2015-10-29 61 views
0

我需要将标题标记插入到我的index.html中,该标记从json中提取标题。我仍然有一个形象和链接。从json获取标题标记

<a class="my-blog" id="post2" href="#"> 
     <div > 
     //title goes here 
     </div> 
and that's my javascript 




function loadBlogWidget() { 
    $.get("http://blog.lawgeex.com/wp-json/posts").done(function (cResult) { 

    for (var i = 0; i < cResult.length; i++) { 
    var cPost = cResult[i]; 
    var strLink = cPost.link; 
    var strImageSrc=cPost.featured_image.attachment_meta.sizes.thumbnail.url; 
    $("#post" + i).attr("href", strLink); 
    $("#post" + i).css("background-image", "url(" + strImageSrc + ")"); 


} 
+0

显示您在cResult变量 –

+0

中获得的信息嗨@Ann欢迎来到Stack Overflow,您是否可以使用任何错误消息更新您的问题或描述当前正在发生的事情,这将帮助我们更清楚地回答您的问题。 – BMac

+0

是的,我添加了这一行$(“#post”+ i).append('');它只是返回一个标签的名称。 – <span class="text-secondary"> <small> <a rel="noopener" target="_blank" href="https://stackoverflow.com/users/5502054/">nundel</a></span> <span></span> </small> </span> </p> </div> </div> </div> </div> </div> </article> </div> <div class="answer-title"> <span class="text-logo margin-top-sm">A</span> <h2 class="title h4">回答</h2> </div> <div class="item-description text-md markdown-body margin-bottom-40 voidso"> <article class="board-top-1 padding-top-10"> <div class="post-col vote-info"> <span class="count">0<i class="fa fa-thumbs-up"></i></span> </div> <div class="post-offset"> <div class="answer fmt"> <p>我看到了json响应,它发出了一个帖子列表。我不太确定问题是否为 1.如何更改显示帖子列表的html的标题 或 2.如何更改用户导航的html标题为</p> <p>就我而言,我认为是您的问题,您可以使用document.title = cPost.title,但是,正如我所提到的那样,响应是一个列表,所以我假设您希望标题成为第一篇文章的标题或东西。</p> <pre><code class="prettyprint-override"><html> <body> <a class="my-blog" id="post2" href="#"> <div > //title goes here </div> </a> </body> <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script> <script> function loadBlogWidget() { $.get("http://blog.lawgeex.com/wp-json/posts").done(function (cResult) { var title_changed = false; for (var i = 0; i < cResult.length; i++) { var cPost = cResult[i]; var strLink = cPost.link; var strImageSrc=cPost.featured_image.attachment_meta.sizes.thumbnail.url; $("#post" + i).attr("href", strLink); $("#post" + i).css("background-image", "url(" + strImageSrc + ")"); if(!title_changed){ document.title = cPost.title; title_changed = true; } } }); } loadBlogWidget(); </script> </html> </code></pre> <p>然而,予以警告,这是一个坏主意,因为爬虫将不能够对网页进行索引正确,因为它不能兑现的动态名称的更改。也许,你可以做的是有一个默认的标题,你希望抓取工具跟踪和只有用户的可见性有这个动态标题改变的东西。</p> <p>希望这会有所帮助!</p> </div> <div class="post-info"> <div class="post-meta row"> <p class="text-secondary col-lg-6"> <span class="source"> <a rel="noopener" target="_blank" href="https://stackoverflow.com/q/33411379">来源</a> </span> </p> <p class="text-secondary col-lg-6"> <span class="float-right date"> <span>2015-10-29 10:13:37</span> </p> <p class="col-12"></p> <p class="col-12"></p></div> </div> </div> </article> <div> <script async src="https://pagead2.googlesyndication.com/pagead/js/adsbygoogle.js"></script> <ins class="adsbygoogle" style="display:block" data-ad-client="ca-pub-6208739752673518" data-ad-slot="1038284119" data-ad-format="auto" data-full-width-responsive="true"></ins> <script> (adsbygoogle = window.adsbygoogle || []).push({}); </script> </div> </div> <div class="clearfix"> </div> <div class="relative-box"> <div class="relative">相关问题</div> <ul class="relative_list"> <li> 1. <a href="http://www.uwenku.com/question/p-bsvtqyba-bgn.html" target="_blank" title="Rails从HTML获取标题标记"> Rails从HTML获取标题标记 </a> </li> <li> 2. <a href="http://www.uwenku.com/question/p-qgcpciga-bev.html" target="_blank" title="获取JSON对象标题"> 获取JSON对象标题 </a> </li> <li> 3. <a href="http://www.uwenku.com/question/p-zfnaqwha-bhn.html" target="_blank" title="JSON - 获取Chrome标签标题"> JSON - 获取Chrome标签标题 </a> </li> <li> 4. <a href="http://www.uwenku.com/question/p-usvhtntd-yc.html" target="_blank" title="按标题(或标记)获取ContentControl"> 按标题(或标记)获取ContentControl </a> </li> <li> 5. <a href="http://www.uwenku.com/question/p-rsrjjghv-eg.html" target="_blank" title="如何获取URL“标题”标记值?"> 如何获取URL“标题”标记值? 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