attributeError：'list'对象没有属性'项目'

Question

我一直在尝试编写一个工具，以查看断开链接的指定路径中的Word文档。 我放弃了让它搜索文件夹，以为我只需要让它首先做文档即可。我的技能有限，阅读并尝试了Copilot的一些建议。 （将其分解为单个任务）我将其放在一起：

import docx
import os
import url

doc = docx.Document('C:/Users/demo.docx')
allText = []

def find_hyperlinks(doc):
    hyperlinks = []
    rels = doc.part.rels
    for rel in rels:
        if "hyperlink" in rels[rel].target_ref:
            hyperlinks.append(rels[rel].target_ref)
    return hyperlinks

def find_broken_links_in_docx(doc):
    broken_links = []
    hyperlinks = find_hyperlinks(doc)
    for url in hyperlinks:
        try:
            response = requests.head(url, allow_redirects=True)
            if response.status_code >= 400:
                broken_links.append(url)
        except requests.RequestException:
            broken_links.append(url)
    return broken_links

def write_report(report, output_file):
    with open(output_file, 'w') as f:
        for file_path, links in report.items():
            f.write(f"File: {file_path}\n")
            for link in links:
                f.write(f"  Broken link: {link}\n")
            f.write("\n")

if __name__ == "__main__":
    
    output_file = "C:/Results/broken_links_report.txt"
    report = find_broken_links_in_docx(doc)
    write_report(report, output_file)
    print(f"Report written to {output_file}")

错误是错误：

 File "c:\Users\Scripts\playground\openinganddocx.py", line 41, in <module>
    write_report(report, output_file)
  File "c:\Users\Scripts\playground\openinganddocx.py", line 31, in write_report
    for file_path, links in report.items():
    AttributeError: 'list' object has no attribute 'items'

为参考：

线31

f.write(f"File: {file_path}\n")

• 线41
print(f"Report written to {output_file}")

问题是您将A

Answer 1

（报告）视为

dict

（报告），事实并非如此。这就是为什么您要得到

AttributeError: list has no attribute 'items'

。

您想要的是具有结构

{'filepath': [<urls>]}

的字典。所以从那里开始：

def find_hyperlinks(doc_path: str):
    doc = docx.Document(doc_path)
    hyperlinks = []
    rels = doc.part.rels
    for rel in rels:
        if "hyperlink" in rels[rel].target_ref:
            hyperlinks.append(rels[rel].target_ref)

    # here is an example of how I might return that value
    return {doc_path: hyperlinks}

# from here, prune the hyperlinks that work
broken_links = {}

for doc_path, links in links_dict.items():
    broken = []
    for link in links:
        if link_works(link):
           continue

        broken.append(link)
    broken_links[doc_path] = broken

# etc

这不是完美的，但会让您走上成功之路