os.walk（）python：目录结构的xml表示，递归

我建议您不要使用

os.walk()

os.listdir()

os.path.join()

os.path.isdir()

import osfrom xml.sax.saxutils import escape as xml_escapedef DirAsXML(path):    result = '<dir>n<name>%s</name>n' % xml_escape(os.path.basename(path))    dirs = []    files = []    for item in os.listdir(path):        itempath = os.path.join(path, item)        if os.path.isdir(itempath): dirs.append(item)        elif os.path.isfile(itempath): files.append(item)    if files:        result += '  <files>n'  + 'n'.join('    <file>n      <name>%s</name>n    </file>' % xml_escape(f) for f in files) + 'n  </files>n'    if dirs:        for d in dirs: x = DirAsXML(os.path.join(path, d)) result += 'n'.join('  ' + line for line in x.split('n'))    result += '</dir>'    return resultif __name__ == '__main__':    print '<structure>n' + DirAsXML(os.getcwd()) + 'n</structure>'

就个人而言，我建议使用一种较为简单的XML模式，将名称放入属性中并摆脱该

<files>

import osfrom xml.sax.saxutils import quoteattr as xml_quoteattrdef DirAsLessXML(path):    result = '<dir name=%s>n' % xml_quoteattr(os.path.basename(path))    for item in os.listdir(path):        itempath = os.path.join(path, item)        if os.path.isdir(itempath): result += 'n'.join('  ' + line for line in      DirAsLessXML(os.path.join(path, item)).split('n'))        elif os.path.isfile(itempath): result += '  <file name=%s />n' % xml_quoteattr(item)    result += '</dir>'    return resultif __name__ == '__main__':    print '<structure>n' + DirAsLessXML(os.getcwd()) + 'n</structure>'

这给出了如下输出：

<structure><dir name="local">  <dir name=".hg">    <file name="00changelog.i" />    <file name="branch" />    <file name="branch.cache" />    <file name="dirstate" />    <file name="hgrc" />    <file name="requires" />    <dir name="store">      <file name="00changelog.i" />

等等

如果

os.walk()

expat