Python to get onclick values - python

I'm using Python and BeautifulSoup to scrape a web page for a small project of mine. The webpage has multiple entries, each separated by a table row in HTML. My code partially works however a lot of the output is blank and it won't fetch all of the results from the web page or even gather them into the same line.
<html>
<head>
<title>Sample Website</title>
</head>
<body>
<table>
<td class=channel>Artist</td><td class=channel>Title</td><td class=channel>Date</td><td class=channel>Time</td></tr>
<tr><td>35</td><td>Lorem Ipsum</td><td>FooWorld</td><td>12/10/2014</td><td>2:53:17 PM</td></tr>
</table>
</body>
</html>
I want to only extract the values from the onclick action 'searchDB', so for example 'LoremIpsum' and 'FooWorld' are the only two results that I want.
Here is the code that I've written. So far, it properly pulls some of the write values, but sometimes the values are empty.
response = urllib2.urlopen(url)
html = response.read()
soup = bs4.BeautifulSoup(html)
properties = soup.findAll('a', onclick=True)
for eachproperty in properties:
print re.findall("'([a-zA-Z0-9]*)'", eachproperty['onclick'])
What am I doing wrong?

try like this:
>>> import re
>>> for x in soup.find_all('a'): # will give you all a tag
... try:
... if re.match('searchDB',x['onclick']): # if onclick attribute exist, it will match for searchDB, if success will print
... print x['onclick'] # here you can do your stuff instead of print
... except:pass
...
searchDB('LoremIpsum','FooWorld')
instead of print you can save it to some variable like
>>> k = x['onclick']
>>> re.findall("'(\w+)'",k)
['LoremIpsum', 'FooWorld']
\w is equivalent to [a-zA-Z0-9]

Try this
or row in rows[1:]:
cols = row.findAll('td')
link = cols[1].find('a').get('onclick')

Related

Python `beautifulsoup` extraction on urls lacking `class`, other attributes?

Quick question [I am not very familiar with Python's BeautifulSoup()] If I have the following element,
how can I extract/get "1 comment" (or, "2 comments", etc.)? There is no class (or id, or other attributes) in that "a" tag.
<td class="subtext">
1 comment
</td>
How about the following, test with local html file
from bs4 import BeautifulSoup
url = "D:\\Temp\\example.html"
with open(url, "r") as page:
contents = page.read()
soup = BeautifulSoup(contents, 'html.parser')
element = soup.select('td.subtext')
value = element[0].get_text()
print(value)
example.html
<html>
<head></head>
<body>
<td class="subtext">
1 comment
</td>
</body>
</html>
You can use select method to apply a querySelect into your html, and then take the contents of the elements you found:
elements = soup.select(".subtext a")
[x.contents for x in elements]

Beautiful Soup - Get all text, but preserve link html?

