Check below example <br>
To loop pages with page=x you need for loop like this
import requests
from bs4 import BeautifulSoup
url = 'http://www.housingcare.org/housing-care/results.aspx?ath=1%2c2%2c3%2c6%2c7&stp=1&sm=3&vm=list&rp=10&page='
for page in range(10):
print('---', page, '---')
r = requests.get(url + str(page))
soup = BeautifulSoup(r.content, "html.parser")
# String substitution for HTML
for link in soup.find_all("a"):
print("<a href='>%s'>%s</a>" % (link.get("href"), link.text))
# Fetch and print general data from title class
general_data = soup.find_all('div', {'class' : 'title'})
for item in general_data:
print(item.contents[0].text)
print(item.contents[1].text.replace('.',''))
print(item.contents[2].text)
Every page can be different and better solution needs more inforamtion about page. Sometimes you can get link to last page and then you can use this information instead 10 in range(10)
Or you can use while True to loop and break to leave loop if there is no link to next page. But first you have to show this page (url to real page) in question.
EDIT: example how to get link to next page and then you get all pages - not only 10 pages as in previous version.
import requests
from bs4 import BeautifulSoup
# link to first page - without `page=`
url = 'http://www.housingcare.org/housing-care/results.aspx?ath=1%2c2%2c3%2c6%2c7&stp=1&sm=3&vm=list&rp=10'
# only for information, not used in url
page = 0
while True:
print('---', page, '---')
r = requests.get(url)
soup = BeautifulSoup(r.content, "html.parser")
# String substitution for HTML
for link in soup.find_all("a"):
print("<a href='>%s'>%s</a>" % (link.get("href"), link.text))
# Fetch and print general data from title class
general_data = soup.find_all('div', {'class' : 'title'})
for item in general_data:
print(item.contents[0].text)
print(item.contents[1].text.replace('.',''))
print(item.contents[2].text)
# link to next page
next_page = soup.find('a', {'class': 'next'})
if next_page:
url = next_page.get('href')
page += 1
else:
break # exit `while True`
To loop pages with `page=x` you need `for` loop like this>
import requests
from bs4 import BeautifulSoup
url = 'http://www.housingcare.org/housing-care/results.aspx?ath=1%2c2%2c3%2c6%2c7&stp=1&sm=3&vm=list&rp=10&page='
for page in range(10):
print('---', page, '---')
r = requests.get(url + str(page))
soup = BeautifulSoup(r.content, "html.parser")
# String substitution for HTML
for link in soup.find_all("a"):
print("<a href='>%s'>%s</a>" % (link.get("href"), link.text))
# Fetch and print general data from title class
general_data = soup.find_all('div', {'class' : 'title'})
for item in general_data:
print(item.contents[0].text)
print(item.contents[1].text.replace('.',''))
print(item.contents[2].text)
Every page can be different and better solution needs more inforamtion about page. Sometimes you can get link to last page and then you can use this information instead `10` in `range(10)`
Or you can use `while True` to loop and `break` to leave loop if there is no link to next page. But first you have to show this page (url to real page) in question.
---
**EDIT:** example how to get link to next page and then you get all pages - not only 10 pages as in previous version.
import requests
from bs4 import BeautifulSoup
# link to first page - without `page=`
url = 'http://www.housingcare.org/housing-care/results.aspx?ath=1%2c2%2c3%2c6%2c7&stp=1&sm=3&vm=list&rp=10'
# only for information, not used in url
page = 0
while True:
print('---', page, '---')
r = requests.get(url)
soup = BeautifulSoup(r.content, "html.parser")
# String substitution for HTML
for link in soup.find_all("a"):
print("<a href='>%s'>%s</a>" % (link.get("href"), link.text))
# Fetch and print general data from title class
general_data = soup.find_all('div', {'class' : 'title'})
for item in general_data:
print(item.contents[0].text)
print(item.contents[1].text.replace('.',''))
print(item.contents[2].text)
# link to next page
next_page = soup.find('a', {'class': 'next'})
if next_page:
url = next_page.get('href')
page += 1
else:
break # exit `while True`
