決議XML時出現型別錯誤

我有一個關于論文的元資料的 XML 檔案，我試圖將作者姓名作為單個字串獲取。XML 中的名稱如下所示：

      <DISS_name>
        <DISS_surname>Clark</DISS_surname>
        <DISS_fname>Brian</DISS_fname>
        <DISS_middle/>
        <DISS_suffix/>
      </DISS_name>

所有名字都有名字和姓氏，但只有一些名字有中間名和/或后綴。這是我的代碼：

    author_surname = record.find('DISS_authorship/DISS_author/DISS_name/DISS_surname').text.strip().title()
    author_fname = record.find('DISS_authorship/DISS_author/DISS_name/DISS_fname').text.strip().title()
    author_mname = record.find('DISS_authorship/DISS_author/DISS_name/DISS_middle')
    author_suffix = record.find('DISS_authorship/DISS_author/DISS_name/DISS_suffix')
    if author_mname is not None and author_suffix is not None:
        author_name = author_surname   ', '   author_fname   author_mname.text   ', '   author_suffix.text
    if author_mname is not None and author_suffix is None:
        author_name = author_surname   ', '   author_fname   author_mname.text
    if author_mname is None and author_suffix is None:
        author_name = author_surname   ', '   author_fname

為什么我會得到這個輸出，我該如何解決？

Traceback (most recent call last):
  File "C:\Users\bpclark2\pythonProject3\prqXML-to-dcCSV.py", line 185, in <module>
    author_name = author_surname   ', '   author_fname   author_mname.text   author_suffix.text
TypeError: can only concatenate str (not "NoneType") to str

修改后的代碼：

    author_surname = record.find('DISS_authorship/DISS_author/DISS_name/DISS_surname').text.strip().title()
    author_fname = record.find('DISS_authorship/DISS_author/DISS_name/DISS_fname').text.strip().title()
    author_mname = record.find('DISS_authorship/DISS_author/DISS_name/DISS_middle').text or ''
    author_suffix = record.find('DISS_authorship/DISS_author/DISS_name/DISS_suffix').text or ''
    author_name = author_surname   ', '   author_fname   ' '   str(author_mname.strip().title())   str(', '   author_suffix.strip().title())
    row.append(author_name)

這得到了我正在尋找的輸出：

    author_surname = record.find('DISS_authorship/DISS_author/DISS_name/DISS_surname').text.strip().title()
    author_fname = record.find('DISS_authorship/DISS_author/DISS_name/DISS_fname').text.strip().title()
    author_mname = record.find('DISS_authorship/DISS_author/DISS_name/DISS_middle').text or ''
    author_suffix = record.find('DISS_authorship/DISS_author/DISS_name/DISS_suffix').text or ''
    author_name = author_surname   ', '   author_fname   ' '   author_mname.strip().title()   ', '   author_suffix.strip().title()
    if author_mname != '' and author_suffix != '':
        author_name = author_surname   ', '   author_fname   ' '   author_mname.strip().title()   ', '   author_suffix.strip().title()
        row.append(author_name)
    if author_mname != '' and author_suffix == '':
        author_name = author_surname   ', '   author_fname   ' '   author_mname.strip().title()
        row.append(author_name)
    if author_mname == '' and author_suffix != '':
        author_name = author_surname   ', '   author_fname   ', '   author_suffix.strip().title()
        row.append(author_name)
    if author_mname == '' and author_suffix == '':
        author_name = author_surname   ', '   author_fname
        row.append(author_name)

uj5u.com熱心網友回復：

把你的代碼改成這樣怎么樣：

author_mname = record.find('DISS_authorship/DISS_author/DISS_name/DISS_middle') or ''
author_suffix = record.find('DISS_authorship/DISS_author/DISS_name/DISS_suffix') or ''

你也可以添加str像這樣的演員表：

...   str(author_suffix.text)

如果您使用的是新 Python，請使用 f-strings！有了他們，生活要輕松得多。

uj5u.com熱心網友回復：

我會通過對代碼進行少量編輯來保持一切簡單。您可以使用 XPath.//DISS_name查找所有<DISS_name>節點，然后將其解壓縮為具有相應名稱的單獨變數。代碼：

import xml.etree.ElementTree as ET

data = """\
<DISS_authorship>
    <DISS_author>
        <DISS_name>
            <DISS_surname>Clark</DISS_surname>
            <DISS_fname>Brian</DISS_fname>
            <DISS_middle/>
            <DISS_suffix/>
        </DISS_name>
    </DISS_author>
</DISS_authorship>"""

root = ET.fromstring(data)
row = []
for name_node in root.iterfind(".//DISS_name"):
    surname, fname, middle, suffix = name_node  # 4 child nodes in this order
    name_str = surname.text   ", "   fname.text
    if middle.text:
        name_str  = " "   middle.text
    if suffix.text:
        name_str  = ", "   suffix.text
    row.append(name_str)

或者更短：

import xml.etree.ElementTree as ET

data = ...

root = ET.fromstring(data)
row = []
for (surname, fname, middle, suffix) in root.iterfind(".//DISS_name"):
    name_str = surname.text   ", "   fname.text
    if middle.text:
        name_str  = " "   middle.text
    if suffix.text:
        name_str  = ", "   suffix.text
    row.append(name_str)

uj5u.com熱心網友回復：

下面是一個簡短的概念

import xml.etree.ElementTree as ET


xml = '''<r><DISS_name>
        <DISS_surname>Clark</DISS_surname>
        <DISS_fname>Brian</DISS_fname>
        <DISS_middle/>
        <DISS_suffix/>
      </DISS_name>
      <DISS_name>
        <DISS_surname>Jack</DISS_surname>
        <DISS_fname>Brian</DISS_fname>
        <DISS_middle>Smith</DISS_middle>
        <DISS_suffix/>
      </DISS_name>
</r>'''
      
root = ET.fromstring(xml)
for name in root.findall('.//DISS_name'):
  parts = [name.find(f'DISS_{f}').text for f in ['surname','fname','middle','suffix'] if name.find(f'DISS_{f}').text is not None ]
  print(", ".join(parts))

輸出

Clark, Brian
Jack, Brian, Smith

標籤：Python xml

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