我是 xml 解析新手。这个xml文件 http://ratings.food.gov.uk/OpenDataFiles/FHRS408en-GB.xml有以下树:
FHRSEstablishment
|--> Header
| |--> ...
|--> EstablishmentCollection
| |--> EstablishmentDetail
| | |-->...
| |--> Scores
| | |-->...
|--> EstablishmentCollection
| |--> EstablishmentDetail
| | |-->...
| |--> Scores
| | |-->...
但是当我使用 ElementTree 访问它并查找child
标签和属性,
import xml.etree.ElementTree as ET
import urllib2
tree = ET.parse(
file=urllib2.urlopen('http://ratings.food.gov.uk/OpenDataFiles/FHRS408en-GB.xml' % i))
root = tree.getroot()
for child in root:
print child.tag, child.attrib
我只得到:
Header {}
EstablishmentCollection {}
我认为这意味着它们的属性是空的。为什么会这样,我怎样才能访问嵌套在里面的孩子EstablishmentDetail
and Scores
?
EDIT
感谢下面的答案,我可以进入树内部,但是如果我想检索诸如中的值Scores
,这失败了:
for node in root.find('.//EstablishmentDetail/Scores'):
rating = node.attrib.get('Hygiene')
print rating
并产生
None
None
None
这是为什么?
您必须对根进行迭代()。
that is root.iter()
就可以了!
import xml.etree.ElementTree as ET
import urllib2
tree =ET.parse(urllib2.urlopen('http://ratings.food.gov.uk/OpenDataFiles/FHRS408en-GB.xml'))
root = tree.getroot()
for child in root.iter():
print child.tag, child.attrib
Output:
FHRSEstablishment {}
Header {}
ExtractDate {}
ItemCount {}
ReturnCode {}
EstablishmentCollection {}
EstablishmentDetail {}
FHRSID {}
LocalAuthorityBusinessID {}
...
- 获取里面的所有标签
EstablishmentDetail
您需要找到该标签,然后循环遍历它的子标签!
也就是说,例如。
for child in root.find('.//EstablishmentDetail'):
print child.tag, child.attrib
Output:
FHRSID {}
LocalAuthorityBusinessID {}
BusinessName {}
BusinessType {}
BusinessTypeID {}
RatingValue {}
RatingKey {}
RatingDate {}
LocalAuthorityCode {}
LocalAuthorityName {}
LocalAuthorityWebSite {}
LocalAuthorityEmailAddress {}
Scores {}
SchemeType {}
NewRatingPending {}
Geocode {}
你所做的就是,它将获得第一Scores
标签,当您调用时,该标签将具有 Hygiene、ConfidenceInManagement、Structural 标签作为子标签for each in root.find('.//Scores'):rating=child.get('Hygiene')
。也就是说,显然三个孩子都不具备该元素!
你需要先
- 找到所有Scores
标签。
- 寻找Hygiene
在找到的每个标签中!
for each in root.findall('.//Scores'):
rating = each.find('.//Hygiene')
print '' if rating is None else rating.text
Output:
5
5
5
0
5
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