仅列出与document.links的未访问链接

Question

我列出了使用document.links（带有用户名）的某些网站底部的所有链接。

访问过的和未访问过的url-s一起出现。我可以为它们设置不同的颜色，这样很容易看出哪些链接是新的，哪些链接是我已经访问过的。但我希望看到仅新链接。

有没有办法让a:link仅列出未访问的（新的）visibility:hidden;）链接？

（如果不可能，那我怎么能隐藏访问过的链接？我在display:none;上尝试了a:visited和//Show Histogram CVPixelBufferRef pixelBuffer = CMSampleBufferGetImageBuffer(sampleBuffer); CFDictionaryRef attachments = CMCopyDictionaryOfAttachments(kCFAllocatorDefault, sampleBuffer, kCMAttachmentMode_ShouldPropagate); CIImage *ciImage = [[CIImage alloc] initWithCVPixelBuffer:pixelBuffer options:(__bridge NSDictionary *)attachments]; NSUInteger count = 256; count = count <= 256 ? count : 256; count = count >= 1 ? count : 1; NSDictionary *params = @{kCIInputImageKey: ciImage, kCIInputExtentKey: [CIVector vectorWithCGRect:[ciImage extent]], @"inputCount": @(256), @"inputScale": @(200) }; CIFilter *filter = [CIFilter filterWithName:@"CIAreaHistogram" withInputParameters:params]; CIImage *outImage = [filter outputImage]; //--------------------------------------------- CIContext *context = [CIContext contextWithOptions:nil]; NSDictionary *params2 = @{ kCIInputImageKey: outImage }; CIFilter *filter2 = [CIFilter filterWithName:@"CIHistogramDisplayFilter" withInputParameters:params2]; CIImage *outputImage = [filter2 outputImage]; CGRect outExtent = [outputImage extent]; CGImageRef cgImage = [context createCGImage:outputImage fromRect:outExtent]; UIImage *outImage2 = [UIImage imageWithCGImage:cgImage]; CGImageRelease(cgImage); // resize UIImage *resized = [self resizeImage:outImage2 withQuality:kCGInterpolationNone rate:2.5]; //Remove the default grey background resized = [self removeColorFromImage:resized grayLevel:137]; dispatch_async(dispatch_get_main_queue(), ^{ self.histogramView.image = resized; }); ，但没有工作。发现它是{{{ 3}}但我不想触摸原始页面上的链接，只能链接到我的链接集合中。“

Answer 1

可能不可能

我不完全确定due to the security issue similar to the one you mentioned in this similar article会导致$('.scIcon').data('itemid') 选择器受到严格限制的可能性如何，这使得在现代浏览器中几乎不可能进行编程访问。

通过:visited

解决方法

我想you could use a workaround similar to the one mentioned in this blog post，它使用localStorage明确存储点击的链接并保留“访问”属性，以便您可以识别那些已被触摸的链接：

localStorage

这种方法会将“访问”类附加到元素中，并允许您使用纯CSS方法显式删除它们：

function check_visited_links(){
    // Access all of the elements that have been visited (from local storage)
    var visited_links = JSON.parse(localStorage.getItem('visited_links')) || [];
    // Iterate through your links
    var links = document.getElementsByTagName('a');
    for (var i = 0; i < links.length; i++) {
        var that = links[i];
        // When the link is clicked, store a reference in localStorage to it
        that.onclick = function () {
            var clicked_url = this.href;
            if (visited_links.indexOf(clicked_url)==-1) {
                visited_links.push(clicked_url);
                localStorage.setItem('visited_links', JSON.stringify(visited_links));
            }
        } 
        // Indicate the link was visited by setting it's class
        if (visited_links.indexOf(that.href)!== -1) { 
            that.className += ' visited';
        }
    }
}

或者使用类似于您最初使用的Javascript技术：

a.visited { display: none; }

使用document.querySelectorAll('a.visited'); 技术的示例

您可以see an example that uses the localStorage technique here（仅在按下每个按钮后重新加载）以及下面的输出示例：