在 selenium 中迭代 webelements 总是只返回表格的第一行

Question

这是我的代码，首先我在 all_trows 变量中获取表的所有行，在迭代它时，我期望不同的输出，因为每一行都有不同的文本，但我总是得到第一个 tr 的值表。

url = 'https://www.un.org/Depts/ptd/eoi?field_commodity_group_eoi_rfi_tid=All&items_per_page=5'

driver.get(url)
driver.implicitly_wait(5)
WebDriverWait(driver,5).until(EC.visibility_of_element_located((By.XPATH,"//*[@id='block-system-main']/div/div[contains(@class,'view-content')] /table/tbody")))
WebDriverWait(driver,5).until(EC.visibility_of_element_located((By.XPATH,"//*[@id='block-system-main']/div/div[contains(@class,'view-footer')]/div[contains(@class,'page_result')]")))

page = 0
while True:
    time.sleep(5)
    driver.implicitly_wait(10)
    current_page_flag = driver.find_element(By.XPATH , "//*[@id='block-system-main']/div/div[@class ='view-footer'] / div[@class ='page_result']").text.strip()
    WebDriverWait(driver,2).until(EC.visibility_of_element_located((By.XPATH, "//*[@id='block-system-main']/div//tbody")))
    all_trows = driver.find_elements_by_xpath("//*[@id='block-system-main']/div/div[@class = 'view-content'] / table / tbody / tr")

    for eachRow in all_trows:
        op_link = eachRow.find_element_by_xpath('//td[position() = 1]/a').get_attribute('href').strip()
        print(op_link)
        
    try:
        WebDriverWait(driver,2).until(EC.visibility_of_element_located((By.XPATH, "//*[@id='block-system-main']/div/div[@class ='item-list']/ul/li")))
        driver.find_element_by_xpath("//*[@id='block-system-main']/div/div[@class = 'item-list']/ul/li[contains(@class, 'pager-next')]/a").click()
        driver.implicitly_wait(5)
        WebDriverWait(driver, 10).until(lambda driver: driver.find_element_by_xpath("//*[@id='block-system-main']/div/div[@class ='view-footer'] / div[@class ='page_result']").text.strip()!= current_page_flag)
        page = page + 1
        print(page)
    except NoSuchElementException:
        print('no next button')
        # driver.quit()
        break

这是我的输出

https://www.un.org/Depts/ptd/business-opportunities/EOI/eoiunmiss17882
https://www.un.org/Depts/ptd/business-opportunities/EOI/eoiunmiss17882
https://www.un.org/Depts/ptd/business-opportunities/EOI/eoiunmiss17882
https://www.un.org/Depts/ptd/business-opportunities/EOI/eoiunmiss17882
https://www.un.org/Depts/ptd/business-opportunities/EOI/eoiunmiss17882
1
https://www.un.org/Depts/ptd/business-opportunities/EOI/eoiunficyp17877
https://www.un.org/Depts/ptd/business-opportunities/EOI/eoiunficyp17877
https://www.un.org/Depts/ptd/business-opportunities/EOI/eoiunficyp17877
https://www.un.org/Depts/ptd/business-opportunities/EOI/eoiunficyp17877
https://www.un.org/Depts/ptd/business-opportunities/EOI/eoiunficyp17877
2
https://www.un.org/Depts/ptd/business-opportunities/EOI/eoiescwa17908
https://www.un.org/Depts/ptd/business-opportunities/EOI/eoiescwa17908
https://www.un.org/Depts/ptd/business-opportunities/EOI/eoiescwa17908
https://www.un.org/Depts/ptd/business-opportunities/EOI/eoiescwa17908
https://www.un.org/Depts/ptd/business-opportunities/EOI/eoiescwa17908

我希望在迭代不同的行时我必须得到不同的 href 值。

Answer 1

你需要改变这一行

op_link = eachRow.find_element_by_xpath('//td[position() = 1]/a').get_attribute('href').strip()

当你迭代行时，它总是选择 DOM 树中的第一项。

要从 DOM 树中为每一行选择直接子级，您需要使用 . 进行更改

所以你的代码应该是

for eachRow in all_trows:
        op_link = eachRow.find_element_by_xpath('.//td[position() = 1]/a').get_attribute('href').strip()
        print(op_link)