使用 ElementTree 的递归 XML 解析 python

Question

我正在尝试使用 Python ElementTree 解析下面的 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。但是，这有点棘手，因为类别元素可能有也可能没有属性，并且类别元素内部可能有一个类别元素。

我在本主题中提到过上一个问题，但它们不包含同名的嵌套元素

我的代码： http://pastebin.com/Fsv2Xzqf

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>

输出应该如下：

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

任何指向正确方向的指针都会有所帮助。

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 5 分钟内最接近的时间。您应该只递归调用一个处理器函数，这样就可以了。你可以从这一点改进:)

---

您还可以为每个标签定义处理函数并将它们全部放在字典中以便于查找。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，否则只是继续盲目打印。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，而不是在代码中添加大量 if/else。

Answer 2

如果你想要一个准系统 XML 递归树解析器 sn-p：

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)

Answer 3

如果你想要一种通用的 xml 导入器，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,[],'','', [],[],[])