Question

我正在尝试搜索Etsy并访问所有链接。在Python中，我知道如何做到这一点（使用BeautifulSoup）但今天我想知道我是否可以使用PhantomJS做同样的事情。我没有走得太远。

此脚本应在Etsy上搜索“hello kitty”并返回所有产品 <a class="listing-thumb" href=...></a>并在控制台中打印它们。理想情况下，我稍后会访问它们并获取我需要的信息。现在它只是冻结了。有什么想法吗？

var page = require('webpage').create();
var url = 'http://www.etsy.com/search?q=hello%20kitty';

page.open(url, function(status){
    // list all the a.href links in the hello kitty etsy page
    var link = page.evaluate(function() {
        return document.querySelectorAll('a.listing-thumb');
    });
    for(var i = 0; i < link.length; i++){ console.log(link[i].href); }
    phantom.exit();
});

我玩过CasperJS玩具，可能更适合这个。

Answer 1

PhantomJS evaluate()无法序列化并返回像HTMLElements或NodeLists这样的复杂对象，因此您必须先将它们映射到可序列化的东西：

var page = require('webpage').create();
var url = 'http://www.etsy.com/search?q=hello%20kitty';

page.open(url, function(status) {
    // list all the a.href links in the hello kitty etsy page
    var links = page.evaluate(function() {
        return [].map.call(document.querySelectorAll('a.listing-thumb'), function(link) {
            return link.getAttribute('href');
        });
    });
    console.log(links.join('\n'));
    phantom.exit();
});

注意：我们在此使用[].map.call()将NodeList视为标准Array。

Answer 2

您的代码唯一的问题是您不了解phantomjs范围。你有幻像和页面范围。您试图从页面范围（page.evaluate在页面范围中运行）返回JavaScript DOM对象引用（那些无法序列化）到幻像主范围。我认为这是不可能的。以下代码有效：

var page = require('webpage').create();
var url = 'http://www.etsy.com/search?q=hello%20kitty';

// for debug (to see if page returns status code 200)
page.onResourceReceived = function(response) {
    if (response.url === url) {
        console.log('Resorce: "' + response.url + '" status: '  + response.status);

        if (response.status === 200) {
            console.log(response.url);
            for (var i = 0; i < response.headers.length; i++) {
                console.log(response.headers[i].name + ': ' + response.headers[i].value);
            }
        }
    }
};

page.onLoadFinished = function(status){
    console.log('Status: ' + status);

    console.log('Starting evaluate...');
    var links = page.evaluate(function() {
        var nodes = [],
            matches = document.querySelectorAll("a.listing-thumb");

            for(var i = 0; i < matches.length; ++i) {
                nodes.push(matches[i].href);
            }

            return nodes;
    });
    console.log('Done evaluate... count: ' + links.length);

    if (links && links.length > 0) {
        for(var i = 0; i < links.length; ++i) {
            console.log('(' + i + ') ' + links[i]);
        }
    } else {
        console.log("No match found!");
    }

    phantom.exit(0);
};

page.open(url);

Answer 3

以下是我最近编写的使用PhantomJs抓取网址的一些代码，如果您只提供一个网址，它会在页面上显示所有网址，如果您提供的参数为class|id，后跟一个＆＃34; class / id name＆＃34;它只会显示class / id的urls。

////////////////////////////////////////////////////////// 
/////  PhantomJS URL Scraper v.1.3 ///// 
// 
// Copyrighted by +A.M.Danischewski  2016+ (c)
// This program may be reutilized without limits, provided this 
// notice remain intact. 
// 
// Usage: phantomjs phantom_urls.js <URL> [["class"|"id"] [<query id/class name>]]
//
//   Argument 1: URL -- "https://www.youtube.com/watch?v=8TniRMwL2Vg" 
//   Argument 2: "class" or "id" 
//   Argument 3: If Argument 2 was provided, "class name" or "id name" 
// 
// By default this program will display ALL urls from a user supplied URL.  
// If a class name or id name is provided then only URL's from the class 
// or id are displayed.  
//  
/////////////////////////////////// 

var page = require('webpage').create(), 
    system = require('system'),
    address;

if (system.args.length === 1) {
  console.log(' Usage: phantomjs phantom_urls.js <URL> [["class"|"id"] [<query id/class name>]]');
  phantom.exit();
}

address = system.args[1];
querytype= system.args[2];
queryclass = system.args[3];
page.open(address, function(status) {
  if (status !== 'success') {
    console.log('Error loading address: '+address);
  } else {
   //console.log('Success! In loading address: '+address);   
  }
});

page.onConsoleMessage = function(msg) {
  console.log(msg);
}

page.onLoadFinished = function(status) {
   var dynclass="function() { window.class_urls = new Array(); window.class_urls_next=0; var listings = document.getElementsByClassName('"+queryclass+"'); for (var i=0; i < listings.length; i++) { var el = listings[i]; var ellnks=[].map.call(el.querySelectorAll('a'),function(link) {return link.getAttribute('href');}); var elhtml=el.innerHTML; window.class_urls.push(ellnks.join('\\n')); }; return window.class_urls;}"; 
   var    dynid="function() { window.id_urls = new Array(); window.id_urls_next=0; var listings = document.getElementById('"+queryclass+"'); var ellnks=[].map.call(listings.querySelectorAll('a'),function(link) {return link.getAttribute('href');}); var elhtml=listings.innerHTML; window.id_urls.push(ellnks.join('\\n'));  return window.id_urls;}";  
   var  allurls="function() { var links = page.evaluate(function() { return [].map.call(document.querySelectorAll('a'), function(link) { return link.getAttribute('href'); };); };); console.log(links.join('\\n')); }"; 
   var page_eval_function="";  
   if (querytype === "class") {
   console.log(page.evaluate(dynclass).toString().replace(/,/g, "\n")); 
   } else if (querytype === "id") {
   console.log(page.evaluate(dynid).toString().replace(/,/g, "\n")); 
   } else { 
   var links = page.evaluate(function() {
        return [].map.call(document.querySelectorAll('a'), function(link) {
            return link.getAttribute('href');
        });
    });    
       console.log(links.join('\n'));
   }             
   phantom.exit();
};

如何用phantomjs刮取链接

3 个答案: