Question

function getHTMLSource(url) {
      return $http.get(url).then(function(response) {
        var html = response.data;
        url = getDetailPage(html)
        return url ;
    }
)};

getHTMLSource('http://www.footpatrol.co.uk/s:282524/search=282524/')

每当我尝试上面的代码时，我在控制台GET http://localhost:9000/templates/footpatrol.co.uk/_assets/images/content/footpatrol_logo.png 404 (Not Found)中收到以下错误消息。

图片似乎存在于http://www.footpatrol.co.uk/templates/footpatrol.co.uk/_assets/images/content/footpatrol_logo.png，但由于我正在使用Chrome的Allow-Control-Allow-Origin插件在localhost上运行scipt，因此它看起来效果不佳。我不想获取我只想要源代码的图像，还有这个吗？

更新：我认为可能是我的解析器导致问题，因为此处会抛出错误消息

function getDetailPage(html) {
  var temp = document.createElement('div');
  temp.innerHTML = html;
  var a = temp.querySelector('a[class*=\'fp-product-thumb-link\']');
  var partOfUrl = a.href;
  var splitUrl = partOfUrl.split('/');
  var url = 'http://www.footpatrol.co.uk/' + splitUrl[3] + '/' + splitUrl[4];
  var url = 'http://www.footpatrol.co.uk/' + splitUrl[3] + '/' + splitUrl[4];
  $log.debug('Detail page url found: ' + url);
  return url;
}

Answer 1

使用DOMParser API:

，而不是创建元素

function getDetailPage(html) {
      //var temp = document.createElement('div');
      //temp.innerHTML = html;
      var parser = new DOMParser();
      var temp = parser.parseFromString(html, "text/html");
      var a = temp.querySelector('a[class*=\'fp-product-thumb-link\']');
      //var partOfUrl = a.href;
      //console.log(partOfUrl);
      //var splitUrl = partOfUrl.split('/');
      //var url = 'http://www.footpatrol.co.uk/' + splitUrl[3] + '/' + splitUrl[4];
      var elem = angular.element(a);
      var url = 'http://www.footpatrol.co.uk/' + elem.attr('href');
      console.debug('Detail page url found: ' + url);
      return url;
}

找到详细信息页面网址：http://www.footpatrol.co.uk//footwear/282524-air-retro-15-obsidian.html

当我不需要解析图像时，http.get尝试检索图像

1 个答案: