Python：具有多个根树子节点的大型 XML 解析 [重复]

Question

需要用python（xml.etree.ElementTree）解析大的XML文件来处理和生成报告，如预期部分所示。

我无法弄清楚如何在第 4 级之前了解某些细节，然后再从第 5 级了解相关/相关数据。在哪里循环以及如何引用孩子，是我的问题。有什么建议请大家多多指教，谢谢。

Input XML File: raw_data.xml

<?xml version="1.0" encoding="ISO-8859-1"?>
<FirstLevel Flevel="my1">
    <SecondLevel Slevel="my2">
        <ThirdLevel Tlevel="my3">
            <FourthLevel test="1" mydata="Needed1">
                <FifthLevel associated="Required for Needed1"/>
            </Fourthlevel>  
            <FourthLevel test="2" mydata="Needed2">
                <FifthLevel associated="Required for Needed2"/>
            </Fourthlevel>  
            <FourthLevel test="3" mydata="Needed3">
                <FifthLevel associated="Required for Needed3-1"/>
                <FifthLevel associated="Required for Needed3-2"/>
            </Fourthlevel>  
            <FourthLevel test="4" mydata="Needed4">
                <FifthLevel associated="Required for Needed4-1"/>
                <FifthLevel associated="Required for Needed4-2"/>
            </Fourthlevel>  
        </ThirdLevel>
    </SecondLevel>
</FirstLevel>
-----------------------------------------------------------

My Code:

    import xml.etree.ElementTree as ET
    tree = ET.parse('raw_data.xml')
    root=tree.getroot()
    mylevel=root.findall('.//FourthLevel')
    for i in mylevel:
        print ("mydata=",i.get('mydata'),"\t")
        assoc=root.findall('.//FifthLevel') ### assoc: Temporary variable for associated data
        for j in assoc:
             print ("associated=",j.get('associated'),"\n")




Output: final_output.txt

mydata=Needed1  associated=Required for Needed1
mydata=Needed2  associated=Required for Needed2
mydata=Needed3  associated=Required for Needed3-1
mydata=Needed3  associated=Required for Needed3-1
mydata=Needed4  associated=Required for Needed4-1
mydata=Needed4  associated=Required for Needed4-1

Answer 1

您已经在迭代 root 的子节点，名称为 ".//FourthLevel"。您只需对每个名为 "FifthLevel" 的子项及其子项应用相同的原则（请注意缺少 斜杠es）。

翻译成代码，只需要换行：

assoc=root.findall('.//FifthLevel')

作者：

assoc = i.findall("FifthLevel")

因为您只需要 当前 节点（第 4^th 级）的第 5^th 级子节点，而不是整个树。更多详情请查看[Python 3.Docs]: xml.etree.ElementTree - The ElementTree XML API。