4

我已经成功使用 xml.etree.ElementTree 来解析 xml,搜索内容,然后将其写入不同的 xml。但是,我只是在单个标签内处理文本。

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”标签之间的所有内容并替换为上面的内容。

基本上,我之前在做什么,但忽略“attr”标签之间的子元素、属性等......并将其视为文本。

谢谢!!

请多多包涵,这个论坛和我以前的(发帖)有点不同!

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)

得到这个工作,见下文,甚至想出了如何导出到新的 .xml 但是,如果 attr 的 # 是差异。从源到目标,我收到以下错误,有什么建议吗?

节点=替换.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"

2011 年 5 月 16 日更新 重组了一些东西来修复上面提到的“IndexError:从空列表中弹出”错误。意识到替换“attr”标签并不总是一对一的替换。例如。有时源 .xml 有 20 个 attr,而目标 .xml 有 25 个 attr。在这种情况下,一对一的替换会窒息。

无论如何,下面将删除所有 attr,然后替换为源 attr。它还检查另一个标签,如果存在“子类型”,它会将它们添加到 attr 之后,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"
4

2 回答 2

1

这是一个lxml用于执行此操作的示例。我不确定您希望如何替换<attr/>节点,但是这个示例应该提供一个可以重用的模式。

更新-我将其更改为将 tree2 中的每个替换<attr>为 tree1 中的相应节点,按文档顺序:

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>
于 2011-05-13T15:53:48.630 回答
0

这听起来像是 XSL-T 转换的目的。你试过吗?

我还推荐一个像 Beautiful Soup 这样的库来解析和操作 XML。

于 2011-05-13T14:23:10.777 回答