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Python Search multiple html files in a variable

I have used Selenium driver to crawl through many site pages. Every time I get a new page I append the html to a variable called "All_APP_Pages". The variable All_APP_Pages is a variable holding html for many pages. Did not post code because its long and no relevant to issue. Python list "All_APP_Pages" as being of type bytes.
from lxml import html
from lxml import etree
import xml.etree.ElementTree as ET
from selenium.webdriver.common.by import By
dom = etree.HTML(All_APP_Pages)
xp = "//tr[.//span[contains(.,'Product Data Solutions (UHC MR)')] and .//td[contains(.,'SQLServer')] and .//td[contains(.,'MR')]]//a"
link = dom.xpath(xp)
print(link)
Once all pages have been scanned I need to get the link from this xpath
"//tr[.//span[contains(.,'Product Data Solutions (ABC MR)')] and .//td[contains(.,'SQLServer')] and .//td[contains(.,'MR')]]//a"
The xpath listed here works. However it only works with the selenium driver if driver is on the page where this link exists. That is why all page are in one variable since I dont know what page the link will be on. The print shows this result
[<Element a at 0x1c39dea1180>]
How do I get this value from link I so can check if value is correct?

You need to iterate the list and get the href value
dom = etree.HTML(All_APP_Pages)
xp = "//tr[.//span[contains(.,'Product Data Solutions (UHC MR)')] and .//td[contains(.,'SQLServer')] and .//td[contains(.,'MR')]]//a"
link = dom.xpath(xp)
hrefs=[l.attrib["href"] for l in link]
print(hrefs)

I can't seem to scrape hrefs or text from web

I've tried various methods including searching through full xpath and finding via class then search for the href via the a tag. I've been at it for hours and I just cant seem to get it. Any help would be greatly appreciated.
from selenium import webdriver
import time
namelist=[]
driver=webdriver.Chrome()
driver.get("https://waxpeer.com/")
time.sleep(15)
when i use .text it just doesn't work
search=driver.find_elements_by_xpath('//*[#id="container"]/div/div/a')
print(search)
namelist.append(search)

Hey you can use time module give delay for loading the elements so it can find the associate tag
As you mention from this code you can extract href links from div tag container
Code:
from selenium import webdriver
import time
path="C:\Program Files (x86)\chromedriver.exe"
driver=webdriver.Chrome(path)
driver.get("https://waxpeer.com/")
time.sleep(10)
main_div=driver.find_elements_by_xpath('//*[#id="container"]')
for divs in main_div:
links=div.find_elements_by_tag_name("a")
for link in links:
print(link.get_attribute('href'))
Output:
https://waxpeer.com/sport-gloves-vice-field-tested/item/21642893513
...

Unable to print the href?

I have been searching across the site in the hope of finding an answer, however, every question I view doesn't have heavily nested HTML code like the page I am trying to scrape.
I am really hoping someone will spot my obvious error. I have the following code which is pulling the category headers and but annoyingly not the href that goes with each one. When run, the code currently returns 'None' for all the href's but I cannot decipher why. I think it may be because I am targeting the wrong element, tag or class in the HTML but cannot correctly identify which one it should be.
from selenium import webdriver
import time
# The website to scrape
url = "https://www.jtinsight.com/JTIRA/JTIRA.aspx#!/full-category-list"
# Creating the WebDriver object using the ChromeDriver
driver = webdriver.Chrome()
# Directing the driver to the defined url
driver.get(url)
# driver.implicitly_wait(5)
time.sleep(1)
# Locate the categories
categories = driver.find_elements_by_xpath('//div[#class="subCatEntry ng-scope"]')
# Print out all categories on current page
num_page_items = len(categories)
print(num_page_items)
for headers in range(num_page_items):
print(categories[headers].text)
for elem in categories:
print(elem.get_attribute("a.divLink[href='*']"))
# Clean up (close browser once task is completed)
time.sleep(1)
driver.close()
I would really appreciate if anyone can point out my error.

Try this below code.
for elem in categories:
print(elem.find_element_by_css_selector("a.divLink").get_attribute('href'))

You are passing the CSS selector for the get_attribute method. That wouldn't work. You have to provide the attribute name only. If the web element elem has an attribute named href then it would print the value of that attribute.
First, get the anchor <a> element. All the subcategory anchors have class divLink. For getting anchor elements try this,
categories = driver.find_elements_by_class_name('divLink')
Second, Print the attribute value by passing the attribute name in the get_ttribute. Try this,
print(elem.get_attribute("href"))
This way you'll be able to print all the href values.

