我用Python编写一些代码与硒组合。我打算从网页解析表格。我有它的工作。但是,当我尝试点击下一页按钮时出现问题。刮板只从第一页,而不是点击下一步按钮它退出而不引发任何错误解析表。所以,我不明白我错过了什么。麻烦点击按钮,下一个
这是给你考虑全码:
from selenium import webdriver
from selenium.webdriver.common.by import By
from selenium.webdriver.support.wait import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
driver = webdriver.Chrome()
wait = WebDriverWait(driver, 10)
driver.get("https://toolkit.financialexpress.net/santanderam")
wait.until(EC.visibility_of_element_located((By.CSS_SELECTOR, 'table.fe-datatable')))
tab_data = driver.find_element_by_css_selector('table.fe-datatable')
while True:
wait.until(EC.visibility_of_element_located((By.CSS_SELECTOR, 'tr')))
list_rows = [[cell.text for cell in row.find_elements_by_css_selector('td')]
for row in tab_data.find_elements_by_css_selector('tr')]
for data in list_rows:
print(data)
try:
driver.find_element_by_css_selector('a.ui-paging-next').click()
except:
break
driver.quit()
元素中的下一个页面按钮存在:
<div class="pagination ui-widget"><span class="ui-paging-current ui-state-default ui-state-disabled ui-corner-all ui-paging-prev">Prev</span><span class="ui-paging-current ui-state-default ui-state-disabled ui-state-highlight ui-corner-all">1</span><a class="ui-paging-button ui-state-default ui-corner-all" href="#">2</a><a class="ui-paging-button ui-state-default ui-corner-all" href="#">3</a><a class="ui-paging-button ui-state-default ui-corner-all" href="#">4</a><span class="ui-state-default ui-corner-all ui-state-disabled ui-paging-ellipse">...</span><a class="ui-paging-button ui-state-default ui-corner-all ep" href="#">7</a><a class="ui-paging-button ui-state-default ui-corner-all ui-paging-next" href="#">Next</a></div>
尝试driver.find_element_by_css_selector(“一[类* = 'UI-寻呼下一']“)。单击()或find_element_by_link_text( '下一步')。单击() – Grasshopper
的Gr感谢蚂蚱,为你答案。它没有使用CSS选择器的技巧,但它似乎与链接文本。测试后会回复你。谢谢。 – SIM
它确实点击链接,但会引发另一个错误。 “raise exception_class(message,screen,stacktrace) selenium.common.exceptions.StaleElementReferenceException:消息:陈旧的元素引用:元素没有附加到页面文档中” – SIM