BeautifulSoup-xml-find_next：限制為一個屬性

任何幫助表示贊賞！使用下面的示例 XML 檔案，我得到了不正確的輸出。

Incorrect output: 
Emp_F_Name: Jill
Emp_M_Name: H
Emp_L_Name: Jones

Desired output:
Emp_F_Name: Jill
Emp_M_Name: None or NULL
Emp_L_Name: Jones

我不確定為什么 find_next 函式會超出宣告的屬性（員工）。

   <?xml version="1.0" encoding="utf-8"?>
    <org value="Tech">
    <employee>
    <name>
    <family>Jones</family>
    <given>Jill</given>
    </name>
    </employee>
    <manager>
    <name>
    <family>Fisher</family>
    <given>Junior</given>
    <given>H</given>
    </name>
    </manager>
    </org>

這是我正在使用的代碼。

employee = soup.find("employee")

for i in employee.find_all('name'):
    fname = employee.find('given')
    print("Emp_F_Name: ", fname.get_text())
    
    mname = fname.find_next('given')
    print("Emp_M_Name: ", mname.get_text())
    
    lname = employee.find('family')
    print("Emp_L_Name: ", lname.get_text())

當我為經理運行相同的代碼時，它似乎有效。

manager = soup.find("manager")

uj5u.com熱心網友回復：

如果結構幾乎相同，您可以嘗試'find_all()' 的所有元素given并檢查是否只有一兩個。

given= i.find_all('given')
fname = given[0]
print("Emp_F_Name: ", fname.get_text())
    
mname = given[1].get_text() if len(given) > 1 else None
print("Emp_M_Name: ", mname)

認為沒有必要迭代，employee但如果是這樣，你應該使用你的i

例子

import requests
from bs4 import BeautifulSoup

xml='''<?xml version="1.0" encoding="utf-8"?>
    <org value="Tech">
    <employee>
    <name>
    <family>Jones</family>
    <given>Jill</given>
    </name>
    </employee>
    <manager>
    <name>
    <family>Fisher</family>
    <given>Junior</given>
    <given>H</given>
    </name>
    </manager>
    </org>'''

soup = BeautifulSoup(xml, 'lxml')

employee = soup.find("employee")

for i in employee.find_all('name'):
    given= i.find_all('given')
    fname = given[0]
    print("Emp_F_Name: ", fname.get_text())
    
    mname = given[1].get_text() if len(given) > 1 else None
    print("Emp_M_Name: ", mname)
    
    lname = i.find('family')
    print("Emp_L_Name: ", lname.get_text())

輸出

Emp_F_Name:  Jill
Emp_M_Name:  None
Emp_L_Name:  Jones

另類

隔離employee為單獨的樹以進行操作find_next()：

employee = BeautifulSoup(str(soup.find("employee")), 'lxml')

for i in employee.find_all('name'):
    fname = i.find('given')
    print("Emp_F_Name: ", fname.get_text())
    
    mname = fname.find_next('given').get_text() if fname.find_next('given') else None
    print("Emp_M_Name: ", mname)
    
    lname = i.find('family')
    print("Emp_L_Name: ", lname.get_text())

uj5u.com熱心網友回復：

使用 XML 決議器：（不需要任何外部庫）

import xml.etree.ElementTree as ET


xml = '''<?xml version="1.0" encoding="UTF-8"?>
<org value="Tech">
   <employee>
      <name>
         <family>Jones</family>
         <given>Jill</given>
      </name>
   </employee>
   <manager>
      <name>
         <family>Fisher</family>
         <given>Junior</given>
         <given>H</given>
      </name>
   </manager>
</org>'''



attrs = {'Emp_F_Name':'given',
         'Emp_L_Name':'family',
         'Emp_M_Name': None}

root = ET.fromstring(xml)
name = root.find('.//name')
for k,v in attrs.items():
  print(f'{k}: {name.find(v).text if v else None}')

輸出

Emp_F_Name: Jill
Emp_L_Name: Jones
Emp_M_Name: None

標籤：Python 数组 xml 美汤 属性

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