Develop Reference

Web Scraping - Extract list of text from multiple pages

I want to extract a list of names from multiple pages of a website.
The website has over 200 pages and i want to save all the names to a text file. I have wrote some code but it's giving me index error.
CODE:
import requests
from bs4 import BeautifulSoup as bs
URL = 'https://hamariweb.com/names/muslim/boy/page-'
#for page in range(1, 203):
page = 1
req = requests.get(URL + str(page))
soup = bs(req.text, 'html.parser')
row = soup.find('div', attrs={'class', 'row'})
books = row.find_all('a')
for book in books:
data = book.find_all('b')[0].get_text()
print(data)
OUTPUT:
Aabbaz
Aabid
Aabideen
Aabinus
Aadam
Aadeel
Aadil
Aadroop
Aafandi
Aafaq
Aaki
Aakif
Aalah
Aalam
Aalamgeer
Aalif
Traceback (most recent call last):
File "C:\Users\Mujtaba\Documents\names.py", line 15, in <module>
data = book.find_all('b')[0].get_text()
IndexError: list index out of range
>>>

The reason for getting the error is since it can't find a <b> tag.
Try this code to request each page and save the data to a file:
import requests
from bs4 import BeautifulSoup as bs
MAIN_URL = "https://hamariweb.com/names/muslim/boy/"
URL = "https://hamariweb.com/names/muslim/boy/page-{}"
with open("output.txt", "a", encoding="utf-8") as f:
for page in range(203):
if page == 0:
req = requests.get(MAIN_URL.format(page))
else:
req = requests.get(URL.format(page))
soup = bs(req.text, "html.parser")
print(f"page # {page}, Getting: {req.url}")
book_name = (
tag.get_text(strip=True)
for tag in soup.select(
"tr.bottom-divider:nth-of-type(n+2) td:nth-of-type(1)"
)
)
f.seek(0)
f.write("\n".join(book_name) + "\n")

I suggest to change your parser to html5lib #pip install html5lib. I just think it's better. Second It's better NOT to do a .find() from your soup object DIRECTLY since it might cause some problems where the tags and classes might have duplicates. SO you might be finding data on a html tag where your data isn't even there. So it's better to check everything and inspect element the the tags you want to get and see on what block of code they might be in cause it is easier that way to scrape, also to avoid more errors.
What I did there is I inspected the elements first and FIND the BLOCK of code where you want to get your data and I found that it is on a div and its class is mb-40 content-box that is where all the names you are trying to get are. Luckily the class is UNIQUE and there are no other elements with the same tag and class so we can just directly .find() it.
Then the value of trs are simply the tr tags inside of that block
(Take note also that those <tr> tags are inside of a <table> tag but the good thing is those are the only <tr> tags that exist so there wouldn't be much of a problem like if there would be another <table> tag with the same class value)
which the <tr> tags contains the names you want to get. You may ask why is there [1:] it's because to start at index 1 to NOT include the Header from the table on the website.
Then just loop through those tr tags and get the text. With regards to your error on why is it happening it is simply because of index out of range you are trying to access a .find_all() result list item where it is out of bounds and this might happen if cases that there are no such data that is being found and that also might happen if you DIRECTLY do a .find() function on your soup variable, because there would be times that there are tags and their respective class values are the same BUT! WITH DIFFERENT CONTENT WITHIN IT. So what happens is you're expecting to scrape that particular part of the website but what actually happening is you're scraping a different part, that's why you might not get any data and wonder why it is happening.
import requests
from bs4 import BeautifulSoup as bs
URL = 'https://hamariweb.com/names/muslim/boy/page-'
#for page in range(1, 203):
page = 1
req = requests.get(URL + str(page))
soup = bs(req.content, 'html5lib')
div_container = soup.find('div', class_='mb-40 content-box')
trs = div_container.find_all("tr",class_="bottom-divider")[1:]
for tr in trs:
text = tr.find("td").find("a").text
print(text)

