如何从相关文章中排除当前 post，即使它具有相同的标签？

Question

我在我的博主博客中使用以下代码来显示基于标签的相关文章：

<div id='related-posts'>
        <b:if cond='data:post.labels'>
          <b:loop values='data:post.labels' var='label'>
            <b:if cond='data:label.isLast == &quot;true&quot;'>
              <data:label.name/>
            </b:if>
          </b:loop>
        </b:if>
      </div>

问题是当前的post显然是因为有标签而被收录在相关文章中。我想排除当前的 post，但我不知道该怎么做。

我认为这是 JavaScript 的：

$("#related-posts").each(function() {
    var v = $(this).text();
    $.ajax({
        url: "/feeds/posts/default/-/" + v + "?alt=json-in-script&max-results=" + related_number,
        type: 'get',
        dataType: "jsonp",
        success: function(e) {
            var u = "";
            var h = '<div class="related-wrap">';
            for (var i = 0; i < e.feed.entry.length; i++) {
                for (var j = 0; j < e.feed.entry[i].link.length; j++) {
                    if (e.feed.entry[i].link[j].rel == "alternate") {
                        u = e.feed.entry[i].link[j].href;
                        break
                    }
                }
                var g = e.feed.entry[i].title.$t;
                var s = e.feed.entry[i].category[0].term;
                var y = e.feed.entry[i].author[0].name.$t;
                var c = e.feed.entry[i].content.$t;
                var $c = $('<div>').html(c);
                if (c.indexOf("//www.youtube.com/embed/") < -1) {
                    var p = e.feed.entry[i].media$thumbnail.url.replace('/default.jpg', '/mqdefault.jpg');
                    var k = p
                } else if (c.indexOf("<img") > -1) {
                    var q = $c.find('img:first').attr('src').replace('s72-c', 's1600');
                    var k = q
                } else {
                    var k = no_image
                }
                h += '<li><div class="related-thumb"><a class="related-img" href="' + u + '" style="background:url(' + k + ') no-repeat center center;background-size: cover"><span class="related-overlay"/></a></div><div class="related-content"><div class="related-tag"><a class="icon ' + s + '" href="/search/label/' + s + '">' + s + '</a></div><h3 class="related-title"><a href="' + u + '">' + g + '</a></h3></div></li>'
            }
            h += '</div><div class="clear"/>';
            $("#related-posts").html(h);
            $(this).find('.related-img').each(function() {
                $(this).attr('style', function(i, src) {
                    return src.replace('/default.jpg', '/mqdefault.jpg')
                }).attr('style', function(i, src) {
                    return src.replace('s72-c', 's1600')
                })
            })
        }
    })
});

Answer 1

将data-postId添加到html代码

<div id='related-posts' expr:data-postId='data:view.postId'>
  <b:if cond='data:post.labels'>
    <b:loop values='data:post.labels' var='label'>
      <b:if cond='data:label.isLast == &quot;true&quot;'>
        <data:label.name/>
      </b:if>
    </b:loop>
  </b:if>
</div>

然后在您的 Javascript 代码中再添加几行

$("#related-posts").each(function() {
    var v = $(this).text();
    
    // 1) get the current post id
    var postId = $(this).attr('data-postId');

    $.ajax({
        url: "/feeds/posts/default/-/" + v + "?alt=json-in-script&max-results=" + related_number,
        type: 'get',
        dataType: "jsonp",
        success: function(e) {
            var u = "";
            var h = '<div class="related-wrap">';
            for (var i = 0; i < e.feed.entry.length; i++) {

                // 2) skip if it's the current post
                if(e.feed.entry[i].id.$t.match(/\d+$/)[0] === postId){
                    continue
                }

                for (var j = 0; j < e.feed.entry[i].link.length; j++) {
                    if (e.feed.entry[i].link[j].rel == "alternate") {
                        u = e.feed.entry[i].link[j].href;
                        break
                    }
                }
                var g = e.feed.entry[i].title.$t;
                var s = e.feed.entry[i].category[0].term;
                var y = e.feed.entry[i].author[0].name.$t;
                var c = e.feed.entry[i].content.$t;
                var $c = $('<div>').html(c);
                if (c.indexOf("//www.youtube.com/embed/") < -1) {
                    var p = e.feed.entry[i].media$thumbnail.url.replace('/default.jpg', '/mqdefault.jpg');
                    var k = p
                } else if (c.indexOf("<img") > -1) {
                    var q = $c.find('img:first').attr('src').replace('s72-c', 's1600');
                    var k = q
                } else {
                    var k = no_image
                }
                h += '<li><div class="related-thumb"><a class="related-img" href="' + u + '" style="background:url(' + k + ') no-repeat center center;background-size: cover"><span class="related-overlay"/></a></div><div class="related-content"><div class="related-tag"><a class="icon ' + s + '" href="/search/label/' + s + '">' + s + '</a></div><h3 class="related-title"><a href="' + u + '">' + g + '</a></h3></div></li>'
            }
            h += '</div><div class="clear"/>';
            $("#related-posts").html(h);
            $(this).find('.related-img').each(function() {
                $(this).attr('style', function(i, src) {
                    return src.replace('/default.jpg', '/mqdefault.jpg')
                }).attr('style', function(i, src) {
                    return src.replace('s72-c', 's1600')
                })
            })
        }
    })
});

如何从相关文章中排除当前 post，即使它具有相同的标签？

How to exclude the current post from related articles even though it has the same label?

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