Python:访问使用ElementTree解析的xml文件中的嵌套子代

Question

我是xml解析的新手.此xml文件具有以下树:

FHRSEstablishment
 |--> Header
 |    |--> ...
 |--> EstablishmentCollection
 |    |--> EstablishmentDetail
 |    |    |-->...
 |    |--> Scores
 |    |    |-->...
 |--> EstablishmentCollection
 |    |--> EstablishmentDetail
 |    |    |-->...
 |    |--> Scores
 |    |    |-->...

但是当我使用ElementTree访问它并查找child标签和属性时,

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 {}

我认为这意味着他们的属性是空的.为什么会这样,我怎么能访问内部嵌套孩子EstablishmentDetail和Scores？

编辑

感谢下面的答案,我可以进入树内,但如果我想要检索诸如此类的值,则会Scores失败:

for node in root.find('.//EstablishmentDetail/Scores'):
    rating = node.attrib.get('Hygiene')
    print rating 

并生产

None
None
None

这是为什么？

Answer 1

哟必须iter()超过你的根!

那就是root.iter()诀窍!

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

输出:

FHRSEstablishment {}
Header {}
ExtractDate {}
ItemCount {}
ReturnCode {}
EstablishmentCollection {}
EstablishmentDetail {}
FHRSID {}
LocalAuthorityBusinessID {}
...

• 要获取所有标签,EstablishmentDetail您需要找到该标签然后循环其子项!

那是,

for child in root.find('.//EstablishmentDetail'):
    print child.tag, child.attrib

输出:

FHRSID {}
LocalAuthorityBusinessID {}
BusinessName {}
BusinessType {}
BusinessTypeID {}
RatingValue {}
RatingKey {}
RatingDate {}
LocalAuthorityCode {}
LocalAuthorityName {}
LocalAuthorityWebSite {}
LocalAuthorityEmailAddress {}
Scores {}
SchemeType {}
NewRatingPending {}
Geocode {}

• 为了得到Hygiene你在评论中提到的分数,

你所做的是,它会得到第一个Scores标签,当你打电话时,它会有卫生,信心管理,结构标签for each in root.find('.//Scores'):rating=child.get('Hygiene').也就是说,显然所有三个孩子都没有这个元素!

你需要先 - 找到所有Scores标签.- 找到Hygiene每个标签!

for each in root.findall('.//Scores'):
    rating = each.find('.//Hygiene')
    print '' if rating is None else rating.text

输出:

5
5
5
0
5