我在XML中有很多行,我试图获得一个特定节点属性的实例。

<foo>
   <bar>
      <type foobar="1"/>
      <type foobar="2"/>
   </bar>
</foo>

我如何访问属性foobar的值?在这个例子中,我想要“1”和“2”。


当前回答

simplified_scrapy:一个新的库,我使用后就爱上了它。我向你推荐。

from simplified_scrapy import SimplifiedDoc
xml = '''
<foo>
   <bar>
      <type foobar="1"/>
      <type foobar="2"/>
   </bar>
</foo>
'''

doc = SimplifiedDoc(xml)
types = doc.selects('bar>type')
print (len(types)) # 2
print (types.foobar) # ['1', '2']
print (doc.selects('bar>type>foobar()')) # ['1', '2']

这里有更多的例子。这个库很容易使用。

其他回答

#If the xml is in the form of a string as shown below then
from lxml  import etree, objectify
'''sample xml as a string with a name space {http://xmlns.abc.com}'''
message =b'<?xml version="1.0" encoding="UTF-8"?>\r\n<pa:Process xmlns:pa="http://xmlns.abc.com">\r\n\t<pa:firsttag>SAMPLE</pa:firsttag></pa:Process>\r\n'  # this is a sample xml which is a string


print('************message coversion and parsing starts*************')

message=message.decode('utf-8') 
message=message.replace('<?xml version="1.0" encoding="UTF-8"?>\r\n','') #replace is used to remove unwanted strings from the 'message'
message=message.replace('pa:Process>\r\n','pa:Process>')
print (message)

print ('******Parsing starts*************')
parser = etree.XMLParser(remove_blank_text=True) #the name space is removed here
root = etree.fromstring(message, parser) #parsing of xml happens here
print ('******Parsing completed************')


dict={}
for child in root: # parsed xml is iterated using a for loop and values are stored in a dictionary
    print(child.tag,child.text)
    print('****Derving from xml tree*****')
    if child.tag =="{http://xmlns.abc.com}firsttag":
        dict["FIRST_TAG"]=child.text
        print(dict)


### output
'''************message coversion and parsing starts*************
<pa:Process xmlns:pa="http://xmlns.abc.com">

    <pa:firsttag>SAMPLE</pa:firsttag></pa:Process>
******Parsing starts*************
******Parsing completed************
{http://xmlns.abc.com}firsttag SAMPLE
****Derving from xml tree*****
{'FIRST_TAG': 'SAMPLE'}'''

你可以使用BeautifulSoup:

from bs4 import BeautifulSoup

x="""<foo>
   <bar>
      <type foobar="1"/>
      <type foobar="2"/>
   </bar>
</foo>"""

y=BeautifulSoup(x)
>>> y.foo.bar.type["foobar"]
u'1'

>>> y.foo.bar.findAll("type")
[<type foobar="1"></type>, <type foobar="2"></type>]

>>> y.foo.bar.findAll("type")[0]["foobar"]
u'1'
>>> y.foo.bar.findAll("type")[1]["foobar"]
u'2'

这里有一个使用cElementTree的非常简单但有效的代码。

try:
    import cElementTree as ET
except ImportError:
  try:
    # Python 2.5 need to import a different module
    import xml.etree.cElementTree as ET
  except ImportError:
    exit_err("Failed to import cElementTree from any known place")      

def find_in_tree(tree, node):
    found = tree.find(node)
    if found == None:
        print "No %s in file" % node
        found = []
    return found  

# Parse a xml file (specify the path)
def_file = "xml_file_name.xml"
try:
    dom = ET.parse(open(def_file, "r"))
    root = dom.getroot()
except:
    exit_err("Unable to open and parse input definition file: " + def_file)

# Parse to find the child nodes list of node 'myNode'
fwdefs = find_in_tree(root,"myNode")

这是来自“python xml解析”。

import xml.etree.ElementTree as ET
data = '''<foo>
           <bar>
               <type foobar="1"/>
               <type foobar="2"/>
          </bar>
       </foo>'''
tree = ET.fromstring(data)
lst = tree.findall('bar/type')
for item in lst:
    print item.get('foobar')

这将打印foobar属性的值。

我推荐ElementTree。同样的API还有其他兼容的实现,比如lxml和Python标准库中的cElementTree;但是,在这种情况下,他们主要增加的是更快的速度——编程的容易程度取决于ElementTree定义的API。

首先从XML中构建一个Element实例根,例如使用XML函数,或者通过解析文件,例如:

import xml.etree.ElementTree as ET
root = ET.parse('thefile.xml').getroot()

或者在ElementTree中显示的许多其他方法中的任何一种。然后这样做:

for type_tag in root.findall('bar/type'):
    value = type_tag.get('foobar')
    print(value)

输出:

1
2