Python AttributeError: 'str' object has no attribute 'row'

Question

Wenn ich dieses Script aufrufe

import openpyxl
import requests
from bs4 import BeautifulSoup
from urllib.parse import urlparse

excel_file_path = r"test.xlsx"
workbook = openpyxl.load_workbook(excel_file_path)
sheet = workbook.active

for row in sheet.iter_rows(min_row=2, min_col=3, max_col=3, values_only=True):
    search_term = row[0]
    search_url = "https://example.com/wiki/doku.php?id={search_term}"
    response = requests.get(search_url)
    soup = BeautifulSoup(response.text, "html.parser")
    first_word = soup.get_text().split()[0]    
    sheet.cell(row=row[0].row, column=4).value = first_word

updated_excel_file_path = r"test_new.xlsx"
workbook.save(updated_excel_file_path)

bekomme ich diesen Fehler

Traceback (most recent call last):
  File "C:\Users\tom\Desktop\test.py", line 20, in <module>
    sheet.cell(row=row[0].row, column=4).value = first_word
AttributeError: 'str' object has no attribute 'row'

Wie kann ich das fixen?

Answer 1

Der Fehler tritt auf, weil row[0] ein String ist, und Strings haben keine Eigenschaft namens "row". Du musst die Zeilennummer verfolgen um auf die jeweilige Zeile zuzugreifen. Versuch es damit

for row_number, row in enumerate(sheet.iter_rows(min_row=2, min_col=3, max_col=3, values_only=True), start=2):
    search_term = row[0]
    search_url = f"https://example.com/doku/doku.php?id={search_term}"
    response = requests.get(search_url)
    soup = BeautifulSoup(response.text, "html.parser")
    first_word = soup.get_text().split()[0]    
    sheet.cell(row=row_number, column=4).value = first_word