我想创建一个函数来修改XML内容而不改变格式。我设法改变了文本,但我不能在不改变XML格式的情况下做到这一点.所以现在,我想做的是在XML文件中的CDATA之前和之后添加空间。
默认的XML文件。
<?xml version="1.0" encoding="utf-8"?>
<Mapsxmlns="http://www.semi.org">
<Map>
<Device>
<ReferenceDevice/>
<Bin>
<Bin Bin="001"/>
</Bin>
<Data>
<Row> <![CDATA[001 001 001]]> </Row>
</Data>
</Device>
</Map>
</Maps>
我得到了这个结果。
<?xml version="1.0" encoding="utf-8"?>
<Mapsxmlns="http://www.semi.org">
<Map>
<Device>
<ReferenceDevice/>
<Bin>
<Bin Bin="001"/>
</Bin>
<Data>
<Row><![CDATA[001 001 099]]></Row>
</Data>
</Device>
</Map>
</Maps>
然而,我希望新的xml是这样的。
<?xml version="1.0" encoding="utf-8"?>
<Mapsxmlns="http://www.semi.org">
<Map>
<Device>
<ReferenceDevice/>
<Bin>
<Bin Bin="001"/>
</Bin>
<Data>
<Row> <![CDATA[001 001 099]]> </Row>
</Data>
</Device>
</Map>
</Maps>
这是我的代码
from lxml import etree as ET
def xml_new(f,fpath,newtext,xmlrow):
xmlrow = 19
parser = ET.XMLParser(strip_cdata=False)
tree = ET.parse(f, parser)
root = tree.getroot()
for child in root:
value = child[0][2][xmlrow].text
text = ET.CDATA("001 001 099")
child[0][2][xmlrow] = ET.Element('Row')
child[0][2][xmlrow].text = text
child[0][2][xmlrow].tail = "\n"
ET.register_namespace('A', "http://www.semi.org")
tree.write(fpath,encoding='utf-8',xml_declaration=True)
return value
有谁能帮帮我吗?
我不太明白你想做什么。这里有一个例子给你。我不知道它是否能满足你的需求。
from simplified_scrapy import SimplifiedDoc,req,utils
html ='''<?xml version="1.0" encoding="utf-8"?>
<Mapsxmlns="http://www.semi.org">
<Map>
<Device>
<ReferenceDevice/>
<Bin>
<Bin Bin="001"/>
</Bin>
<Data>
<Row> <![CDATA[001 001 001]]> </Row>
</Data>
</Device>
</Map>
</Maps>'''
doc = SimplifiedDoc(html)
row = doc.Data.Row # Get the node you want to modify.
row.setContent(" "+row.html+" ") # Modify the node content.
print (doc.html)
结果。
<?xml version="1.0" encoding="utf-8"?>
<Mapsxmlns="http://www.semi.org">
<Map>
<Device>
<ReferenceDevice />
<Bin>
<Bin Bin="001" />
</Bin>
<Data>
<Row> <![CDATA[001 001 001]]> </Row>
</Data>
</Device>
</Map>
</Maps>
谢谢你的帮助 我已经找到了另一种方法来实现我想要的结果。
这是代码。
# what you want to change
replaceby = '020]]> </Row>\n'
# row you want to change
row = 1
# col you want to change based on list
col = 3
file = open(file,'r')
line = file.readlines()
i = 0
editedXML=[]
for l in line:
if 'cdata' in l.lower():
i=i+1
if i == row:
oldVal = l.split(' ')
newVal = []
for index, old in enumerate(oldVal):
if index == col:
newVal.append(replaceby)
else:
newVal.append(old)
editedXML.append(' '.join(newVal))
else:
editedXML.append(l)
else:
editedXML.append(l)
file2 = open(newfile,'w')
file2.write(''.join(editedXML))
file2.close()