我有一个简单的XML文档,如下所示:
<Lineup>
<Player>
<GameID>20150010</GameID>
<GameDate>2015-04-06T00:00:00-04:00</GameDate>
<DH>0</DH>
<TeamId>16</TeamId>
<PlayerId>458913</PlayerId>
<LineupPosition>1</LineupPosition>
<DefensivePosition>8</DefensivePosition>
</Player>
<Player>
<GameID>20150010</GameID>
<GameDate>2015-04-06T00:00:00-04:00</GameDate>
<DH>0</DH>
<TeamId>16</TeamId>
<PlayerId>607054</PlayerId>
<LineupPosition>2</LineupPosition>
<DefensivePosition>4</DefensivePosition>
</Player>
<Player>
<GameID>20150010</GameID>
<GameDate>2015-04-06T00:00:00-04:00</GameDate>
<DH>0</DH>
<TeamId>16</TeamId>
<PlayerId>455976</PlayerId>
<LineupPosition>3</LineupPosition>
<DefensivePosition>9</DefensivePosition>
</Player>
</Lineup>
我尝试遍历每个播放器对象并单独访问每个子节点及其各自的值。我正在尝试以下内容:
xml_doc.xpath("//Lineup/Player").each do |lineup|
puts "lineup: #{lineup.inspect}"
end
我不完全确定如何访问lineup
这里的单个子元素。使用nokogiri最好的方法是什么?
答案 0 :(得分:2)
尝试使用XPath /Lineup/Player/*
检查每个“Player”节点的每个子元素:
doc = Nokogiri::XML(File.read('my.xml'))
doc.xpath('/Lineup/Player/*').each do |node|
puts "#{node.name}: #{node.text}"
end
# GameID: 20150010
# GameDate: 2015-04-06T00:00:00-04:00
# DH: 0
# ...etc...
或者,您可以选择每个“播放器”并迭代其element children (by using #elements
or #element_children
):
doc.xpath('/Lineup/Player').each do |player|
puts "-- NEXT PLAYER --"
player.elements.each do |node|
puts "#{node.name}: #{node.text}"
end
end