I have to process a large archive of extremely messy HTML full of extraneous tables, spans and inline styles into markdown.
I am trying to use Beautiful Soup to accomplish this task, and my goal is basically the output of the get_text() function, except to preserve anchor tags with the href intact.
As an example, I would like to convert:
<td>
<font><span>Hello</span><span>World</span></font><br>
<span>Foo Bar <span>Baz</span></span><br>
<span>Example Link: Google</span>
</td>
Into:
Hello World
Foo Bar Baz
Example Link: Google
My thought process so far was to simply grab all the tags and unwrap them all if they aren't anchors, but this causes the text to be repeated several times as soup.find_all(True) returns recursively nested tags as individual elements:
#!/usr/bin/env python
from bs4 import BeautifulSoup
example_html = '<td><font><span>Hello</span><span>World</span></font><br><span>Foo Bar <span>Baz</span></span><br><span>Example Link: Google</span></td>'
soup = BeautifulSoup(example_html, 'lxml')
tags = soup.find_all(True)
for tag in tags:
if (tag.name == 'a'):
print("<a href='{}'>{}</a>".format(tag['href'], tag.get_text()))
else:
print(tag.get_text())
Which returns multiple fragments/duplicates as the parser moves down the tree:
HelloWorldFoo Bar BazExample Link: Google
HelloWorldFoo Bar BazExample Link: Google
HelloWorldFoo Bar BazExample Link: Google
HelloWorld
Hello
World
Foo Bar Baz
Baz
Example Link: Google
<a href='https://google.com'>Google</a>
One of the possible ways to tackle this problem would be to introduce some special handling for a elements when it comes to printing out a text of an element.
You can do it by overriding _all_strings() method and returning a string representation of an a descendant element and skip a navigable string inside an a element. Something along these lines:
from bs4 import BeautifulSoup, NavigableString, CData, Tag
class MyBeautifulSoup(BeautifulSoup):
def _all_strings(self, strip=False, types=(NavigableString, CData)):
for descendant in self.descendants:
# return "a" string representation if we encounter it
if isinstance(descendant, Tag) and descendant.name == 'a':
yield str(descendant)
# skip an inner text node inside "a"
if isinstance(descendant, NavigableString) and descendant.parent.name == 'a':
continue
# default behavior
if (
(types is None and not isinstance(descendant, NavigableString))
or
(types is not None and type(descendant) not in types)):
continue
if strip:
descendant = descendant.strip()
if len(descendant) == 0:
continue
yield descendant
Demo:
In [1]: data = """
...: <td>
...: <font><span>Hello</span><span>World</span></font><br>
...: <span>Foo Bar <span>Baz</span></span><br>
...: <span>Example Link: <a href="https://google.com" target="_blank" style="mso-line-height-rule: exactly;-ms-text-size-adjust: 100%;-webkit-text-size-adjust: 100%;color: #395c99;font-weight: normal;tex
...: t-decoration: underline;">Google</a></span>
...: </td>
...: """
In [2]: soup = MyBeautifulSoup(data, "lxml")
In [3]: print(soup.get_text())
HelloWorld
Foo Bar Baz
Example Link: Google
To only consider direct children set recursive = False then you need to process each 'td' and extract the text and anchor link individually.
#!/usr/bin/env python
from bs4 import BeautifulSoup
example_html = '<td><font><span>Some Example Text</span></font><br><span>Another Example Text</span><br><span>Example Link: Google</span></td>'
soup = BeautifulSoup(example_html, 'lxml')
tags = soup.find_all(recursive=False)
for tag in tags:
print(tag.text)
print(tag.find('a'))
If you want the text printed on separate lines you will have to process the spans individually.
for tag in tags:
spans = tag.find_all('span')
for span in spans:
print(span.text)
print(tag.find('a'))

Using Python + BeautifulSoup to extract tags in tandem, creating a list of lists

I'm a bit new to python/BeautifulSoup, and was wondering if I could get some direction on how to get the following accomplished.
I have html from a webpage, that is structured as follows:
1) block of code contained within a tag that contains all image names (Name1, Name2, Name3.
2) block of code contained within a tag that has image urls.
3) a date, that appears one on the webpage. I put it into 'date' variable (this has already been extracted)
From the code, I'm trying to extract a list of lists that will contain [['image1','url1', 'date'], ['image2','url2','date']], which i will later convert into a dictionary (via dict(zip(labels, values)) function), and insert into a mysql table.
All I can come up with is how to extract two lists that contain all images , and all url's. Any idea on how to get what i'm trying to do accomplished?
Few things to keep in mind:
1) number of images always changes, along with names (1:1)
2) date always appears once.
P.S. Also, if there is a more elegant way to extract the data via bs4, please let me know!
from bs4 import BeautifulSoup
name = []
url = []
date = '2017-10-12'
text = '<div class="tabs"> <ul><li> NAME1</li><li> NAME2</li><li> NAME3</li> </ul> <div><div><div class="img-wrapper"><img alt="" src="www.image1.com/1.jpg" title="image1.jpg"></img> </div> <center><a class="button print" href="javascript: w=window.open("www.image1.com/1.jpg); w.print();"> Print</a> </center></div><div> <div class="img-wrapper"><img alt="" src="www.image2.com/2.jpg" title="image2.jpg"></img> </div> <center><a class="button print" href="javascript: w=window.open("www.image2.com/2.jpg"); w.print();">Print</a> </center></div><div> <div class="img-wrapper"><img alt="" src="www.image1.com/3.jpg" title="image3.jpg"></img></div> <center><a class="button print" href="javascript: w=window.open("www.image1.com/3.jpg"); w.print();"> Print</a> </center></div> </div></div>'
soup = BeautifulSoup(text, 'lxml')
#print soup.prettify()
#get names
for imgz in soup.find_all('div', attrs={'class':'img-wrapper'}):
for imglinks in imgz.find_all('img', src = True):
#print imgz
url.append((imglinks['src']).encode("utf-8"))
#3 get ad URLS
for ultag in soup.find_all('ul'):
for litag in ultag.find_all('li'):
name.append((litag.text).encode("utf-8")) #dump all urls into a list
print url
print name
Here's another possible route to pulling the urls and names:
url = [tag.get('src') for tag in soup.find_all('img')]
name = [tag.text.strip() for tag in soup.find_all('li')]
print(url)
# ['www.image1.com/1.jpg', 'www.image2.com/2.jpg', 'www.image1.com/3.jpg']
print(name)
# ['NAME1', 'NAME2', 'NAME3']
As for ultimate list creation, here's something that's functionally similar to what #t.m.adam has suggested:
print([pair + [date] for pair in list(map(list, zip(url, name)))])
# [['www.image1.com/1.jpg', 'NAME1', '2017-10-12'],
# ['www.image2.com/2.jpg', 'NAME2', '2017-10-12'],
# ['www.image1.com/3.jpg', 'NAME3', '2017-10-12']]
Note that map is pretty infrequently used nowadays and its use is outright discouraged in some places.
Or:
n = len(url)
print(list(map(list, zip(url, name, [date] * n))))
# [['www.image1.com/1.jpg', 'NAME1', '2017-10-12'], ['www.image2.com/2.jpg', 'NAME2', '2017-10-12'], ['www.image1.com/3.jpg', 'NAME3', '2017-10-12']]

