我正在尝试解析一些XML,但遇到的问题是强制它只选择请求标记(如果它是父标记)。例如,我的部分XML是:
<Messages>
<Message ChainCode="LI" HotelCode="5501" ConfirmationID="5501">
<MessageContent>
<OTA_HotelResNotifRQ TimeStamp="2014-01-24T21:02:43.9318703Z" Version="4" ResStatus="Book">
<HotelReservations>
<HotelReservation>
<RoomStays>
<RoomStay MarketCode="CC" SourceOfBusiness="CRS">
<RoomRates>
<RoomRate EffectiveDate="2014-02-04" ExpireDate="2014-02-06" RoomTypeCode="12112" NumberOfUnits="1" RatePlanCode="RAC">
<Rates>
<Rate EffectiveDate="2014-02-04" ExpireDate="2014-02-06" RateTimeUnit="Day" UnitMultiplier="3">
<Base AmountBeforeTax="749.25" CurrencyCode="USD" />
<Total AmountBeforeTax="749.25" CurrencyCode="USD" />
</Rate>
</Rates>
</RoomRate>
</RoomRates>
<Total AmountBeforeTax="2247.75" CurrencyCode="USD">
<Taxes Amount="0.00" />
</Total>
</RoomStay>
</RoomStays>
</HotelReservation>
</HotelReservations>
</OTA_HotelResNotifRQ>
</MessageContent>
</Message>
</Messages>
除了“Total”标签之外,我已经解析了我需要它的全部内容。
我想要获得的总标签是:
<Total AmountBeforeTax="2247.75" CurrencyCode="USD">
<Taxes Amount="0.00" />
</Total>
发生了什么,是否正在返回RoomRates \ RoomRate \ Rates \ Rate的子级“Total”标记。我正在试图弄清楚如何指定它只返回RoomStays \ RoomStay \ Total标签。我现在拥有的是:
soup = bs(response, "xml")
messages = soup.find_all('Message')
for message in messages:
hotel_code = message.get('HotelCode')
reservations = message.find_all('HotelReservation')
for reservation in reservations:
uniqueid_id = reservation.UniqueID.get('ID')
uniqueid_idcontext = reservation.UniqueID.get('ID_Context')
roomstays = reservation.find_all('RoomStay')
for roomstay in roomstays:
total = roomstay.Total
有关如何指定我想要提取的确切标记的任何想法?如果有人想知道for循环,那是因为通常有多个“消息”,“酒店预订”,“房间住宿”等标签,但我已将它们删除只显示一个。有时候还有多个Rate \ Rates标签,所以我不能要求它给我第二个“Total”标签。
希望我已经解释过这个。
答案 0 :(得分:1)
有时候还有多个Rate \ Rates标签,所以我不能要求它给我第二个“Total”标签。
为什么不迭代所有Total
标签并跳过那些没有Taxes
孩子的标签?
reservations = message.find_all('HotelReservation')
for reservation in reservations:
totals = reservation.find_all('Total')
for total in totals:
if total.find('Taxes'):
# do stuff
else:
# these aren't the totals you're looking for
如果您更希望消除那些没有子节点的那些,您可以执行以下任一操作:
if next(total.children, None):
# it's a parent of something
if total.contents:
# it's a parent of something
或者你可以use a function instead of a string as your filter:
total = reservation.find(lambda node: node.name == 'Total' and node.contents)
或者您可以通过其他方式查找此标记:它是RoomStay
的直接子项,而不仅仅是后代;它不是Rate
的后代;它是Taxes
下的最后一个RoomStay
后代;所有这些都可以轻松完成。
话虽这么说,这似乎是XPath的完美工作,BeautifulSoup
不支持,但ElementTree
和lxml
做...