python根据文件名打标签

文件名打标签"/>

python根据文件名打标签

我已经更改了我的

Python 2.7例程以接受文件路径作为例程的参数,因此我不必通过在方法内插入多个文件路径来复制代码.

当我的方法被调用时,我收到以下错误：

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？