLWithString:srcOfImage];
        if (imgsrcURL && imgsrcURL.scheme && imgsrcURL.host) {
            isImage = TRUE;
        }
    }
    return YES;
}

- (BOOL)webView:(UIWebView *)webView shouldStartLoadWithRequest:(NSURLRequest *)request navigationType:(UIWebViewNavigationType)navigationType {
    if ((navigationType == UIWebViewNavigationTypeLinkClicked) && (isImage)) {
        return NO;
        isImage = FALSE;
    } else {
        return YES;
    }
}

拦截UIWebView中图像中的链接

1 个答案: