我成功使用xml.etree.ElementTree来解析xml，搜索内容，然后将其写入不同的xml。但是，我只是在一个标签内部处理文本。
import os, sys, glob, xml.etree.ElementTree as ET
path = r"G:\\63D RRC GIS Data\\metadata\\general\\2010_contract"
for fn in os.listdir(path):
filepaths = glob.glob(path + os.sep + fn + os.sep + "*overall.xml")
for filepath in filepaths:
(pa, filename) = os.path.split(filepath)
####use this section to grab element text from old, archived metadata files; this text then gets put into current, working .xml###
root = ET.parse(pa + os.sep + "archive" + os.sep + "base_metadata_overall.xml").getroot()
iterator = root.getiterator()
for item in iterator:
if item.tag == "abstract":
correct_abstract = item.text
root2 = ET.parse(pa + os.sep + "base_metadata_overall.xml").getroot()
iterator2 = root2.getiterator("descript")
for item in iterator2:
if item.tag == "abstract":
old_abstract = item.find("abstract")
old_abstract_text = old_abstract.text
item.remove(old_abstract)
new_symbol_abstract = ET.SubElement(item, "title")
new_symbol_abstract.text = correct_abstract
tree = ET.ElementTree(root2)
tree.write(pa + os.sep + "base_metadata_overall.xml")
print "created --- " + filename + " metadata"
但现在，我需要：
1）搜索xml并抓取“attr”标签之间的所有内容，下面是示例：
<attr><attrlabl Sync="TRUE">OBJECTID</attrlabl><attalias Sync="TRUE">ObjectIdentifier</attalias><attrtype Sync="TRUE">OID</attrtype><attwidth Sync="TRUE">4</attwidth><atprecis Sync="TRUE">0</atprecis><attscale Sync="TRUE">0</attscale><attrdef Sync="TRUE">Internal feature number.</attrdef></attr>
2）现在，我需要打开一个不同的xml并搜索相同“attr”标记之间的所有内容，并替换为上面的内容。
基本上，我以前做过的，但忽略了“attr”标签之间的子元素，属性等等，并将其视为文本。
谢谢！
请耐心等待，这个论坛有点不同（发帖）然后我习惯了！
这是我到目前为止所拥有的：
import os, sys, glob, re, xml.etree.ElementTree as ET
from lxml import etree
path = r"C:\\temp\\python\\xml"
for fn in os.listdir(path):
filepaths = glob.glob(path + os.sep + fn + os.sep + "*overall.xml")
for filepath in filepaths:
(pa, filename) = os.path.split(filepath)
xml = open(pa + os.sep + "attributes.xml")
xmltext = xml.read()
correct_attrs = re.findall("<attr> (.*?)</attr>",xmltext,re.DOTALL)
for item in correct_attrs:
correct_attribute = "<attr>" + item + "</attr>"
xml2 = open(pa + os.sep + "base_metadata_overall.xml")
xmltext2 = xml2.read()
old_attrs = re.findall("<attr>(.*?)</attr>",xmltext,re.DOTALL)
for item2 in old_attrs:
old_attribute = "<attr>" + item + "</attr>"
old = etree.fromstring(old_attribute)
replacement = new.xpath('//attr')
for attr in old.xpath('//attr'):
attr.getparent().replace(attr, copy.deepcopy(replacement))
print lxml.etree.tostring(old)
让这个工作，见下文，甚至想出如何导出到新的.xml 但是，如果attr的数量是不同的。从source到dest，我得到以下错误，有什么建议吗？
node = replacements.pop（）
IndexError：从空列表中弹出
import os, sys, glob, re, copy, lxml, xml.etree.ElementTree as ET
from lxml import etree
path = r"C:\\temp\\python\\xml"
for fn in os.listdir(path):
filepaths = glob.glob(path + os.sep + fn + os.sep + "*overall.xml")
for filepath in filepaths:
xmlatributes = open(pa + os.sep + "attributes.xml")
xmlatributes_txt = xmlatributes.read()
xmltarget = open(pa + os.sep + "base_metadata_overall.xml")
