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社区首页 >问答首页 >Python Selenium:如何在更改表页时点击表内容

Python Selenium:如何在更改表页时点击表内容
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Stack Overflow用户
提问于 2018-06-25 03:31:15
回答 2查看 1.2K关注 0票数 0
代码语言:javascript
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from selenium import webdriver
from selenium.webdriver.common.keys import Keys
from selenium.webdriver.support.wait import WebDriverWait
from bs4 import BeautifulSoup
import time

url = "https://www.bungol.ca/"
driver = webdriver.Firefox(executable_path ='/usr/local/bin/geckodriver')
driver.get(url)

#Select toronto by default
driver.find_element_by_xpath("""/html/body/section/div[2]/div/div[1]/form/div/select/optgroup[1]/option[1]""").click()
time.sleep(1)
driver.find_element_by_xpath("""/html/body/section/div[2]/div/div[1]/form/div/button""").click()
driver.find_element_by_xpath("""/html/body/nav/div[1]/ul[1]/li[3]/select/option[8]""").click()
#select last 2 years
driver.find_element_by_xpath("""//*[@id="activeListings"]""").click()


#opening sold listing in that area
driver.find_element_by_xpath("""/html/body/div[5]/i""").click() #closes property type slide
driver.find_element_by_xpath("""//*[@id="navbarDropdown"]""").click()
driver.find_element_by_xpath("""//*[@id="listViewToggle"]""").click()


def data_collector():
    hidden_next = driver.find_element_by_class_name("nextPaginate")
    #inputs in textbox
    inputElement = driver.find_element_by_id('navbarSearchAddressInput')
    inputElement.send_keys('M3B2B6')
    time.sleep(1)
    #inputElement.send_keys(Keys.ENTER)
    row_count = 3
    table = driver.find_elements_by_css_selector("""#listViewTableBody""")

    while hidden_next.is_displayed(): #while there is a next page button to be pressed

        time.sleep(3) #delay for table refresh

        #row_count = len(driver.find_elements_by_css_selector("""html body#body div#listView.table-responsive table#listViewTable.table.table-hover.mb-0 tbody#listViewTableBody tr.mb-2"""))           
        for row in range(row_count): #loop through the rows found

            #alternate row by changing the tr index
            driver.find_element_by_xpath("""/html/body/div[8]/table/tbody/tr[""" + str(row + 1) + """]/td[1]""").click()
            time.sleep(2)
            print(driver.find_element_by_css_selector("""#listingStatus""").text) #sold price

            #closes the pop up after getting the data
            driver.find_element_by_css_selector('.modal-xl > div:nth-child(1) > div:nth-child(1) > button:nth-child(1)').click()
            time.sleep(1)

        #clicks next page button for the table    
        driver.find_element_by_xpath("""//*[@id="listViewNextPaginate"]""").click()    


if __name__ == "__main__":
    data_collector()

代码遍历第一个表中的所有行(当前设置为3用于测试),单击每一行-弹出窗口显示,获取信息并关闭弹出窗口。但是当它点击到下一页时,它不会点击第二页的任何行。它也不会显示找不到行xpath的错误。而是显示弹出窗口关闭按钮的错误,因为由于没有按下行以显示弹出窗口,所以没有打开弹出窗口。

当表格翻到下一页时,我如何让它单击行?

对于表格引用:

https://www.bungol.ca/map/location/toronto/

关闭左侧的属性滑块

单击tool ->打开列表

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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/51013308

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