Question

在这段HTML代码中：

<div class="item">

    <div class="thumb">
        <a href="http://www.mp3crank.com/wolf-eyes/lower-demos-121866" rel="bookmark" lang="en" title="Wolf Eyes - Lower Demos album downloads">
        <img width="100" height="100" alt="Mp3 downloads Wolf Eyes - Lower Demos" title="Free mp3 downloads Wolf Eyes - Lower Demos" src="http://www.mp3crank.com/cover-album/Wolf-Eyes-–-Lower-Demos.jpg" /></a>
    </div>

    <div class="release">
        <h3>Wolf Eyes</h3>
        <h4>
        <a href="http://www.mp3crank.com/wolf-eyes/lower-demos-121866" title="Wolf Eyes - Lower Demos">Lower Demos</a>
        </h4>
        <script src="/ads/button.js"></script>
    </div>

    <div class="release-year">
        <p>Year</p>
        <span>2013</span>
    </div>

    <div class="genre">
        <p>Genre</p>
        <a href="http://www.mp3crank.com/genre/rock" rel="tag">Rock</a>
        <a href="http://www.mp3crank.com/genre/pop" rel="tag">Pop</a>
    </div>

</div>

我知道如何以其他方式解析它，但我想使用HTMLAgilityPack库检索此信息：

Title : Wolf Eyes - Lower Demos
Cover : http://www.mp3crank.com/cover-album/Wolf-Eyes-–-Lower-Demos.jpg
Year  : 2013
Genres: Rock, Pop
URL   : http://www.mp3crank.com/wolf-eyes/lower-demos-121866

这些html行是什么：

Title : title="Wolf Eyes - Lower Demos"
Cover : src="http://www.mp3crank.com/cover-album/Wolf-Eyes-–-Lower-Demos.jpg"
Year  : <span>2013</span>
Genre1: <a href="http://www.mp3crank.com/genre/rock" rel="tag">Rock</a>
Genre2: <a href="http://www.mp3crank.com/genre/pop" rel="tag">Pop</a>
URL   : href="http://www.mp3crank.com/wolf-eyes/lower-demos-121866"

这就是我正在尝试的，但在尝试选择单个节点时，我总是遇到object reference not set异常，抱歉，我是HTML的新手，我试图按照这个问题的步骤进行HtmlAgilityPack basic how to get title and link?

Public Class Form1

    Private htmldoc As HtmlAgilityPack.HtmlDocument = New HtmlAgilityPack.HtmlDocument
    Private htmlnodes As HtmlAgilityPack.HtmlNodeCollection = Nothing

    Private Title As String = String.Empty
    Private Cover As String = String.Empty
    Private Genres As String() = {String.Empty}
    Private Year As Integer = -0
    Private URL as String = String.Empty

    Private Sub Test() Handles MyBase.Shown

        ' Load the html document.
        htmldoc.LoadHtml(IO.File.ReadAllText("C:\source.html"))

        ' Select the (10 items) nodes.
        htmlnodes = htmldoc.DocumentNode.SelectNodes("//div[@class='item']")

        ' Loop trough the nodes.
        For Each node As HtmlAgilityPack.HtmlNode In htmlnodes

            Title = node.SelectSingleNode("//div[@class='release']").Attributes("title").Value
            Cover = node.SelectSingleNode("//div[@class='thumb']").Attributes("src").Value
            Year = CInt(node.SelectSingleNode("//div[@class='release-year']").Attributes("span").Value)
            Genres = ¿select multiple nodes?
            URL = node.SelectSingleNode("//div[@class='release']").Attributes("href").Value

        Next

    End Sub

End Class

Answer 1

你在这里的错误是尝试从你找到的那个中访问一个子节点的属性。

当您调用node.SelectSingleNode("//div[@class='release']")时，您会返回正确的div，但调用.Attributes只会返回div标记本身的属性，而不会返回任何内部HTML元素。

可以编写选择子节点的XPATH查询，例如： //div[@class='release']/a - 有关XPATH的更多信息，请参阅http://www.w3schools.com/xpath/xpath_syntax.asp。虽然示例适用于XML，但大多数原则应适用于HTML文档。

