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javascript - 如何简单地显示 YQL 的 xml 输出或将 JSON 输出到 html

转载 作者:太空宇宙 更新时间:2023-11-04 15:30:53 25 4
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所以我一直在研究一种从页面抓取数据并显示它的方法(格式与源代码大致相同)。我找到了 YQL,我发现它很棒,除了我无法弄清楚如何只显示整个输出而没有什么特别的(除了基本格式)

YQL输入代码为:

select * from html where url="http://directory.vancouver.wsu.edu/anthropology" and xpath="//div[@id='facdir']"

使用它返回 JSON:

http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20html%20where%20url%3D%22http%3A%2F%2Fdirectory.vancouver.wsu.edu%2Fanthropology%22%20and%20xpath%3D%22%2F%2Fdiv%5B%40id%3D'facdir'%5D%22&format=json&callback=anthropology

我遵循了雅虎教程,并创建了新闻小部件等,但没有一个教程涵盖了基本 View (也不需要链接,只需要段落设置)。

像这样:

Name
Title
Phone:(###)###-####
Location: Building and Room #
email@vancouver.wsu.edu

这是我从 http://christianheilmann.com 得到的输出, 但它没有做任何事情(显然她的教程都没有用,每个教程都试过了):

<html>
<head>
<script src="http://code.jquery.com/jquery-latest.js"></script>
</head>
<body>
<p>
<b>Copied:</b>
</p>
<div>
<script>
function anthropology (0) {
// get the DIV with the ID $
var info = document.getElementById('facdir');
// add a class for styling
info.className = 'js';
// if it exists
if(info){
// get the info data returned from YQL
var data = o.query.results.span;
var link = info.getElementsByTagName('a')[0];
link.innerHTML = '(see all info)';
// to the main container DIV
var out = document.createElement('span');
out.className = 'info';
info.insertBefore(out,link.parentNode);
}
}
</script>
<script src='http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20html%20where%20url%3D%22http%3A%2F%2Fdirectory.vancouver.wsu.edu%2Fanthropology%22%20and%20xpath%3D%22%2F%2Fdiv%2Fdiv%2Fdiv%2Fdiv%2Fdiv%2Fdiv%2Fdiv%2Fdiv%2Fdiv%2Fdiv%2Fdiv%22&format=json&callback=anthropology'></script>
</div>

最佳答案

我最近用几个 jsFiddle 完成了一个教程,并解释了如何使用 YQLXPATH 和 jQuery .ajax() 对于不同的 SO 问题,这将为您指明方向。你可以看到 SO Answer here .

为了让您的问题获得可接受的答案,我整理了一个工作演示,向您展示从您请求的网页中抓取数据是多么容易。

jsFiddle Demo 包含大量注释和console.log() 消息以了解工作流程。确保激活浏览器控制台并使用 Firebug例如。 HTMLCSS 用于构建教师成员(member)箱 模仿原始网站的那些,包括图片、姓名、电子邮件中的链接和网页主题也是。

演示:

jsFiddle Data Scraping XML: Dynamic Webpage Building

修改!!! 除了修改上面的jsFiddle,查看相关

jsFiddle Tutorial: Creating Dynamic Div's (Now Improved!)

HTML:

<div id="results"></div>

jQuery:

var directoryName = 'child-development-program';

$.ajax({
type: 'GET',
url: "http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20html%20where%20url%3D%22http%3A%2F%2Fdirectory.vancouver.wsu.edu%2F" + directoryName + "%22%20and%20xpath%3D%22%2F%2Fdiv%5B%40id%3D'content-inner'%5D%2Fdiv%2Fdiv%2Fdiv%2Fdiv%2Fdiv%5B2%5D%22",
dataType: 'xml',
success: function(data) {

if (data) {

// Show in console the jQuery Object.
console.info('Here is the returned query');
console.log( $(data).find('query') );

// Show in console the results in inner-html text.
var textResults = $(data).find('results').text();
console.log( textResults );

// Parse the list of faculty members. Variable indexFM is not used for indexed faculty member.
$(data).find('results').find('.views-row').each(function(indexFM){

// This variable will store the current faculty member.
var facultyMember = this;
console.info('Faculty jQuery DIV Object shown on next lines.');
console.log( facultyMember );

// Parse the contents of each faculty member. Variable indexFC is not used for indexed faculty content.
$(facultyMember).each(function(indexFC){

// Get Thumbnail Image of Faculty Member
var facultyMemberImage = $(this).find('.views-field-field-profile-image-fid #directoryimage a img').attr('src');
console.log( facultyMemberImage );

// Get Title (Name) of Faculty Member
var facultyMemberTitle = $(this).find('.views-field-field-professional-title-value #largetitle').text();
console.log( facultyMemberTitle );
// Get relative URL fragment.

//
// Stackoverflow Edit: Much more extraction in this section, see jsFiddle link above.
//

// Get Email of Faculty Member
var facultyMemberEmail = $(this).find('.views-field-field-email-value span').text();

// Simple dashed line to separate faculty members as seen in browser console.
console.log('--------');

var divObject = '<div class="dynamicResults"><div class="dynamicThumb"><a href="' + facultyMemberUrl + '"><img src="' + facultyMemberImage + '" alt=""></a></div><div class="dynamicInfo"><div class="dynamicText"><a href="' + facultyMemberUrl + '" class="dynamicName">' + facultyMemberTitle + '</a></div><div class="dynamicText">' + facultyMemberPosition + '</div><div class="dynamicText">Phone: ' + facultyMemberPhone + '</div><div class="dynamicText">Location: ' + facultyMemberBuilding + ' <span>' + facultyMemberRoom + '</span></div><div class="dynamicText"><a href="' + facultyMemberEmailUrl + '" class="dynamicEmail">' + facultyMemberEmail + '</a><span class="dynamicEmailpic"></span></div></div></div><div class="clear"></div>';

// Build webpage with dynamic data.
$('#results').append( divObject );

});

});

}
}
});

屏幕截图: 照片中的缩略图为 100px x 100px Revised Photo for Revised jsFiddle!!


但是在真正查看您的问题时,我想尝试一些新的和简单的东西……但是结果是可以接受的。这一次,数据抓取技术是使用网页原生 CSS 文件作为 jsFiddle 中的 Assets ,同时还将返回的数据直接用于 DOM

此方法使用与上述相同的原理,除了它使用 html 作为 .ajax() dataType 以获得可用的 原始网页的近似克隆。唯一的缺点是需要整个 CSS 文件,但您可以解析原始文件以删除多余的样式和不需要的选择器(重要的是不要打破 IE 中的 4096 CSS 选择器障碍)。

演示:

jsFiddle Data Scraping HTML: Clone That Webpage

HTML

<link type="text/css" rel="stylesheet" media="all" href="http://directory.vancouver.wsu.edu/sites/directory.vancouver.wsu.edu/files/css/css_f9f00e4e3fa0bf34a1cb2b226a5d8344.css" />

<div id="facultyAnthropology"></div>

jQuery:

var directoryName = 'anthropology';

$.ajax({
type: 'GET',
url: "http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20html%20where%20url%3D%22http%3A%2F%2Fdirectory.vancouver.wsu.edu%2F"+directoryName+"%22%20and%20xpath%3D%22%2F%2Fdiv%5B%40id%3D'content-area'%5D%22",
dataType: 'html',
success: function(data) {
$('#facultyAnthropology').append($(data).find('results'));
}
});

屏幕截图: 如上,照片中的缩略图为 100px x 100px

关于javascript - 如何简单地显示 YQL 的 xml 输出或将 JSON 输出到 html,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/14048943/

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