The issue you're having with the IndexError means that in this case the b-tag you found doesn't contains the information that you are looking for.
You can simply wrap that piece of code in a try-except clause.
for book in books:
try:
data = book.find_all('b')[0].get_text()
print(data)
# Add data to the all_titles list
all_titles.append(data)
except IndexError:
pass # There was no element available
This will catch you error and move on. But not break the code.
Below I have also added some extra lines to save your title to a text-file.
Take a look at the inline comments.
import requests
from bs4 import BeautifulSoup as bs
URL = 'https://hamariweb.com/names/muslim/boy/page-'
# Theres is where your titles will be saved. Changes as needed
PATH = '/tmp/title_file.txt'
page = 1
req = requests.get(URL + str(page))
soup = bs(req.text, 'html.parser')
row = soup.find('div', attrs={'class', 'row'})
books = row.find_all('a')
# Here your title will be stored before writing to file
all_titles = []
for book in books:
try:
# Add strip() to cleanup the input
data = book.find_all('b')[0].get_text().strip()
print(data)
# Add data to the all_titles list
all_titles.append(data)
except IndexError:
pass # There was no element available
# Open path to write
with open(PATH, 'w') as f:
# Write all titles on a new line
f.write('\n'.join(all_titles))

Print out all article titles

I'm new to Python. Here are some lines of coding in Python to print out all article titles on http://www.nytimes.com/.
import requests
from bs4 import BeautifulSoup
base_url = 'http://www.nytimes.com'
r = requests.get(base_url)
soup = BeautifulSoup(r.text)
for story_heading in soup.find_all(class_="story-heading"):
if story_heading.a:
print(story_heading.a.text.replace("\n", " ").strip())
else:
print(story_heading.contents[0].strip())
What do .a and .text mean?
Thank you very much.

First, let's see what printing one story_heading alone gives us:
>>> story_heading
<h2 class="story-heading">Mortgage Calculator</h2>
To extract only the a tag, we access it using story_heading.a:
>>> story_heading.a
Mortgage Calculator
To get only the text inside the tag itself, and not it's attributes, we use .text:
>>> story_heading.a.text
'Mortgage Calculator'

Here,
.a gives you the first anchor tag
.text gives you the text within the tag

Scraping data with python

I'm working on a program to scrape address data from companies from a website. For this, I already have built a list with the links where this data can be found. On each of those links, the page source of the data I need looks like this:
http://imgur.com/a/V0kBK
For some reason though, I cannot seem to fetch the first line of the address, in this case "'t walletje 104 101"
All the other information comes through fine, as you can see here:
http://imgur.com/a/aUmSI
This is my code:
from urllib.request import urlopen
from bs4 import BeautifulSoup
import requests
import string
import urllib
urllist=[]
url1="http://www.bisy.be/"
fname = "test2.txt"
fh = open(fname)
for line in fh:
line = line.rstrip()
if line.startswith(" <tr class="):
words= line.split()
url2 = words[6]
url3 = url1 + url2[1:48]
urllist.append(url3)
for link in urllist:
document = urllib.request.urlopen(link)
html = document.read()
soup = BeautifulSoup(html,"html.parser")
name = soup.find("br")
name2 = name.text.strip()
print (name2)
This code is doing the basics so far. Once everything works I will clean it and finetune it a bit. Is there anyone who can help me?
An example link for the people who want to check out the page source: http://www.bisy.be/?q=nl/bedrijf_fiche&id=KNO01C2015&nr=2027420160
Is there anyone who can point me in the right direction?

this is a workaround by finding p tag with specific font size:
elements = soup.find_all("p")
for tag in elements:
try:
if "9pt" in tag["style"]:
details = tag.text
print(details)
except:
pass

your can't select text in br it new line not like p, span div or others. you can try using BeautifulSoup above or use regex below which is faster
import re
for link in urllist:
document = urllib.request.urlopen(link)
html = document.read().decode('utf-8')
name = re.compile(r"9pt; font-family: Arial;'>(.*?)<a", re.DOTALL).findall(html)
# clean html tag
name = re.sub('<[^>]*>', '', name[0])
print(name)

Python - How do I find a link on webpage that has no class?

I am a beginner python programmer and I am trying to make a webcrawler as practice.
Currently I am facing a problem that I cannot find the right solution for. The problem is that I am trying to get a link location/address from a page that has no class, so I have no idea how to filter that specific link.
It is probably better to show you.
The page I am trying to get the link from.
As you can see, I am trying to get what is inside of the href attribute of the "Historical prices" link. Here is my python code:
import requests
from bs4 import BeautifulSoup
def find_historicalprices_link(url):
source = requests.get(url)
text = source.text
soup = BeautifulSoup(text, 'html.parser')
link = soup.find_all('li', 'fjfe-nav-sub')
href = str(link.get('href'))
find_spreadsheet(href)
def find_spreadsheet(url):
source = requests.get(url)
text = source.text
soup = BeautifulSoup(text, 'html.parser')
link = soup.find('a', {'class' : 'nowrap'})
href = str(link.get('href'))
download_spreadsheet(href)
def download_spreadsheet(url):
response = requests.get(url)
text = response.text
lines = text.split("\\n")
filename = r'google.csv'
file = open(filename, 'w')
for line in lines:
file.write(line + "\n")
file.close()
find_historicalprices_link('https://www.google.com/finance?q=NASDAQ%3AGOOGL&ei=3lowWYGRJNSvsgGPgaywDw')
In the function "find_spreadsheet(url)", I could easily filter the link by looking for the class called "nowrap". Unfortunately, the Historical prices link does not have such a class and right now my script just gives me the following error:
AttributeError: ResultSet object has no attribute 'get'. You're probably treating a list of items like a single item. Did you call find_all() when you meant to call find()?
How do I make sure that my crawler only takes the href from the "Historical prices"?
Thank you in advance.
UPDATE:
I found the way to do it. By only looking for the link with a specific text attached to it, I could find the href I needed.
Solution:
soup.find('a', string="Historical prices")

Does the following code sniplet helps you? I think you can solve your problem with the following code as I hope:
from bs4 import BeautifulSoup
html = """<a href='http://www.google.com'>Something else</a>
<a href='http://www.yahoo.com'>Historical prices</a>"""
soup = BeautifulSoup(html, "html5lib")
urls = soup.find_all("a")
print(urls)
print([a["href"] for a in urls if a.text == "Historical prices"])

Python and BeautifulSoup Opening pages

I am wondering how would I open another page in my list with BeautifulSoup? I have followed this tutorial, but it does not tell us how to open another page on the list. Also how would I open a "a href" that is nested inside of a class?
Here is my code:
# coding: utf-8
import requests
from bs4 import BeautifulSoup
r = requests.get("")
soup = BeautifulSoup(r.content)
soup.find_all("a")
for link in soup.find_all("a"):
print link.get("href")
for link in soup.find_all("a"):
print link.text
for link in soup.find_all("a"):
print link.text, link.get("href")
g_data = soup.find_all("div", {"class":"listing__left-column"})
for item in g_data:
print item.contents
for item in g_data:
print item.contents[0].text
print link.get('href')
for item in g_data:
print item.contents[0]
I am trying to collect the href's from the titles of each business, and then open them and scrape that data.

I am still not sure where you are getting the HTML from, but if you are trying to extract all of the href tags, then the following approach should work based on the image you have posted:
import requests
from bs4 import BeautifulSoup
r = requests.get("<add your URL here>")
soup = BeautifulSoup(r.content)
for a_tag in soup.find_all('a', class_='listing-name', href=True):
print 'href: ', a_tag['href']
By adding href=True to the find_all(), it ensures that only a elements that contain an href attribute are returned therefore removing the need to test for it as an attribute.
Just to warn you, you might find some websites will lock you out after one or two attempts as they are able to detect that you are trying to access a site via a script, rather than as a human. If you feel you are not getting the correct responses, I would recommend printing the HTML you are getting back to ensure it it still as you expect.
If you then want to get the HTML for each of the links, the following could be used:
import requests
from bs4 import BeautifulSoup
# Configure this to be your first request URL
r = requests.get("http://www.mywebsite.com/search/")
soup = BeautifulSoup(r.content)
for a_tag in soup.find_all('a', class_='listing-name', href=True):
print 'href: ', a_tag['href']
# Configure this to the root of the above website, e.g. 'http://www.mywebsite.com'
base_url = "http://www.mywebsite.com"
for a_tag in soup.find_all('a', class_='listing-name', href=True):
print '-' * 60 # Add a line of dashes
print 'href: ', a_tag['href']
request_href = requests.get(base_url + a_tag['href'])
print request_href.content
Tested using Python 2.x, for Python 3.x please add parentheses to the print statements.

I had the same problem and I will like to share my findings because I did try the answer, for some reasons it did not work but after some research I found something interesting.
You might need to find the attributes of the "href" link itself:
You will need the exact class which contains the href link in your case, I am thinking="class":"listing__left-column" and equate it to a variable say "all" for example:
from bs4 import BeautifulSoup
all = soup.find_all("div", {"class":"listing__left-column"})
for item in all:
for link in item.find_all("a"):
if 'href' in link.attrs:
a = link.attrs['href']
print(a)
print("")
I did this and I was able to get into another link which was embedded in the home page