如何使用 Selenium Python 通过已知 src 属性查找元素

发布于 2025-01-09 10:59:28 字数 425 浏览 0 评论 0原文

我试图通过使用 Selenium 和 python，通过 src 属性的已知值来查找元素。该元素是 Google 地图中某个位置的地址左侧的点图片。

即我尝试选择的元素的 html：

<img alt="" jstcache="935" src="//www.gstatic.com/images/icons/material/system_gm/1x/place_gm_blue_24dp.png" class="Liguzb" jsan="7.Liguzb,0.alt,8.src">

如何通过使用链接搜索来选择给定元素：

www.gstatic.com/images/icons/material/system_gm/1x/place_gm_blue_24dp.png

谢谢。

原文

I am trying to find an element by using Selenium with python, by a known value of src attribute. The element is the point picture on the left of the address of a location in Google Maps.

Namely the html for the element I'm trying to select:

<img alt="" jstcache="935" src="//www.gstatic.com/images/icons/material/system_gm/1x/place_gm_blue_24dp.png" class="Liguzb" jsan="7.Liguzb,0.alt,8.src">

How can I select the given element by searching for it by using the link:

www.gstatic.com/images/icons/material/system_gm/1x/place_gm_blue_24dp.png

Thanks.

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守望孤独 2025-01-16 10:59:28

要定位元素，因为您知道 src 属性的值，您可以使用以下任一定位器策略：

使用css_selector：

element = driver.find_element(By.CSS_SELECTOR, "img.Liguzb[src*='gstatic.com/images/icons/material/system_gm/1x/place_gm_blue_24dp']")

使用xpath：

element = driver.find_element(By.XPATH, "//img[@class='Liguzb' and contains(@src, 'gstatic.com/images/icons/material/system_gm/1x/place_gm_blue_24dp' )]")

定位可见元素而不是presence_of_element_ located()你需要诱导WebDriverWait 的 visibility_of_element_ located( ) 并且您可以使用以下任一定位器策略：

使用CSS_SELECTOR：

element = WebDriverWait(driver, 20).until(EC.visibility_of_element_ located((By.CSS_SELECTOR, "img.Liguzb[src*='gstatic.com/images/icons/material/system_gm/1x/place_gm_blue_24dp) ']")))

使用XPATH：

element = WebDriverWait(driver, 20).until(EC.visibility_of_element_ located((By.XPATH, "//img[@class='Liguzb' and contains(@src, 'gstatic.com/images/图标/材质/system_gm/1x/place_gm_blue_24dp')]")))

注意：您必须添加以下导入：

从 selenium.webdriver.support.ui 导入 WebDriverWait
从 selenium.webdriver.common.by 导入
从 selenium.webdriver.support 导入预期条件作为 EC

To locate the element as the value of src attribute is know to you, you can use either of the following Locator Strategies:

Using css_selector:

element = driver.find_element(By.CSS_SELECTOR, "img.Liguzb[src*='gstatic.com/images/icons/material/system_gm/1x/place_gm_blue_24dp']")

Using xpath:

element = driver.find_element(By.XPATH, "//img[@class='Liguzb' and contains(@src, 'gstatic.com/images/icons/material/system_gm/1x/place_gm_blue_24dp')]")

To locate a visible element instead of presence_of_element_located() you need to induce WebDriverWait for the visibility_of_element_located() and you can use either of the following locator strategies:

Using CSS_SELECTOR:

element = WebDriverWait(driver, 20).until(EC.visibility_of_element_located((By.CSS_SELECTOR, "img.Liguzb[src*='gstatic.com/images/icons/material/system_gm/1x/place_gm_blue_24dp']")))

Using XPATH:

element = WebDriverWait(driver, 20).until(EC.visibility_of_element_located((By.XPATH, "//img[@class='Liguzb' and contains(@src, 'gstatic.com/images/icons/material/system_gm/1x/place_gm_blue_24dp')]")))

Note : You have to add the following imports :

from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.common.by import By
from selenium.webdriver.support import expected_conditions as EC

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