Selenium(Python) – SELECT
现在我的脚本转到页面并从下拉列表中打开第二个对象“Vijesti”,然后才收到错误消息. 这是错误:
来自Selenium网站:
我想要选择每个对象,然后打开它. 这是来自url的SELECT部分: <select id="kategorija" name="kategorija"> <option value="0">Kategorija</option> <option value="12">Vijesti</option> <option value="8">Biznis</option> <option value="5">Sport</option> <option value="2">Magazin</option> <option value="7">Lifestyle</option> <option value="3">Scitech</option> <option value="6">Auto</option> </select> 码: from selenium import webdriver from selenium.webdriver.common.keys import Keys from selenium.webdriver.support.ui import Select import time driver = webdriver.Firefox() driver.get("http://www.klix.ba/") assert "Klix" in driver.title elem = driver.find_element_by_name("q") elem.send_keys("test") elem.send_keys(Keys.RETURN) select = Select(driver.find_element_by_name('kategorija')) all_options = [o.get_attribute('value') for o in select.options] for x in all_options: select.select_by_value(x) time.sleep(3) 这是我做测试的url. 解决方法从下拉列表中选择项目时,页面会自动刷新.您需要“重新”选择每个选项上的select元素: select = Select(driver.find_element_by_name('kategorija')) for index in range(len(select.options)): select = Select(driver.find_element_by_name('kategorija')) select.select_by_index(index) # grab the results (编辑:大庆站长网) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |