linkScrape | Enumerates employee names from LinkedIn.com | Portal library
kandi X-RAY | linkScrape Summary
kandi X-RAY | linkScrape Summary
Enumerates employee names from LinkedIn.com based off company search results.
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Top functions reviewed by kandi - BETA
- Name of linkScrape - data .
- Launch linkScrape wizard .
- Connects to the linked site .
- Help for linkScrape . py .
- Write linkScrape data to file .
- Clear the system .
linkScrape Key Features
linkScrape Examples and Code Snippets
Community Discussions
Trending Discussions on linkScrape
QUESTION
I am attempting to scrape sports schedule from multiple links on a site. The URL's are being found and printed correctly, but only data from the last scraped URL is being outputted to console and text file.
My code is below:
...ANSWER
Answered 2020-Aug-01 at 04:43You are correct the problem does lie in this line of code:
QUESTION
I am currently using the code below to scrape data from sports schedule sites and output the information to text files. Currently with the code I have, the data correctly prints to the console and data from the first URL (https://sport-tv-guide.live/live/darts) is outputted to the text file as expected.
The problem is that the content from the second URL (https://sport-tv-guide.live/live/boxing/) is not outputted to the expected text file( the text file is created but there is no content in it).
The code I am using is below:
...ANSWER
Answered 2020-Jul-24 at 06:24Found the problem. In your code, for boxing
url - https://sport-tv-guide.live/live/boxing/
there are no extra channels. Hence, the control won't go inside the loop and there is no output written to file.
You can collect all the extra channels in a list and then write to file
QUESTION
I am attempting to receive a list of urls that are on the following page
https://sport-tv-guide.live/live/tennis
When these URLS are gathered, I then need to pass each URL to a scrape function to scrape and output the relevant match data.
The data is correctly outputted if there is only one match on a specific page such as - https://sport-tv-guide.live/live/darts (see output below)
The issue occurs when I use a page with more than one link present such as - https://sport-tv-guide.live/live/tennis , it appears that the URLs are being scraped correctly (confirmed with using print, to print URLS) but they don't seem to be passed correctly for the content to be scraped, as the script just fails silently (see output below )
The code is below:
...ANSWER
Answered 2020-Jul-19 at 06:36After analysing the links, the 2 links point to different pages with different layouts.
https://sport-tv-guide.live/live/tennis - Using this link when you get all the links, they point to different page layout.
https://sport-tv-guide.live/live/darts - the links in this page point to this layout.
If you need to scrape the data from all the links from https://sport-tv-guide.live/live/tennis, the following script works.
QUESTION
I am using Scrapy and am having trouble with the script. It works fine with the shell:
scrapy shell "www.redacted.com"
I use response.xpath("//li[@a data-urltype()"]).extract
I am able to scrape 200 or so links from the page.
Here is the code from the webpage I am trying to scrape:
...ANSWER
Answered 2019-Oct-23 at 03:05if you are going to scrape data-val
from a
. use below xpath.
QUESTION
I've written a script in python using two different links (one has pagination but the other doesn't)
to see whether my script can fetch all the next page links. It is necessary that the script must print this No pagination found
line if there is no pagination option.
I've applied @check_pagination
decorator to check for the existance of pagination and I want to keep this decorator within my scraper.
I've already achieved what I've described above complying the following:
...ANSWER
Answered 2018-Dec-05 at 19:38Simply apply the decorator to get_base
:
QUESTION
I am trying to scrape some links with headless-chrome/puppeteer while scrolling down like this:
...ANSWER
Answered 2018-Feb-02 at 08:00I can find the following possible reasons why your interval would not get stopped:
- You are never getting to the
stop
condition. - You are overwriting the
interval
variable somehow so the actual interval you want to stop is no longer saved. - You are getting a rejected promise.
There does not appear to be any reason why the interval
variable needs to be outside the linkScraper
function and putting it inside the function will prevent it from getting overwritten in any way.
With this many await
calls, it seems wise to add a try/catch to catch any rejected promises and stop the interval if there's an error.
If you see the STOPPING
being logged, then you are apparently hitting the stop condition so it appears it would have to be an overwritten interval
variable.
Here's a version that cannot overwrite the interval
variable and makes a few other changes for code cleanliness:
Community Discussions, Code Snippets contain sources that include Stack Exchange Network
Vulnerabilities
No vulnerabilities reported
Install linkScrape
You can use linkScrape like any standard Python library. You will need to make sure that you have a development environment consisting of a Python distribution including header files, a compiler, pip, and git installed. Make sure that your pip, setuptools, and wheel are up to date. When using pip it is generally recommended to install packages in a virtual environment to avoid changes to the system.
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