我正在开发一个新闻聚合器网站,在其中添加了24个发布者,所有发布者数据都存在xml文件中,因此我使用ajax对其进行调用。但是加载该页面需要30秒。我的网页中有四个部分(针对您,技术,政治,电影/电视),每个部分都有一个外部JS文件。这些文件正在请求数据。 PLZ帮助我减少页面加载时间,并提出一些策略。
//technology JS file code sample
$.when($.get('https://cors-anywhere.herokuapp.com/https://feed.cnet.com/feed/topics/sci-tech'), $.get('https://cors-anywhere.herokuapp.com/https://in.mashable.com/tech.xml')).then(function(r1, r2) {
// console.log(r1[0]);
// console.log(r2[0]);
// cnet article
$(r1[0]).find('item').each(function(i, j) {
title = $(j).find('title').text();
description = $(j).find('description').text();
link = $(j).find('link').text();
thumbnail = $(j).find('[url]').attr('url');
newsname = "Cnet";
logo = "http://i.i.cbsi.com/cnwk.1d/i/ne/gr/prtnr/CNET_Logo_150.gif";
articlestech(i, title, description, link, thumbnail, newsname, logo);
});
//Mashable articles
$(r2[0]).find('item').each(function(i, j) {
title = $(j).find('title').text();
description = $(j).find('description').text();
//pos3 = description.indexOf("</a>");
// des = description.substring(pos3 + 2);
// thumbnail = $(j).find('image').text();
link = $(j).find('link').text();
var pos2 = description.indexOf("width");
var pos1 = description.indexOf("src=");
thumbnail = description.substring(pos1 + 5, pos2 - 2);
newsname = "Mashable India Tech";
logo = "https://media.glassdoor.com/sqll/255718/mashable-squarelogo-1430326903133.png";
des = "";
articlestech(i, title, des, link, thumbnail, newsname, logo);
});}) ;
//movies/tv shows code sample
$.ajax({
type: "GET",
url: "https://cors-anywhere.herokuapp.com/https://www.cinemablend.com/rss/topic/reviews/movies",
dataType: "xml",
success: function(xml) {
$(r1[0]).find('item').each(function(i, j) {
title = $(j).find('title').text();
description = $(j).find('description').text() + "...";
link = $(j).find('link').text();
//console.log(link);
thumbnail = $(j).find('enclosure').attr('url');
//console.log(thumbnail);
newsname = "Cinemablend";
logo = "https://s3.amazonaws.com/media.muckrack.com/groups/icons/cinemablend.jpeg";
articlestech(i, title, description, link, thumbnail, newsname, logo);
});
}
});
$.ajax({
type: "GET",
url: "https://cors-anywhere.herokuapp.com/https://www.cinejosh.com/rss-feed/1/review.html",
dataType: "xml",
success: function(xml) {
$(xml).find('item').each(function(i, j) {
title = $(j).find('title').text();
//des = $(j).find('description').text();
link = $(j).find('link').text();
description = "";
thumbnail = $(j).find('[url]').attr('url');;
console.log(thumbnail);
newsname = "CineJosh";
logo = "https://www.cinejosh.com/gallereys/others/normal/cinejosh_logo_1902160805/cinejosh_logo_1902160805_01.jpg";
articlestech(i, title, description, link, thumbnail, newsname, logo);
});
}
});
其余两个JS文件具有相同的代码结构。如何减少加载时间。
问题是您正在等待所有六个请求返回以开始对响应进行处理。正如我从代码片段中看到的那样,您提供的请求彼此之间并不相互依赖,因此您可以将所有请求分隔开,一旦第一个请求完成,用户便可以在屏幕上看到结果。
所以,这个:
$.when(
$.get('https://cors-anywhere.herokuapp.com/https://feed.cnet.com/feed/topics/sci-tech'),
$.get('https://cors-anywhere.herokuapp.com/https://in.mashable.com/tech.xml'),
$.get('https://www.theverge.com/rss/tech/index.xml'),
$.get('https://cors-anywhere.herokuapp.com/https://www.buzzfeed.com/tech.xml'),
$.get('https://cors-anywhere.herokuapp.com/https://www.cnbc.com/id/19854910/device/rss/rss.html'),
$.get('https://cors-anywhere.herokuapp.com/https://www.engadget.com/rss.xml'))
需要像这样的东西:
$.when($.get('https://cors-anywhere.herokuapp.com/https://feed.cnet.com/feed/topics/sci-tech')).then(function(r1) {
$(r1[0]).find('item').each(function(i, j) {
title = $(j).find('title').text();
description = $(j).find('description').text();
link = $(j).find('link').text();
thumbnail = $(j).find('[url]').attr('url');
newsname = "Cnet";
logo = "http://i.i.cbsi.com/cnwk.1d/i/ne/gr/prtnr/CNET_Logo_150.gif";
articlestech(i, title, description, link, thumbnail, newsname, logo);
});
})
$.when($.get('https://cors-anywhere.herokuapp.com/https://in.mashable.com/tech.xml')).then(function(r1) {
$(r2[0]).find('item').each(function(i, j) {
title = $(j).find('title').text();
description = $(j).find('description').text();
//pos3 = description.indexOf("</a>");
// des = description.substring(pos3 + 2);
// thumbnail = $(j).find('image').text();
link = $(j).find('link').text();
var pos2 = description.indexOf("width");
var pos1 = description.indexOf("src=");
thumbnail = description.substring(pos1 + 5, pos2 - 2);
newsname = "Mashable India Tech";
logo = "https://media.glassdoor.com/sqll/255718/mashable-squarelogo-1430326903133.png";
des = "";
articlestech(i, title, des, link, thumbnail, newsname, logo);
});
})
...