How to make python script wait until entering specific URL

Question

I want python script to be triggered by entering specific URLs with specific part of address.

Here is an example:

http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=1234
#1234 can be any random 4 digits Number

Basically I want python script to activate when URL has "http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=" in it.

Here is what I wrote so far:

from selenium import webdriver
from selenium.webdriver.support.ui import WebDriverWait

driver = webdriver.Chrome('./chromedriver')
wait = WebDriverWait(driver, 9999)
desired_url = http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=\d{4}
def wait_for_correct_current_url):
      wait.until(lambda driver: driver.current_url == desired_url)



driver.get("http://www.google.com")
wait_for_correct_current_url(desired_url)
**(Script that activates after entering desired_url)**

I am wondering if regex will do the trick, but I am new to python... so what do i know.

Thanks in advance!

Use this selenium api expected_conditions.url_changes

Kamal
– Kamal

2019-02-21 05:33:15 +00:00
Commented Feb 21, 2019 at 5:33 — Kamal
– Kamal, Commented Feb 21, 2019 at 5:33
I got this: 'WebDriver' object has no attribute 'support'

NulliPy
– NulliPy

2019-02-21 05:45:38 +00:00
Commented Feb 21, 2019 at 5:45 — NulliPy
– NulliPy, Commented Feb 21, 2019 at 5:45

Kamal · Accepted Answer · 2019-02-21 05:51:15Z

1

Use like this,

from selenium.webdriver.support import expected_conditions as EC

WebDriverWait(driver, 1000).until(EC.url_contains("desired_url"))

answered Feb 21, 2019 at 5:51

Kamal

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Comments

undetected Selenium · Accepted Answer · 2019-02-21 07:07:13Z

A bit more information about your usecase would have helped to provide a more canonical answer.

However to invoke a URL with specific part of address to be random, you can identify the range through range() and iterate the range through format() to construct the complete URL and you can use the following solution:

Code Block:

for i in range(1000,1010):
    desired_url = "http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER={}".format(i)
    print(desired_url)

Console Output:

http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=1000
http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=1001
http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=1002
http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=1003
http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=1004
http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=1005
http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=1006
http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=1007
http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=1008
http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=1009

Incase you are looking for random numbers as a part of the URL, you can use randrange() to generate as follows:

Code Block:

import random

for i in range(0,10) :
    desired_url = "http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER={}".format(str(random.randrange(1000, 9999)))
    print(desired_url)

Console Output:

http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=8776
http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=1662
http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=3255
http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=1524
http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=6463
http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=4511
http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=3273
http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=7471
http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=4672
http://11.111.11.11:0000/Menu_EXAMPLE.jsp?NUMBER=2828

NulliPy · Accepted Answer · 2019-02-21 08:10:18Z

0

Tried all answers above with no luck. Thanks anyways. I guess my explanation was quite not enough

This did a trick for me.

def condition(driver):
look_for = ("Any word within expected URL1", "Any word within expected URL2")
url = driver.current_url
for s in look_for:
    if url.find(s) != -1:
        [Any Script I want to activate]

return False

wait.until(condition)

answered Feb 21, 2019 at 8:10

NulliPy

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1 Comment

undetected Selenium Over a year ago

How does this solution answers the question where you mentioned ...script to be triggered by entering specific URLs with specific part of address... and you were looking for ...I am wondering if regex will do the trick, but I am new to python... so what do i know...?

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