Question

我最近开始研究python。我试图解析一个xml文档。请考虑以下xml文件以供参考：

<?xml version="1.0"?>
<catalog>
   <book id="bk101">
      <author>Gambardella, Matthew</author>
      <title>XML Developer's Guide</title>
      <genre>Computer</genre>
      <price>44.95</price>
      <publish_date>2000-10-01</publish_date>
      <description>An in-depth look at creating applications
      with XML.</description>
   </book>
   <book id="bk102">
      <author>Ralls, Kim</author>
      <title>Midnight Rain</title>
      <genre>Fantasy</genre>
      <price>5.95</price>
      <publish_date>2000-12-16</publish_date>
      <description>A former architect battles corporate zombies,
      an evil sorceress, and her own childhood to become queen
      of the world.</description>
   </book>
</catalog>

这里我想要检索第一个book标签及其所有内容，即

<book id="bk101">
  <author>Gambardella, Matthew</author>
  <title>XML Developer's Guide</title>
  <genre>Computer</genre>
  <price>44.95</price>
  <publish_date>2000-10-01</publish_date>
  <description>An in-depth look at creating applications
  with XML.</description>
</book>

我来自scala背景，我可以轻松地使用

执行此操作

val node = scala.xml.XML.loadString(str)
val nodeSeq = node \\ "book"
nodeSeq.head.toString()

我已尝试使用lxml和xpath执行此操作，但它变得复杂（获取嵌套元素的递归内容）以达到上述要求。有没有简单的方法在python中执行此操作？也可以扩展为html吗？

TIA

Answer 1

使用lxml和xpath

from lxml import etree

data = """<?xml version="1.0"?>
<catalog>
   <book id="bk101">
      <author>Gambardella, Matthew</author>
      <title>XML Developer's Guide</title>
      <genre>Computer</genre>
      <price>44.95</price>
      <publish_date>2000-10-01</publish_date>
      <description>An in-depth look at creating applications
      with XML.</description>
   </book>
   <book id="bk102">
      <author>Ralls, Kim</author>
      <title>Midnight Rain</title>
      <genre>Fantasy</genre>
      <price>5.95</price>
      <publish_date>2000-12-16</publish_date>
      <description>A former architect battles corporate zombies,
      an evil sorceress, and her own childhood to become queen
      of the world.</description>
   </book>
</catalog>"""

tree = etree.fromstring(data)
book = tree.xpath("//catalog/book") #or book = tree.xpath("(//catalog/book)[1]")
for i in book[0]:#[0] means first book
    print etree.tostring(i)

输出 -

<book id="bk101">
      <author>Gambardella, Matthew</author>
      <title>XML Developer's Guide</title>
      <genre>Computer</genre>
      <price>44.95</price>
      <publish_date>2000-10-01</publish_date>
      <description>An in-depth look at creating applications
      with XML.</description>
   </book>

Answer 2

这是仅提取第一本书的XPath：

//catalog/book[1]

这是返回所需结果的完整代码：

from lxml import html

XML = """<?xml version="1.0"?>
<catalog>
   <book id="bk101">
      <author>Gambardella, Matthew</author>
      <title>XML Developer's Guide</title>
      <genre>Computer</genre>
      <price>44.95</price>
      <publish_date>2000-10-01</publish_date>
      <description>An in-depth look at creating applications
      with XML.</description>
   </book>
   <book id="bk102">
      <author>Ralls, Kim</author>
      <title>Midnight Rain</title>
      <genre>Fantasy</genre>
      <price>5.95</price>
      <publish_date>2000-12-16</publish_date>
      <description>A former architect battles corporate zombies,
      an evil sorceress, and her own childhood to become queen
      of the world.</description>
   </book>
</catalog>"""

tree = html.fromstring(XML)
first_book = tree.xpath('//catalog/book[1]')[0]
book_id = first_book.xpath('@id')[0]
author = first_book.xpath('.//author/text()')[0]
title = first_book.xpath('.//title/text()')[0]
genre = first_book.xpath('.//genre/text()')[0]
price = first_book.xpath('.//price/text()')[0]
publish_date = first_book.xpath('.//publish_date/text()')[0]
description = first_book.xpath('.//description/text()')[0].replace('\n',' ').replace('  ','')

print """Book Id:\t\t{}
Author:\t\t\t{}
Title:\t\t\t{}
Genre:\t\t\t{}
Price:\t\t\t{}
Publish Date:\t{}
Description:\t{}""".format(book_id,author,title,genre,price,publish_date,description)

<强>输出：

Book Id:        bk101
Author:         Gambardella, Matthew
Title:          XML Developer's Guide
Genre:          Computer
Price:          44.95
Publish Date:   2000-10-01
Description:    An in-depth look at creating applications with XML.

如果您需要从every内的<catalog>本书中获取相同的信息，则只需将//catalog/book[1]更改为//catalog/book，然后将结果循环至提取每本书的字段数据。

xml（和html）在python中解析

2 个答案: