Python 和 selenium - 从网页获取所有链接

Question

I have a script that download all .csv files from a site. It is working ok but I need also to get a list with the urls to .csv files. The part of code for downloading looks like:

# Download files according to Xpath in table
def downloadfiles(Xpath):
    global browser
    time.sleep(10);
    # Click csv img
    try:
        browser.find_element_by_xpath(Xpath).click()
        global downloadcount
        downloadcount = downloadcount + 1
        return
    # Element not found
    except NoSuchElementException as e:
        print("Error enxontró csv")
    return

Here I think i need to do something: browser.find_element_by_xpath(Xpath).click() instead of click I want to get the link. The code is made in python 3.6 and selenium. Xpath is "//*[@id=\\"ctl00_ContentPlaceHolder1_ListViewNodos_ctrl0_ListViewArchivosSIN_ctrl0_linkCSV\\"]"

How can I get the links list to .csv files?

Answer 1

You can get the link using get_attribute("href") use below line of code

link = browser.find_element_by_xpath(Xpath).get_attribute("href")
print link