Extracting an attribute value with beautifulsoup

Question:

I am trying to extract the content of a single "value" attribute in a specific "input" tag on a webpage. I use the following code:

import urllib
f = urllib.urlopen("http://58.68.130.147")
s = f.read()
f.close()

from BeautifulSoup import BeautifulStoneSoup
soup = BeautifulStoneSoup(s)

inputTag = soup.findAll(attrs={"name" : "stainfo"})

output = inputTag['value']

print str(output)

I get TypeError: list indices must be integers, not str

Even though, from the Beautifulsoup documentation, I understand that strings should not be a problem here… but I am no specialist, and I may have misunderstood.

Any suggestion is greatly appreciated!

Asked By: Barnabe

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Source

Answers:

.find_all() returns list of all found elements, so:

input_tag = soup.find_all(attrs={"name" : "stainfo"})

input_tag is a list (probably containing only one element). Depending on what you want exactly you either should do:

output = input_tag[0]['value']

or use .find() method which returns only one (first) found element:

input_tag = soup.find(attrs={"name": "stainfo"})
output = input_tag['value']
Answered By: Łukasz

I would actually suggest you a time saving way to go with this assuming that you know what kind of tags have those attributes.

suppose say a tag xyz has that attritube named “staininfo”..

full_tag = soup.findAll("xyz")

And i wan’t you to understand that full_tag is a list

for each_tag in full_tag:
    staininfo_attrb_value = each_tag["staininfo"]
    print staininfo_attrb_value

Thus you can get all the attrb values of staininfo for all the tags xyz

Answered By: b1tchacked

If you want to retrieve multiple values of attributes from the source above, you can use findAll and a list comprehension to get everything you need:

import urllib
f = urllib.urlopen("http://58.68.130.147")
s = f.read()
f.close()

from BeautifulSoup import BeautifulStoneSoup
soup = BeautifulStoneSoup(s)

inputTags = soup.findAll(attrs={"name" : "stainfo"})
### You may be able to do findAll("input", attrs={"name" : "stainfo"})

output = [x["stainfo"] for x in inputTags]

print output
### This will print a list of the values.
Answered By: Margath

In Python 3.x, simply use get(attr_name) on your tag object that you get using find_all:

xmlData = None

with open('conf//test1.xml', 'r') as xmlFile:
    xmlData = xmlFile.read()

xmlDecoded = xmlData

xmlSoup = BeautifulSoup(xmlData, 'html.parser')

repElemList = xmlSoup.find_all('repeatingelement')

for repElem in repElemList:
    print("Processing repElem...")
    repElemID = repElem.get('id')
    repElemName = repElem.get('name')

    print("Attribute id = %s" % repElemID)
    print("Attribute name = %s" % repElemName)

against XML file conf//test1.xml that looks like:

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<root>
    <singleElement>
        <subElementX>XYZ</subElementX>
    </singleElement>
    <repeatingElement id="11" name="Joe"/>
    <repeatingElement id="12" name="Mary"/>
</root>

prints:

Processing repElem...
Attribute id = 11
Attribute name = Joe
Processing repElem...
Attribute id = 12
Attribute name = Mary
Answered By: amphibient

you can also use this :

import requests
from bs4 import BeautifulSoup
import csv

url = "http://58.68.130.147/"
r = requests.get(url)
data = r.text

soup = BeautifulSoup(data, "html.parser")
get_details = soup.find_all("input", attrs={"name":"stainfo"})

for val in get_details:
    get_val = val["value"]
    print(get_val)
Answered By: Mr.Bones

I am using this with Beautifulsoup 4.8.1 to get the value of all class attributes of certain elements:

from bs4 import BeautifulSoup

html = "<td class='val1'/><td col='1'/><td class='val2' />"

bsoup = BeautifulSoup(html, 'html.parser')

for td in bsoup.find_all('td'):
    if td.has_attr('class'):
        print(td['class'][0])

Its important to note that the attribute key retrieves a list even when the attribute has only a single value.

Answered By: PeterXX

For me:

<input id="color" value="Blue"/>

This can be fetched by below snippet.

page = requests.get("https://www.abcd.com")
soup = BeautifulSoup(page.content, 'html.parser')
colorName = soup.find(id='color')
print(colorName['value'])
Answered By: vijayraj34

You could try to use the new powerful package called requests_html:

from requests_html import HTMLSession
session = HTMLSession()

r = session.get("https://www.bbc.co.uk/news/technology-54448223")
date = r.html.find('time', first = True) # finding a "tag" called "time"
print(date)  # you will have: <Element 'time' datetime='2020-10-07T11:41:22.000Z'>
# To get the text inside the "datetime" attribute use:
print(date.attrs['datetime']) # you will get '2020-10-07T11:41:22.000Z'
Answered By: Yasser M

You can try gazpacho:

Install it using pip install gazpacho

Get the HTML and make the Soup using:

from gazpacho import get, Soup

soup = Soup(get("http://ip.add.ress.here/"))  # get directly returns the html

inputs = soup.find('input', attrs={'name': 'stainfo'})  # Find all the input tags

if inputs:
    if type(inputs) is list:
        for input in inputs:
             print(input.attr.get('value'))
    else:
         print(inputs.attr.get('value'))
else:
     print('No <input> tag found with the attribute name="stainfo")
Answered By: Vthechamp

Here is an example for how to extract the href attrbiutes of all a tags:

import requests as rq 
from bs4 import BeautifulSoup as bs

url = "http://www.cde.ca.gov/ds/sp/ai/"
page = rq.get(url)
html = bs(page.text, 'lxml')

hrefs = html.find_all("a")
all_hrefs = []
for href in hrefs:
    # print(href.get("href"))
    links = href.get("href")
    all_hrefs.append(links)

print(all_hrefs)
Answered By: HackerGprat