点击 python 中 Selenium 的多个链接

Question

I'm trying to web-scrape data from a structure that looks like that:

<div class = "tables">
        <div class = "table1">
            <div class = "row">
                <div class = 'data'>Useful Data</div>
                <a href = "url1"
            </div>
            <div class = "row">
                <div class = 'data'>Useful Data</div>
                <a href = "url1">
            </div>
        </div>
        
        <div class = "table2">
            <div class = "row">
                <div class = 'data'>Useful Data</div>
                <a href = "url3"
            </div>
            <div class = "row">
                <div class = 'data'>Useful Data</div>
                <a href = "url4">
            </div>
        </div>
     </div>

The data that I want is in the div "data", and also on a some other pages accessible by clicking on the urls. I iterate through the 'tables' using BeautifulSoup, and I'm trying to click on the links with Selenium like so:

tables = soup.find_all('div', class_ = 'tables')
 for line in tables:
     row = line.find_all('div', class_ = "row")
     for element in row:
         link = driver.find_element_by_xpath('//a[contains(@href,"href")]')
         #some code

In my script, this line

link = driver.find_element_by_xpath('//a[contains(@href,"href")]')

always return the first url, when I want it to 'follow' BeautifulSoup and return to following hrefs. So is there a way to modify href depending on the url from the source code? I should add that all my urls are pretty similiar, except for the last part. (ex.: url1 = questions/ask/ 1000 , url2 = questions/ask/ 1001 )

I've also tried to find all the href in the page to iterate trough them using

links = self.driver.find_element_by_xpath('//a[@href]')

but that doesn't work either. Since the page contains a lot of links that aren't useful to me, I'm not sure if that's the best way to go.

Answer 1

Seems to be a bit complicated - Why not extracting the href with BeautifulSoup directly?

for a in soup.select('.tables a[href]'):
    link = a['href']

You also can modify it, concat with baseUrl and store in a list to iterate over:

urls = [baseUrl+a['href'] for a in soup.select('.tables a[href]')]

Example

baseUrl = 'http://www.example.com'

html='''
<div class = "tables">
        <div class = "table1">
            <div class = "row">
                <div class = 'data'>Useful Data</div>
                <a href = "/url1"
            </div>
            <div class = "row">
                <div class = 'data'>Useful Data</div>
                <a href = "/url1">
            </div>
        </div>

        <div class = "table2">
            <div class = "row">
                <div class = 'data'>Useful Data</div>
                <a href = "/url3"
            </div>
            <div class = "row">
                <div class = 'data'>Useful Data</div>
                <a href = "/url4">
            </div>
        </div>
     </div>'''
soup = BeautifulSoup(html,'lxml')

urls = [baseUrl+a['href'] for a in soup.select('.tables a[href]')]

for url in urls:
    print(url)#or request the website,....

Output

http://www.example.com/url1
http://www.example.com/url1
http://www.example.com/url3
http://www.example.com/url4