Extracting data from an inconsistent HTML page using BeautifulSoup4 and Python

I’m trying to extract data from this webpage and I'm having some trouble due to inconsistancies within the page's HTML formatting. I have a list of OGAP IDs and I want to extract the Gene Name and any literature information (PMID #) for each OGAP ID I iterate through. Thanks to other questions on here, and the BeautifulSoup documentation, I've been able to consistantly get the gene name for each ID, but I'm having trouble with the literature part. Here's a couple search terms that highlight the inconsitancies.
HTML sample that works
Search term: OG00131
<tr>
<td colspan="4" bgcolor="#FBFFCC" class="STYLE28">Literature describing O-GlcNAcylation:
<br> PMID:
20068230
[CAD, ETD MS/MS]; <br>
<br>
</td>
</tr>
HTML sample that doesn't work
Search term: OG00020
<td align="top" bgcolor="#FBFFCC">
<div class="STYLE28">Literature describing O-GlcNAcylation: </div>
<div class="STYLE28">
<div class="STYLE28">PMID:
16408927
[Azide-tag, nano-HPLC/tandem MS]
</div>
<br>
Site has not yet been determined. Use
OGlcNAcScan
to predict the O-GlcNAc site. </div>
</td>
Here's the code I have so far
import urllib2
from bs4 import BeautifulSoup
#define list of genes
#initialize variables
gene_list = []
literature = []
# Test list
gene_listID = ["OG00894", "OG00980", "OG00769", "OG00834","OG00852", "OG00131","OG00020"]
for i in range(len(gene_listID)):
print gene_listID[i]
# Specifies URL, uses the "%" to sub in different ogapIDs based on a list provided
dbOGAP = "https://wangj27.u.hpc.mssm.edu/cgi-bin/DB_tb.cgi?textfield=%s&select=Any" % gene_listID[i]
# Opens the URL as a page
page = urllib2.urlopen(dbOGAP)
# Reads the page and parses it through "lxml" format
soup = BeautifulSoup(page, "lxml")
gene_name = soup.find("td", text="Gene Name").find_next_sibling("td").text
print gene_name[1:]
gene_list.append(gene_name[1:])
# PubMed IDs are located near the <td> tag with the term "Data and Source"
pmid = soup.find("span", text="Data and Source")
# Based on inspection of the website, need to move up to the parent <td> tag
pmid_p = pmid.parent
# Then we move to the next <td> tag, denoted as sibling (since they share parent <tr> (Table row) tag)
pmid_s = pmid_p.next_sibling
#for child in pmid_s.descendants:
# print child
# Now we search down the tree to find the next table data (<td>) tag
pmid_c = pmid_s.find("td")
temp_lit = []
# Next we print the text of the data
#print pmid_c.text
if "No literature is available" in pmid_c.text:
temp_lit.append("No literature is available")
print "Not available"
else:
# and then print out a list of urls for each pubmed ID we have
print "The following is available"
for link in pmid_c.find_all('a'):
# the <a> tag includes more than just the link address.
# for each <a> tag found, print the address (href attribute) and extra bits
# link.string provides the string that appears to be hyperlinked.
# In this case, it is the pubmedID
print link.string
temp_lit.append("PMID: " + link.string + " URL: " + link.get('href'))
literature.append(temp_lit)
print "\n"
So it seems the element is what is throwing the code for a loop. Is there a way to search for any element with the text "PMID" and return the text that comes after it (and url if there is a PMID number)? If not, would I just want to check each child, looking for the text I'm interested in?
I'm using Python 2.7.10
import requests
from bs4 import BeautifulSoup
import re
gene_listID = ["OG00894", "OG00980", "OG00769", "OG00834","OG00852", "OG00131","OG00020"]
urls = ('https://wangj27.u.hpc.mssm.edu/cgi-bin/DB_tb.cgi?textfield={}&select=Any'.format(i) for i in gene_listID)
for url in urls:
r = requests.get(url)
soup = BeautifulSoup(r.text, 'lxml')
regex = re.compile(r'http://www.ncbi.nlm.nih.gov/pubmed/\d+')
a_tag = soup.find('a', href=regex)
has_pmid = 'PMID' in a_tag.previous_element
if has_pmid :
print(a_tag.text, a_tag.next_sibling, a_tag.get("href"))
else:
print("Not available")
out:
18984734 [GalNAz-Biotin tagging, CAD MS/MS]; http://www.ncbi.nlm.nih.gov/pubmed/18984734
20068230 [CAD, ETD MS/MS]; http://www.ncbi.nlm.nih.gov/pubmed/20068230
20068230 [CAD, ETD MS/MS]; http://www.ncbi.nlm.nih.gov/pubmed/20068230
Not available
16408927 [Azide-tag, nano-HPLC/tandem MS]; http://www.ncbi.nlm.nih.gov/pubmed/16408927
Not available
16408927 [Azide-tag, nano-HPLC/tandem MS] http://www.ncbi.nlm.nih.gov/pubmed/16408927?dopt=Citation
find the first a tag that match the target url, which end with numbers, than check if 'PMID' in it's previous element.
this web is so inconsistancies , and i try many times, hope this would help

