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我收到此错误

[Errno 22] Invalid argument: 'S:\\Job Data$\\Jobs\\14-00060006\\Completed Files\\14-00060006_data.txt'

和其他人试图打开文件。我认为这是因为 \14 和 \13 等，但我不确定。

这是代码：

def crawl(path, phrase=""):
found_files = 0
for root, dirs, files in os.walk(path):
    for f in files:
        if f.endswith(".txt"):
            found_file = os.path.join(root, f)
            if phrase != "":
                try:
                    with open(found_file, mode='r') as text_file:
                        for line in text_file:
                            for part in line.split():
                                if phrase in part:
                                    found_files += 1
                                    print("\n\nFound one:")
                                    print(found_file)
                                    print(line)
                except Exception as e:
                    print(e)

我试图在目录树中找到特定的单词，数千个目录都以年份为前缀，2014 年为 14，依此类推，无法更改，我怎样才能让 python 读取这些文件？

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0 回答 0