Python 文件名，而不是标记。打开这个文件并将文件句柄传递给 Beautiful Soup

Question

我已将 Python 2.7 例程更改为接受文件路径作为例程的参数，因此我不必通过在方法中插入多个文件路径来复制代码。

当我的方法被调用时，我得到以下错误：

looks like a filename, not markup. You should probably open this file and pass the filehandle into Beautiful Soup.
  '"%s" looks like a filename, not markup. You should probably open this file and pass the filehandle into Beautiful Soup.' % markup)

我的方法实现是：

def extract_data_from_report3(filename):
    html_report_part1 = open(filename,'r').read()
    soup = BeautifulSoup(filename, "html.parser")
    th = soup.find_all('th')
    td = soup.find_all('td')

    headers = [header.get_text(strip=True) for header in soup.find_all("th")]
    rows = [dict(zip(headers, [td.get_text(strip=True) for td in row.find_all("td")]))
        for row in soup.find_all("tr")[1:-1]]
    print(rows)
    return rows

调用方法如下：

rows_part1 =  report.extract_data_from_report3(r"E:\test_runners\selenium_regression_test_5_1_1\TestReport\SeleniumTestReport_part1.html")
print "part1 = "
print rows_part1

如何将文件名作为参数传递？

Answer 1

您应该将已阅读文件的实际内容传递给BeautifulSoup：

html_report_part1 = open(filename,'r').read()
soup = BeautifulSoup(html_report_part1, "html.parser")

Answer 2

如果你想传递一个文件句柄，那么不要调用 read，只需传递 open(filename) 或文件句柄而不调用 read ：

def extract_data_from_report3(filename):
    html_report_part1 = open(filename,'r')
    soup = BeautifulSoup( html_report_part1, "html.parser")

或者：

def extract_data_from_report3(filename):
    soup = BeautifulSoup(open(filename), "html.parser")

您可以按照建议调用 read 后传递html_report_part1，但您不需要，BeautifulSoup 可以获取文件对象。