另一种方法是在您找到的节点上使用更多XPATH调用。我修改了你的代码，使其能够使用这种方法：

' Load the html document.
htmldoc.LoadHtml(IO.File.ReadAllText("C:\source.html"))

' Select the (10 items) nodes.
htmlnodes = htmldoc.DocumentNode.SelectNodes("//div[@class='item']")

' Loop through the nodes.
For Each node As HtmlAgilityPack.HtmlNode In htmlnodes

    Dim releaseNode = node.SelectSingleNode(".//div[@class='release']")
    'Assumes we find the node and it has a a-tag
    Title = releaseNode.SelectSingleNode(".//a").Attributes("title").Value
    URL = releaseNode.SelectSingleNode(".//a").Attributes("href").Value

    Dim thumbNode = node.SelectSingleNode(".//div[@class='thumb']")
    Cover = thumbNode.SelectSingleNode(".//img").Attributes("src").Value

    Dim releaseYearNode = node.SelectSingleNode(".//div[@class='release-year']")
    Year = CInt(releaseYearNode.SelectSingleNode(".//span").InnerText)

    Dim genreNode = node.SelectSingleNode(".//div[@class='genre']")
    Dim genreLinks = genreNode.SelectNodes(".//a")
    Genres = (From n In genreLinks Select n.InnerText).ToArray()

    Console.WriteLine("Title : {0}", Title)
    Console.WriteLine("Cover : {0}", Cover)
    Console.WriteLine("Year  : {0}", Year)
    Console.WriteLine("Genres: {0}", String.Join(",", Genres))
    Console.WriteLine("URL   : {0}", URL)

Next

请注意，在此代码中，我们假设文档已正确形成，并且每个节点/元素/属性都存在且正确。您可能希望为此添加大量错误检查，例如If someNode Is Nothing Then ....

编辑：我稍微修改了上面的代码，以确保每个.SelectSingleNode使用“.//”前缀 - 这确保它有效，如果有几个“项目”节点，否则它选择第一个匹配来自文档而不是当前节点。

如果您想要更短的XPATH解决方案，请使用以下方法使用相同的代码：

' Load the html document.
htmldoc.LoadHtml(IO.File.ReadAllText("C:\source.html"))

' Select the (10 items) nodes.
htmlnodes = htmldoc.DocumentNode.SelectNodes("//div[@class='item']")

' Loop through the nodes.
For Each node As HtmlAgilityPack.HtmlNode In htmlnodes

    Title = node.SelectSingleNode(".//div[@class='release']/h4/a[@title]").Attributes("title").Value
    URL = node.SelectSingleNode(".//div[@class='release']/h4/a[@href]").Attributes("href").Value

    Cover = node.SelectSingleNode(".//div[@class='thumb']/a/img[@src]").Attributes("src").Value

    Year = CInt(node.SelectSingleNode(".//div[@class='release-year']/span").InnerText)

    Dim genreLinks = node.SelectNodes(".//div[@class='genre']/a")
    Genres = (From n In genreLinks Select n.InnerText).ToArray()

    Console.WriteLine("Title : {0}", Title)
    Console.WriteLine("Cover : {0}", Cover)
    Console.WriteLine("Year  : {0}", Year)
    Console.WriteLine("Genres: {0}", String.Join(",", Genres))
    Console.WriteLine("URL   : {0}", URL)
    Console.WriteLine()

Next

Answer 2

你离解决方案的距离不远。两个重要的注意事项：

//是一个递归调用。它可能会产生一些严重的性能影响，并且它可能会选择您不想要的节点，所以我建议您只在层次结构很深或复杂或可变时使用它，并且您不想指定整个路径。 / LI>
在名为XmlNode的{{1}}上有一个有用的辅助方法，即使它不存在，您也会得到一个属性（您需要指定默认值）。

以下是一个似乎有用的示例：

GetAttributeValue

使用HTMLAgilityPack库检索属性和跨度

2 个答案: