Read csv file from url

Question

I want to read url but they give me error that URL must be a string is there any possible solution for that kindly recommend me how to make the URL string kindly suggest any method

from selenium import webdriver           
import time
from selenium.webdriver.chrome.service import Service
from selenium.webdriver.common.by import By
from webdriver_manager.chrome import ChromeDriverManager
from bs4 import BeautifulSoup
import pandas as pd
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.chrome.service import Service
from selenium.webdriver.common.by import By
from selenium.webdriver.support.wait import WebDriverWait
import requests
from csv import writer


with open("data.csv") as file:
    start_urls=[line.strip() for line in file]  

    options = webdriver.ChromeOptions()
    options.add_argument("--no-sandbox")
    options.add_argument("--disable-gpu")
    options.add_argument("--window-size=1920x1080")
    options.add_argument("--disable-extensions")
    driver = webdriver.Chrome(service=Service(ChromeDriverManager().install()))

    driver.get(start_urls)

Asked By: Amen Aziz

||

Source

Answer 1

As commented by @JeJe,

you need to iterate through your start_urls list

start_urls is a LIST of strings, and driver.get expects just one url as string – something like

# driver.get(start_urls) # NOT like this
for sUrl in start_urls:
    driver.get(sUrl)
    ### SCRAPE AS NEEDED ###

or, if you want to keep track of progress, something like

suLen = len(start_urls)
for si, sUrl in enumerate(start_urls):
    print(f'[{si} of {suLen}] reading {sUrl}')
    driver.get(sUrl)
    ### SCRAPE AS NEEDED ###

Btw, you don’t need to do everything under with open... – getting start_urls is enough:

with open("data.csv") as file:
    start_urls=[line.strip() for line in file]  

options = webdriver.ChromeOptions()
options.add_argument("--no-sandbox")
options.add_argument("--disable-gpu")
options.add_argument("--window-size=1920x1080")
options.add_argument("--disable-extensions")
driver = webdriver.Chrome(service=Service(ChromeDriverManager().install()))

suLen = len(start_urls)
for si, sUrl in enumerate(start_urls):
    print(f'[{si} of {suLen}] reading {sUrl}')
    driver.get(sUrl)
    ### SCRAPE AS NEEDED ###

but you can only use just start_urls=[line.strip() for line in file] if you csv has only on column with no headers.

Answered By: Driftr95

Answer 2

you could rewrite your code using functions to deal with each step of your code.


def read_file(has_header:bool=True):
    with open("data.csv") as file:
        start_urls = [line.strip() for line in file]

    # assumes there is a header
    if has_header:
        return start_urls[1:]
    else:
        return start_urls

def scrap_urls():
    urls = read_file(has_header=True)

    options = webdriver.ChromeOptions()
    options.add_argument("--no-sandbox")
    options.add_argument("--disable-gpu")
    options.add_argument("--window-size=1920x1080")
    options.add_argument("--disable-extensions")
    driver = webdriver.Chrome(service=Service(ChromeDriverManager().install()))

    for url in urls:
        driver.get(url)

if __name__ == '__main__':
    
    scrap_urls()

Answered By: Je Je

Read csv file from url

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