带有dict元素的Plist XPath查询

Question

我正在尝试通过 Nokigiri 从 plist 的 iTunes 库导出中加载歌曲名称：

doc = Nokogiri::XML(open(file.path))

@songs = Array.new  
doc.xpath(<XPATH_HERE>).each do |n|
  @songs.push(n)  #append data to array
end 

plist 的开头如下所示：

<plist version="1.0">
<dict>
  <key>Major Version</key><integer>1</integer>
  <key>Minor Version</key><integer>1</integer>
  <key>Date</key><date>2014-10-15T22:52:19Z</date>
  <key>Application Version</key><string>11.4</string>
  <key>Features</key><integer>5</integer>
  <key>Show Content Ratings</key><true/>
  <key>Music Folder</key><string>file://localhost/Users/mike/Music/iTunes/iTunes%20Media/</string>
  <key>Library Persistent ID</key><string>280B84572DDCF406</string>
  <key>Tracks</key>
  <dict>
    <key>96</key>
    <dict>
      <key>Track ID</key><integer>96</integer>
      <key>Name</key><string>Get Lucky (Daft Punk cover)</string>
      <key>Artist</key><string>Daughter</string>
      <key>Kind</key><string>MPEG audio file</string>
      <key>Size</key><integer>4716638</integer>
      <key>Total Time</key><integer>294112</integer>
      <key>Date Modified</key><date>2013-11-12T20:54:14Z</date>
      <key>Date Added</key><date>2013-12-18T17:56:09Z</date>
      <key>Bit Rate</key><integer>128</integer>
      <key>Sample Rate</key><integer>44100</integer>
      <key>Persistent ID</key><string>C3B1B6F26134C9C1</string>
      <key>Track Type</key><string>File</string>
      <key>Location</key><string>file://localhost/Users/mike/Music/iTunes/iTunes%20Media/Music/Daughter/Unknown%20Album/Get%20Lucky%20(Daft%20Punk%20cover).mp3</string>
      <key>File Folder Count</key><integer>5</integer>
      <key>Library Folder Count</key><integer>1</integer>
    </dict>
    <key>98</key>
    <dict>
      <key>Track ID</key><integer>98</integer>
      <key>Name</key><string>Swimming in Solace (DJ Fergie Ferg Remash)</string>
      <key>Kind</key><string>MPEG audio file</string>

我希望从每个曲目加载的是名称键后面的曲目名称字符串。我认为应该工作的 XPath 是

/plist/dict[key[. = 'Tracks']/following-sibling::*[1]]/dict[key/following-sibling::*[1]]/dict[key[. = 'Name']/following-sibling::*[1]]/string

XPath 返回：

<string>Get Lucky (Daft Punk cover)</string>
<string>Daughter</string>
<string>MPEG audio file</string>
<string>C3B1B6F26134C9C1</string>
<string>File</string>
<string>file://localhost/Users/mike/Music/iTunes/iTunes%20Media/Music/Daughter/Unknown%20Album/Get%20Lucky%20(Daft%20Punk%20cover).mp3</string>
<string>Swimming in Solace (DJ Fergie Ferg Remash)</string>
<string>MPEG audio file</string>

看起来，虽然我的 XPath 为每个字符串指定了键，但实际上它正在采用所有每个 dict 的“跟随兄弟”。

我该怎么做才能使查询更具体，以便 plist 的这一部分返回：

Get Lucky (Daft Punk cover)

和

Swimming in Solace (DJ Fergie Ferg Remash)

Answer 1

这是一种可能的 XPath ：

/plist/dict[key='Tracks']/dict/dict/key[.='Name']/following-sibling::string[1]

XPath 的开头可能会有所不同，但我认为最重要的部分是最后两个路径步骤 ( key[.='Name']/following-sibling::string[1])。它告诉<string>在每个<key>Name</key>元素之后获取最近的元素。