用Python打印XML的最佳方法(或各种方法)是什么?


当前回答

如果由于某种原因你无法获得其他用户提到的任何Python模块,我建议使用以下Python 2.7解决方案:

import subprocess

def makePretty(filepath):
  cmd = "xmllint --format " + filepath
  prettyXML = subprocess.check_output(cmd, shell = True)
  with open(filepath, "w") as outfile:
    outfile.write(prettyXML)

据我所知,这种解决方案适用于安装了xmllint包的基于unix的系统。

其他回答

另一个解决方案是借用这个缩进函数,用于自2.5以来内置在Python中的ElementTree库。 下面是它的样子:

from xml.etree import ElementTree

def indent(elem, level=0):
    i = "\n" + level*"  "
    j = "\n" + (level-1)*"  "
    if len(elem):
        if not elem.text or not elem.text.strip():
            elem.text = i + "  "
        if not elem.tail or not elem.tail.strip():
            elem.tail = i
        for subelem in elem:
            indent(subelem, level+1)
        if not elem.tail or not elem.tail.strip():
            elem.tail = j
    else:
        if level and (not elem.tail or not elem.tail.strip()):
            elem.tail = j
    return elem        

root = ElementTree.parse('/tmp/xmlfile').getroot()
indent(root)
ElementTree.dump(root)

我试图编辑上面“ade”的回答,但在我最初匿名提供反馈后,Stack Overflow不让我编辑。这是一个bug较少的函数版本,用于漂亮地打印一个ElementTree。

def indent(elem, level=0, more_sibs=False):
    i = "\n"
    if level:
        i += (level-1) * '  '
    num_kids = len(elem)
    if num_kids:
        if not elem.text or not elem.text.strip():
            elem.text = i + "  "
            if level:
                elem.text += '  '
        count = 0
        for kid in elem:
            indent(kid, level+1, count < num_kids - 1)
            count += 1
        if not elem.tail or not elem.tail.strip():
            elem.tail = i
            if more_sibs:
                elem.tail += '  '
    else:
        if level and (not elem.tail or not elem.tail.strip()):
            elem.tail = i
            if more_sibs:
                elem.tail += '  '

如果由于某种原因你无法获得其他用户提到的任何Python模块,我建议使用以下Python 2.7解决方案:

import subprocess

def makePretty(filepath):
  cmd = "xmllint --format " + filepath
  prettyXML = subprocess.check_output(cmd, shell = True)
  with open(filepath, "w") as outfile:
    outfile.write(prettyXML)

据我所知,这种解决方案适用于安装了xmllint包的基于unix的系统。

from lxml import etree
import xml.dom.minidom as mmd

xml_root = etree.parse(xml_fiel_path, etree.XMLParser())

def print_xml(xml_root):
    plain_xml = etree.tostring(xml_root).decode('utf-8')
    urgly_xml = ''.join(plain_xml .split())
    good_xml = mmd.parseString(urgly_xml)
    print(good_xml.toprettyxml(indent='    ',))

它的工作良好的xml与中国!

import xml.dom.minidom

dom = xml.dom.minidom.parse(xml_fname) # or xml.dom.minidom.parseString(xml_string)
pretty_xml_as_string = dom.toprettyxml()