Reputation: 242
I want to create an addon which will load html content of a specific url and save a specific line of that page and then move to that url. I read a lot of things on Mozila.org about content of a web page but I don't understand how to read the html content.
Upvotes: 0
Views: 670
Reputation: 2075
Without knowing the page and URL to find on it I can't create a complete solution, but here's an example Greasemonkey script I wrote that does something similar.
This script is for Java articles on DZone. When an article has a link to the source, it redirects to this source page:
// ==UserScript==
// @name DZone source
// @namespace com.kwebble
// @description Directly go to the source of a DZone article.
// @include http://java.dzone.com/*
// @version 1
// @grant none
// ==/UserScript==
var node = document.querySelector('a[target="_blank"]');
if (node !== null) {
document.location = node.getAttribute('href');
}
Usage:
After saving the script visit the page with the link. Greasemonkey should execute your script and redirect the browser.
[edit] This searches script tags for text like you described and redirects.
// ==UserScript==
// @name Test
// @namespace com.kwebble
// @include your_page
// @version 1
// @grant none
// ==/UserScript==
var nodes = document.getElementsByTagName('script'),
i, matches;
for (i = 0; i < nodes.length; i++) {
if (nodes.item(i).innerHTML !== '') {
matches = nodes.item(i).innerHTML.match(/windows\.location = "(.*?).php";/);
if (matches !== null){
document.location = matches[1];
}
}
}
The regular expression to find the URL might need some tweaking to match the exact page content.
Upvotes: 1
Reputation: 7721
Addon or GreaseMonkey script have a similar approach but addon can use native Firefox APIs. (but it is a lot more complicated than scripts)
Basically, this is the process (without knowing your exact requirements)
Get the content of a remote URL with XMLHttpReques()
Get the data that you need with RegEx or DOMParser()
Change the current URL to that target with location.replace()
Upvotes: 0
Reputation: 37238
Here's a simple snippet that does XHR request, WITHOUT cookies. Don't worry about cross-origin as you are running from privelaged scope, meaning you aren't coding this in a website but as a firefox addon.
var {Cu: utils, Cc: classes, Ci: instances} = Components;
Cu.import('resource://gre/modules/Services.jsm');
function xhr(url, cb) {
let xhr = Cc["@mozilla.org/xmlextras/xmlhttprequest;1"].createInstance(Ci.nsIXMLHttpRequest);
let handler = ev => {
evf(m => xhr.removeEventListener(m, handler, !1));
switch (ev.type) {
case 'load':
if (xhr.status == 200) {
cb(xhr.response);
break;
}
default:
Services.prompt.alert(null, 'XHR Error', 'Error Fetching Package: ' + xhr.statusText + ' [' + ev.type + ':' + xhr.status + ']');
break;
}
};
let evf = f => ['load', 'error', 'abort'].forEach(f);
evf(m => xhr.addEventListener(m, handler, false));
xhr.mozBackgroundRequest = true;
xhr.open('GET', url, true);
xhr.channel.loadFlags |= Ci.nsIRequest.LOAD_ANONYMOUS | Ci.nsIRequest.LOAD_BYPASS_CACHE | Ci.nsIRequest.INHIBIT_PERSISTENT_CACHING;
//xhr.responseType = "arraybuffer"; //dont set it, so it returns string, you dont want arraybuffer. you only want this if your url is to a zip file or some file you want to download and make a nsIArrayBufferInputStream out of it or something
xhr.send(null);
}
var href = 'http://www.bing.com/'
xhr(href, data => {
Services.prompt.alert(null, 'XHR Success', data);
});
Upvotes: 1