python's selenium “send_keys” with chrome driv

问题:

I am using selenium package with Python (https://pypi.python.org/pypi/selenium) with Windows 7. When I try to login to my facebook account I use the send_keys command, e.g.

elem = browser.find_element_by_name("email")
elem.send_keys(email);
elem = browser.find_element_by_name("pass")
elem.send_keys(password);

Login fails apparently because the second send_keys drops the first character of the password (I found this by directly sending the password characters to the email field.

What's going on? Why can't selenium do something so simple as sending keys to an input field? Is this some kind of a protection measure coded by Facebook to reject automatic scripting?

Tried to send the whole alphabet and got this:

abcdefghijklmnopqrstuvwxyzBCDFGIKLNOQSTWX

Notice how many characters are missing...

Update

Apparently the problem has nothing to do with facebook but to the chrome driver.

If I send the following simple code

browser = webdriver.Chrome()
browser.get("https://www.google.com") # Load page
elem = browser.find_element_by_name("q") # Find the query box
query= "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
elem.send_keys(query)

With chrome driver I get BCDFGIKLNOQSTWX Notice how A, E, H ... Y, Z are missing With firefox driver (replacing browser = webdriver.Chrome() with browser = webdriver.Firefox() I get: ABCDEFGHIJKLMNOPQRSTUVWXYZ

回答1:

Looks like there are (were) some bugs in the Chrome webdriver: https://code.google.com/p/chromedriver/issues/detail?id=435

The core of the problem looks to be when either the keyboard is configured for a non-English language, or if the webdriver process and the chrome display are running in different language/environments (e.g., when going through a remote display from one host to another, etc.)

回答2:

I've solved using a custom method for send_keys, which works a little bit lower but fast enough.

from selenium.webdriver.remote.webelement import WebElement

def send_keys(el: WebElement, keys: str):
    for i in range(len(keys)):
        el.send_keys(keys[i])

send_keys(el, keys)

回答3:

Use selenium Ide and export test case in python

from selenium import webdriver
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import Select
from selenium.common.exceptions import NoSuchElementException
import unittest, time, re

class Test1(unittest.TestCase):
    def setUp(self):
        self.driver = webdriver.Firefox()
        self.driver.implicitly_wait(30)
        self.base_url = "https://www.facebook.com/"
        self.verificationErrors = []
        self.accept_next_alert = True

    def test_1(self):
        driver = self.driver
        driver.get(self.base_url + "/")
        driver.find_element_by_id("email").clear()
        driver.find_element_by_id("email").send_keys("username")
        driver.find_element_by_id("pass").clear()
        driver.find_element_by_id("pass").send_keys("password")
        driver.find_element_by_id("u_0_b").click()
        driver.find_element_by_xpath("//div[@id='u_ps_0_1_5']/div/div").click()
        driver.find_element_by_link_text("1 Requests").click()
        driver.find_element_by_id("globalContainer").click()

    def is_element_present(self, how, what):
        try: self.driver.find_element(by=how, value=what)
        except NoSuchElementException, e: return False
        return True

    def is_alert_present(self):
        try: self.driver.switch_to_alert()
        except NoAlertPresentException, e: return False
        return True

    def close_alert_and_get_its_text(self):
        try:
            alert = self.driver.switch_to_alert()
            alert_text = alert.text
            if self.accept_next_alert:
                alert.accept()
            else:
                alert.dismiss()
            return alert_text
        finally: self.accept_next_alert = True

    def tearDown(self):
        self.driver.quit()
        self.assertEqual([], self.verificationErrors)

if __name__ == "__main__":
    unittest.main()