读取单个文件中的多个XML

huangapple
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英文:

Reading multiple XML in a single file

问题

我需要从各种地方收集多个短XML片段，它们都具有相似的格式，在Python中。问题是它们将全部呈现在一个文件中：

  1. <infodump>
  2.    ...
  3. </infodump>
  5. <infodump>
  6.    ...
  7. </infodump>

以此类推。我可以找到许多在多个文件中解析XML的示例（例如，迭代目录中的所有文件），以及合并成单个文件的示例（如使用xmlmerge等），但我还没有找到如何解析同一文件中的多个示例的方法。

我尝试了通常的方法：

  1. tree = ET.parse("id.xml")
  2. root = tree.getroot()

但它在文件中的第二个XML上出现问题：

  1. Traceback (most recent call last):
  2.     File "C:\Users\dlevey\PycharmProjects\reconcile\main.py", line 27, in <module>
  3.       main()
  4.     File "C:\Users\dlevey\PycharmProjects\reconcile\main.py", line 5, in main
  5.       tree = ET.parse("id.xml")
  6.            ^^^^^^^^^^^^^^^^^^^^^
  7.     File "C:\Program Files\Python311\Lib\xml\etree\ElementTree.py", line 1218, in parse
  8.       tree.parse(source, parser)
  9.     File "C:\Program Files\Python311\Lib\xml\etree\ElementTree.py", line 580, in parse
  10.       self._root = parser._parse_whole(source)
  11.                  ^^^^^^^^^^^^^^^^^^^^^^^^^^^
  12.     xml.etree.ElementTree.ParseError: junk after document element: line 25, column 0

我真的不想再次将它们拆分成多个文件，也不想要求它们都作为单独的文件发送，因为其中一些来自源头的文件是作为单个文件发送的，我将一次处理数千个XML样本。

英文:

I need to collect multiple pieces of short XML from a variety of places, all with similar format, in Python. The catch is that they will be presented to me all in one file:

  1. &lt;infodump&gt;
  2.    ...
  3. &lt;/infodump&gt;
  5. &lt;infodump&gt;
  6.    ...
  7. &lt;/infodump&gt;

and so on. I can find many examples of parsing XML in multiple files (iterating over all the files in a directory, for example), and examples of merging into a single file (with such things as xmlmerge), but I have yet found nothing on parsing multiple examples in the same file.

I try the usual

  1. tree = ET.parse(&quot;./id.xml&quot;)
  2. root = tree.getroot()

but it chokes on the second XML in the file:

  1. Traceback (most recent call last):
  2.     File &quot;C:\Users\dlevey\PycharmProjects\reconcile\main.py&quot;, line 27, in &lt;module&gt;
  3.       main()
  4.     File &quot;C:\Users\dlevey\PycharmProjects\reconcile\main.py&quot;, line 5, in main
  5.       tree = ET.parse(&quot;./id.xml&quot;)
  6.            ^^^^^^^^^^^^^^^^^^^^^
  7.     File &quot;C:\Program Files\Python311\Lib\xml\etree\ElementTree.py&quot;, line 1218, in parse
  8.       tree.parse(source, parser)
  9.     File &quot;C:\Program Files\Python311\Lib\xml\etree\ElementTree.py&quot;, line 580, in parse
  10.       self._root = parser._parse_whole(source)
  11.                  ^^^^^^^^^^^^^^^^^^^^^^^^^^^
  12.     xml.etree.ElementTree.ParseError: junk after document element: line 25, column 0

I'd really rather not split them again into multiple files, or request that they are all sent as individual files, as some of them are coming from the source as a single file, and I'll be processing thousands of XML samples at once.

答案1

得分: 1

正如评论中所暗示的，您的输入不是格式良好的（详见此处 了解格式良好与有效 XML 的详细信息），因为它没有单一的根元素。

解决此问题的最简单方法是在解析之前将输入包装在根元素中。

下面是一个简单的示例，它将输入的 XML 片段包装在一个 input 元素中...

  1. import xml.etree.ElementTree as ET
  3. with open("id.xml", "r") as input_file:
  4.     content = input_file.read()
  6. well_formed_input = f"<input>{content}</input>"
  8. root = ET.fromstring(well_formed_input)
  10. for elem in root.findall("infodump"):
  11.     print(elem)

这将打印以下输出：

  1. <Element 'infodump' at 0x0000020B5FFB8400>
  2. <Element 'infodump' at 0x0000020B5FFB84A0>
英文:

Like hinted to in the comments, your input is not well-formed (see here for details on well-formed vs valid xml) because it doesn't have a single root element.

The easiest way to solve this is to wrap the input in a root element before parsing.

Here's a simple example that wraps the input xml fragments in an input element...

  1. import xml.etree.ElementTree as ET
  3. with open(&quot;id.xml&quot;, &quot;r&quot;) as input_file:
  4.     content = input_file.read()
  6. well_formed_input = f&quot;&lt;input&gt;{content}&lt;/input&gt;&quot;
  8. root = ET.fromstring(well_formed_input)
  10. for elem in root.findall(&quot;infodump&quot;):
  11.     print(elem)

This prints the following output:

  1. &lt;Element &#39;infodump&#39; at 0x0000020B5FFB8400&gt;
  2. &lt;Element &#39;infodump&#39; at 0x0000020B5FFB84A0&gt;

答案2

得分: 0

作为对Daniel建议的另一种选择，您可以使用一个HTML解析器：

  1. from html.parser import HTMLParser
  3. class MyHTMLParser(HTMLParser):
  4.     def handle_starttag(self, tag, attrs):
  5.         print("开始标签：", tag)
  7.     def handle_endtag(self, tag):
  8.         print("结束标签：", tag)
  10.     def handle_data(self, data):
  11.         print("一些数据：", data)
  13. parser = MyHTMLParser()
  14. parser.feed('''<infodump>
  15.    ...
  16. </infodump>
  18. <infodump>
  19.    ...
  20. </infodump>''')

输出：

  1. 开始标签： infodump
  2. 一些数据： 
  3.    ...
  5. 结束标签： infodump
  6. 一些数据： 
  9. 开始标签： infodump
  10. 一些数据： 
  11.    ...
  13. 结束标签： infodump
英文:

As an alternative to Daniel’s suggestion you can use a html parser:

  1. from html.parser import HTMLParser
  3. class MyHTMLParser(HTMLParser):
  4.     def handle_starttag(self, tag, attrs):
  5.         print(&quot;Start tag:&quot;, tag)
  7.     def handle_endtag(self, tag):
  8.         print(&quot;End tag :&quot;, tag)
  10.     def handle_data(self, data):
  11.         print(&quot;Some data  :&quot;, data)
  13. parser = MyHTMLParser()
  14. parser.feed(&quot;&quot;&quot;&lt;infodump&gt;
  15.    ...
  16. &lt;/infodump&gt;
  18. &lt;infodump&gt;
  19.    ...
  20. &lt;/infodump&gt;&quot;&quot;&quot;)

Output:

  1. Start tag: infodump
  2. Some data  : 
  3.    ...
  5. End tag : infodump
  6. Some data  : 
  9. Start tag: infodump
  10. Some data  : 
  11.    ...
  13. End tag : infodump

huangapple
  • 本文由 发表于 2023年6月22日 19:40:16
  • 转载请务必保留本文链接：https://go.coder-hub.com/76531507.html
  • python
  • xml
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