Trying to select element by xpath with Selenium but getting error "Unable to locate element"

I am trying to scrape the list of followings for a given instagram user. This requires using Selenium to navigate to the user's Instagram page and then clicking "following". However, I cannot seem to click the "following" button with Selenium.
driver = webdriver.Chrome()
url = 'https://www.instagram.com/beforeeesunrise/'
driver.get(url)
driver.find_element_by_xpath('//*[#id="react-root"]/section/main/article/header/div[2]/ul/li[3]/a').click()
However, this results in a NoSuchElementException. I copied the xpath from the html, tried using the class name, partial link and full link and cannot seem to get this to work! I've also made sure that the above xpath include the element with a "click" event listener.
UPDATE: By logging in I was able to get the above information. However (!), now I cannot get the resulting list of "followings". When I click on the button with the driver, the html does not include the information in the pop up dialog that you see on Instagram. My goal is to get all of the users that the given username is following.

Make sure you are using the correct X Path.
Use the following link to get perfect X Paths to access web elements and then try.
Selenium Command
Hope this helps to solve the problem!

Try a different XPath. I've verified this is unique on the page.
driver.find_element_by_xpath("//a[contains(.,'following')]")

It's not the main goal of selenium to provide rich functionalities, from a web-scraping perspective, to find elements on the page, so the better option is to delegate this task to a specific tool, like BeautifulSoup. After we find what we're looking for, then, we can ask for selenium to interact with the element.
The bridge between selenium and BeautifulSoup will be this amazing function below that I found here. The function gets a single BeautifulSoup element and generates a unique XPATH that we can use on selenium.
import os
import re
from selenium import webdriver
from bs4 import BeautifulSoup as bs
import itertools
def xpath_soup(element):
"""
Generate xpath of soup element
:param element: bs4 text or node
:return: xpath as string
"""
components = []
child = element if element.name else element.parent
for parent in child.parents:
"""
#type parent: bs4.element.Tag
"""
previous = itertools.islice(parent.children, 0, parent.contents.index(child))
xpath_tag = child.name
xpath_index = sum(1 for i in previous if i.name == xpath_tag) + 1
components.append(xpath_tag if xpath_index == 1 else '%s[%d]' % (xpath_tag, xpath_index))
child = parent
components.reverse()
return '/%s' % '/'.join(components)
driver = webdriver.Chrome(executable_path=YOUR_CHROMEDRIVER_PATH)
driver.get(url = 'https://www.instagram.com/beforeeesunrise/')
source = driver.page_source
soup = bs(source, 'html.parser')
button = soup.find('button', text=re.compile(r'Follow'))
xpath_for_the_button = xpath_soup(button)
elm = driver.find_element_by_xpath(xpath_for_the_button)
elm.click()
...and works!
( but you need writing some code to log in with an account)

Parsing HTML Content using BeautifulSoup & Selenium

from selenium import webdriver
from selenium.webdriver.support.ui import Select
from bs4 import BeautifulSoup
import csv
import requests
import re
driver2 = webdriver.Chrome()
driver2.get("http://www.squawka.com/match-results?ctl=10_s2015")
soup=BeautifulSoup(driver2.page_source)
print soup
driver2.quit()
I'm trying to get the HREF of every "td", "Class":"Match Centre" and I need to use selenium to navigate through the pages but im struggling to incorporate the two so I can change the menu options and navigate through the different pages while feeding the links into my other code.
I've researched and tried ('inner-html') and the page.source currently in the code, but it doesn't get any of the web links I need.
Does anyone have a solution to get these links and navigate on the page. Could there be a way to get the XML of this page to get all the links?

Not sure why would you need BeautifulSoup (BS) here. Selenium alone is capable of locating elements and navigating through links on a page. For example, to get all the links to the match details page you can do as follow :
>>> matches = driver.find_elements_by_xpath("//td[#class='match-centre']/a")
>>> print [match.get_attribute("href") for match in matches]
As for navigating through the pages, you can use the following XPath :
//span[contains(#class,'page-numbers')]/following-sibling::a[1]
The above XPath finds link to the next page. To navigate through all the pages, you can try using a while loop; while the link to the next page is found :
perform a click action on the link,
grab all the href from current page,
locate the next page link.