Question

在我的XML文件中，我有一些Windows标签，我想从中以适当的窗口名称逐一提取视点的名称。

我的XML文件：

<windows>
<window class='dashboard' maximized='true' name='Dashboard 1'>
      <viewpoints>
        <viewpoint name='Category sheet'>
          <zoom type='entire-view' />
        </viewpoint>
        <viewpoint name='Segment Sheet'>
          <zoom type='entire-view' />
          <selection-collection>
            <tuple-selection>
              <tuple-reference>
                <tuple-descriptor>
                  <pane-descriptor>
                    <x-fields>
                      <field>[federated.1y3sjvb0pyupci132wn6b0wdgpc3].[none:Segment:nk]</field>
                    </x-fields>
                    <y-fields>
                      <field>[federated.1y3sjvb0pyupci132wn6b0wdgpc3].[sum:Sales:qk]</field>
                    </y-fields>
                  </pane-descriptor>
                  <columns>
                    <field>[federated.1y3sjvb0pyupci132wn6b0wdgpc3].[none:Segment:nk]</field>
                    <field>[federated.1y3sjvb0pyupci132wn6b0wdgpc3].[sum:Sales:qk]</field>
                  </columns>
                </tuple-descriptor>
                <tuple>
                  <value>&quot;Corporate&quot;</value>
                  <value>706146.36680000008</value>
                </tuple>
              </tuple-reference>
            </tuple-selection>
          </selection-collection>
        </viewpoint>
        <viewpoint name='Subcat Sheet'>
          <zoom type='entire-view' />
        </viewpoint>
        <viewpoint name='sub cat grp Sheet'>
          <zoom type='entire-view' />
        </viewpoint>
      </viewpoints>
      <active id='4' />
    </window>
    <window class='dashboard' name='Story 1'>
      <viewpoints />
      <active id='4' />
    </window>
  </windows>

我的尝试：

for win in root.findall('./windows/window'):
                    dashwins = win.find('window[@class="dashboard"]') 
                    if dashwins != None:
                        print(dashwins.attrib)
                        for i in dashwins:
                            view1 = i.find('viewpoint')
                            for j in view1:
                                print(j.get('name'))

当且仅当在window标记内存在class = dashboard属性时，才需要提取视点名称。

Answer 1

您可以扩展xpath表达式以获取目标viewpoint元素下的所有window元素，然后询问他们的name属性。

>>> from xml.etree import ElementTree
>>> tree = ElementTree.parse('sanu.xml')
>>> for el in tree.findall('.//window[@class="dashboard"]//viewpoint'):
...     el.attrib['name']
... 
'Category sheet'
'Segment Sheet'
'Subcat Sheet'
'sub cat grp Sheet'

使用Etree Python提取XML元素

1 个答案: