Python filename, not markup. open this file and pass the filehandle into Beautiful Soup

Question

I have changed my Python 2.7 routine to accept a file path as a parameter for the routine so I don't have to duplicate code by inserting multiple file paths inside the method.

When my method is called I get the following error:

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)

My method implementation is:

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

To call the method is as follows:

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

How can I pass the file name as a parameter?

Answer 1

If you want to pass a file handle then don't call read, just pass open(filename) or the file handle without calling read :

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

Or:

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

You can pass html_report_part1 after calling read as suggested but you don't need to, BeautifulSoup can take a file object.