xmltarget_txt = xmltarget.read()
source = lxml.etree.fromstring(xmlatributes_txt)
dest = lxml.etree.fromstring(xmltarget_txt)
replacements = source.xpath('//attr')
replacements.reverse()
for attr in dest.xpath('//attr'):
node = replacements.pop()
attr.getparent().replace(attr, copy.deepcopy(node))
#print lxml.etree.tostring(dest)
tree = ET.ElementTree(dest)
tree.write (pa + os.sep + "edited_metadata.xml")
print fn + "--- sucessfully edited"
更新2011年5月16日 重组了一些东西来修复上面提到的“IndexError：pop from empty list”错误。意识到更换“attr”标签并不总是一对一的替代品。对于前者有时源.xml有20个attr's，目的地.xml有25个attr's。在这种情况下，1比1的替换会窒息。
无论如何，下面将删除所有的attr，然后用源attr替换。它还检查另一个标签，“子类型”（如果存在），它在attr之后添加它们，但在“详细”标签内。
再次感谢所有帮助过的人。
import os, sys, glob, re, copy, lxml, xml.etree.ElementTree as ET
from lxml import etree
path = r"G:\\63D RRC GIS Data\\metadata\\general\\2010_contract"
#path = r"C:\\temp\python\\xml"
for fn in os.listdir(path):
correct_title = fn.replace ('_', ' ') + " various facilities"
correct_fc_name = fn.replace ('_', ' ')
filepaths = glob.glob(path + os.sep + fn + os.sep + "*overall.xml")
for filepath in filepaths:
print "-----" + fn + "-----"
(pa, filename) = os.path.split(filepath)
xmlatributes = open(pa + os.sep + "attributes.xml")
xmlatributes_txt = xmlatributes.read()
xmltarget = open(pa + os.sep + "base_metadata_overall.xml")
xmltarget_txt = xmltarget.read()
source = lxml.etree.fromstring(xmlatributes_txt)
dest = lxml.etree.fromstring(xmltarget_txt)
replacements = source.xpath('//attr')
replacesubtypes = source.xpath('//subtype')
subtype_true_f = len(replacesubtypes)
attrtag = dest.xpath('//attr')
#print len(attrtag)
num_realatrs = len(replacements)
for n in attrtag:
n.getparent().remove(n)
print n.tag + " removed"
detailedtag = dest.xpath('//detailed')
for n2 in detailedtag:
pos = 0
for realatrs in replacements:
n2.insert(pos + 1, realatrs)
print "attr's replaced"
if subtype_true_f >= 1:
#print subtype_true_f
for realsubtypes in replacesubtypes:
n2.insert(num_realatrs + 1, realsubtypes)
print "subtype's replaced"
tree = ET.ElementTree(dest)
tree.write (pa + os.sep + "base_metadata_overall_v2.xml")
print fn + "--- sucessfully edited"
答案 0 :(得分：1)
以下是使用lxml执行此操作的示例。我不是完全确定您希望如何替换<attr/>个节点，但此示例应提供您可以重复使用的模式。
更新 - 我更改了它，用tree1中的相应节点替换tree2中的每个<attr>，按文档顺序排列：
import copy
import lxml.etree
xml1 = '''<root><attr><chaos foo="0"/></attr><attr><arena foo="1"/></attr></root>'''
xml2 = '''<tree><attr><one/></attr><attr><two/></attr></tree>'''
tree1 = lxml.etree.fromstring(xml1)
tree2 = lxml.etree.fromstring(xml2)
# select <attr/> nodes from tree1, will be used to replace corresponding
# nodes in tree2
replacements = tree1.xpath('//attr')
replacements.reverse()
for attr in tree2.xpath('//attr'):
# replace the attr node in tree2 with 'replacement' from tree1
node = replacements.pop()
attr.getparent().replace(attr, copy.deepcopy(node))
print lxml.etree.tostring(tree2)
结果：
<tree>
<attr><chaos foo="0"/></attr>
<attr><arena foo="1"/></attr>
</tree>
答案 1 :(得分：0)
这听起来像XSL-T转换的内容。你试过了吗？
我还推荐像Beautiful Soup这样的库来解析和操作XML。