使用 ElementTree 递归 XML 解析 python

Question

我正在尝试使用 Python ElementTree 解析以下 XML 以生成如下输出。我正在尝试为顶级元素编写模块来打印它们。然而，这有点棘手，因为类别元素可能有也可能没有属性，而类别元素内部可能有类别元素。

我在本主题中提到了上一个问题，但它们不包含同名的嵌套元素

work.xml:
<suite id="1" name="MainApplication">
<displayNameKey>my Application</displayNameKey>
<displayName>my Application</displayName>
<application id="2" name="Sub Application1">
<displayNameKey>sub Application1</displayNameKey>
<displayName>sub Application1</displayName>
<category id="2423" name="about">
<displayNameKey>subApp.about</displayNameKey>
<displayName>subApp.about</displayName>
<category id="2423" name="comms">
<displayNameKey>subApp.comms</displayNameKey>
<displayName>subApp.comms</displayName>
<property id="5909" name="copyright" type="string_property" width="40">
<value>2014</value>
</property>
<property id="5910" name="os" type="string_property" width="40">
<value>Linux 2.6.32-431.29.2.el6.x86_64</value>
</property>
</category>
<property id="5908" name="releaseNumber" type="string_property" width="40">
<value>9.1.0.3.0.54</value>
</property>
</category>
</application>
</suite>

输出应如下所示：

Suite: MainApplication
    Application: Sub Application1
        Category: about
            property: releaseNumber | 9.1.0.3.0.54
            category: comms
                property: copyright | 2014
                property: os | Linux 2.6.32-431.29.2.el6.x86_64

任何正确方向的指示都会有所帮助。

Answer 1

import xml.etree.ElementTree as ET
tree = ET.ElementTree(file='work.xml')

indent = 0
ignoreElems = ['displayNameKey', 'displayName']

def printRecur(root):
    """Recursively prints the tree."""
    if root.tag in ignoreElems:
        return
    print ' '*indent + '%s: %s' % (root.tag.title(), root.attrib.get('name', root.text))
    global indent
    indent += 4
    for elem in root.getchildren():
        printRecur(elem)
    indent -= 4

root = tree.getroot()
printRecur(root)

输出：

Suite: MainApplication
    Application: Sub Application1
        Category: about
            Category: comms
                Property: copyright
                    Value: 2014
                Property: os
                    Value: Linux 2.6.32-431.29.2.el6.x86_64
            Property: releaseNumber
                Value: 9.1.0.3.0.54

这是我能在 5 分钟内到达的最接近的位置。您应该 递归地调用一个处理器函数 就可以了。您可以从这一点开始改进:)

您还可以为每个标签定义处理函数，并将它们全部放入字典中以便于查找。然后你可以检查你是否有一个适合该标签的处理函数，然后调用它，否则继续盲目打印。例如：

HANDLERS = {
    'property': 'handle_property',
    <tag_name>: <handler_function>
}

def handle_property(root):
    """Takes property root element and prints the values."""
    data = ' '*indent + '%s: %s ' % (root.tag.title(), root.attrib['name'])
    values = []
    for elem in root.getchildren():
        if elem.tag == 'value':
            values.append(elem.text)
    print data + '| %s' % (', '.join(values))

# printRecur would get modified accordingly.
def printRecur(root):
    """Recursively prints the tree."""
    if root.tag in ignoreElems:
        return

    global indent
    indent += 4

    if root.tag in HANDLERS:
        handler = globals()[HANDLERS[root.tag]]
        handler(root)
    else:
        print ' '*indent + '%s: %s' % (root.tag.title(), root.attrib.get('name', root.text))
        for elem in root.getchildren():
            printRecur(elem)

    indent -= 4

上面的输出：

Suite: MainApplication
    Application: Sub Application1
        Category: about
            Category: comms
                Property: copyright | 2014
                Property: os | Linux 2.6.32-431.29.2.el6.x86_64
            Property: releaseNumber | 9.1.0.3.0.54

我发现这非常有用，而不是在代码中放入大量 if/else。

Answer 2

如果您想要一种通用的 xml 导入器，请为每个 xml 元素创建一条记录

import pandas as pd
import xml.etree.ElementTree as ET

tree = ET.parse('file.xml')
root = tree.getroot()


def rij(elem,level,tags,rtag,mtag,keys,rootkey,data):
    otag=mtag
    mtag=elem.tag
    mtag=mtag[mtag.rfind('}')+1:]
    tags.append(mtag)    
    if level==1:
        rtag=mtag
        if elem.keys() is not None:
            mkey=[]
            if len(elem.keys())>1:
                for key in elem.keys():
                    mkey.append(elem.attrib.get(key))
                    rootkey=mkey
            else:
                for key in elem.keys():
                    rootkey=elem.attrib.get(key)
    else:
        if elem.keys() is not None:
                mkey=[]
                lkey=[]
                for key in elem.keys():
                    if len(elem.keys())>1:
                        mkey.append(elem.attrib.get(key))
                        keys=mkey
                else:
                    for key in elem.keys():
                        keys=elem.attrib.get(key)  
                        lkey=key

    if elem.text is not None:
        if elem.text!='\n        ':
            data.append([rootkey,tags,rtag,otag,mtag,lkey,keys,elem.text])
    else:
            data.append([rootkey,tags,rtag,otag,mtag,lkey,keys,''])

            #print(data)
    level+=1
    for chil in elem.getchildren():
            data = rij(chil, level,tags,rtag,mtag, keys,rootkey,data)
    level-=1
    mtag=elem.tag
    mtag=mtag[mtag.rfind('}')+1:]    
    tags.remove(mtag)
    return data


data = rij(root,0,[],'','', [],[],[])

Answer 3

如果你想要一个准系统 XML 递归树解析器片段：

from xml.etree import ElementTree
tree = ElementTree.parse('english_saheeh.xml')
root = tree.getroot()
def walk_tree_recursive(root):
    #do whatever with .tags here
    for child in root:
        walk_tree_recursive(child)
walk_tree_recursive(root)

使用 ElementTree 递归 XML 解析 python

Recursive XML parsing python using ElementTree

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