Reading web pages with Python

I'm trying to read and handle a web-page in Python which has lines like the following in it:
<div class="or_q_tagcloud" id="tag1611"></div></td></tr><tr><td class="or_q_artist"><a title="[Artist916]" href="http://rateyourmusic.com/artist/ac_dc" class="artist">AC/DC</a></td><td class="or_q_album"><a title="[Album374717]" href="http://rateyourmusic.com/release/album/ac_dc/live_f5/" class="album">Live</a></td><td class="or_q_rating" id="rating374717">4.0</td><td class="or_q_ownership" id="ownership374717">CD</td><td class="or_q_tags_td">
I'm currently only interested in the artist name (AC/DC) and album name (Live). I can read and print them with libxml2dom but I can't figure out how I can distinguish between the links because the node value for every link is None.
One obvious way would be to read the input line at a time but is there a more clever way of handling this html file so that I can create either two separate lists where each index matches the other or a struct with this info?
import urllib
import sgmllib
import libxml2dom
def collect_text(node):
"A function which collects text inside 'node', returning that text."
s = ""
for child_node in node.childNodes:
if child_node.nodeType == child_node.TEXT_NODE:
s += child_node.nodeValue
else:
s += collect_text(child_node)
return s
f = urllib.urlopen("/home/x/Documents/rym_list.html")
s = f.read()
doc = libxml2dom.parseString(s, html=1)
links = doc.getElementsByTagName("a")
for link in links:
print "--\nNode " , artist.childNodes
if artist.localName == "artist":
print "artist"
print collect_text(artist).encode('utf-8')
f.close()
Given the small snippit of HTML, I've no idea whether this would be effective on the full page, but here's how to extract 'AC/DC' and 'Live' using lxml.etree and xpath.
>>> from lxml import etree
>>> doc = etree.HTML("""<html>
... <head></head>
... <body>
... <tr>
... <td class="or_q_artist"><a title="[Artist916]" href="http://rateyourmusic.com/artist/ac_dc" class="artist">AC/DC</a></td>
... <td class="or_q_album"><a title="[Album374717]" href="http://rateyourmusic.com/release/album/ac_dc/live_f5/" class="album">Live</a></td>
... <td class="or_q_rating" id="rating374717">4.0</td><td class="or_q_ownership" id="ownership374717">CD</td>
... <td class="or_q_tags_td">
... </tr>
... </body>
... </html>
... """)
>>> doc.xpath('//td[#class="or_q_artist"]/a/text()|//td[#class="or_q_album"]/a/text()')
['AC/DC', 'Live']
See if you can solve the problem in javascript using jQuery style DOM/CSS selectors to get at the elements/text that you want.
If you can then get a copy of BeautifulSoup for python and you should be good to go in a matter of minutes.

Categories