如何使用 selenium 和 python 迭代和访问列表中的每个链接？如何点击 Yelp 页面上的每个链接

Question

我很想访问这个 yelp 页面上每个链接文本中包含饺子的链接。

Picture of List of Links on yelp

这里是代码。我到了第二页，但不确定如何点击每个链接。我希望每个链接都以新的形式打开。但只有饺子。谢谢你。希望这是有道理的。

all_links = []

#yelp page
driver.get("https://yelp.com");

#Selecting the first box
 type_of_food = driver.find_element_by_id('find_desc')

#Enter Chinese Food
type_of_food.send_keys('Chinese Food');

#press enter
type_of_food.send_keys(Keys.ENTER);


driver.implicitly_wait(30)
#Button click

driver.find_element_by_id("header-search-submit").click()


driver.implicitly_wait(9)



driver.find_element_by_partial_link_text('Dumpling')

for link in all_links:
    links.append(link.get_attribute('href'))

Answer 1

我不确定你的代码是否正确。

A.您已在搜索区域输入Chinese Food，然后键入ENTER，然后再次单击Search 按钮。但只有他们中的一个会这样做。两者似乎都不是必需的。

B. for link in all_links: 我不明白您为什么要遍历已声明的空白列表（顶部声明的列表：all_links = []）。它里面没有任何东西。假设您正在尝试链接饺子元素，我已将您的代码重构为以下内容：

all_links = []
#yelp page
driver.get("https://yelp.com")
#Selecting the first box
type_of_food = driver.find_element(By.ID, 'find_desc')
#Enter Chinese Food
type_of_food.send_keys('Chinese Food')
#press enter
# type_of_food.send_keys(Keys.ENTER)
driver.implicitly_wait(30)
#Button click
driver.find_element(By.ID, "header-search-submit").click()
driver.implicitly_wait(9)
dumplings = driver.find_elements(By.PARTIAL_LINK_TEXT, 'Dumpling')
for item in dumplings:
    all_links.append(item.get_attribute('href'))
    # Replace CONTROL with COMMAND if you are on Mac
    ActionChains(driver).key_down(Keys.CONTROL).click(item).key_up(Keys.CONTROL).perform()
    

print(all_links)

输出：

['https://www.yelp.com/biz/dumpling-kitchen-san-francisco?osq=Chinese+Food', 'https://www.yelp.com/biz/dumpling-house-san-francisco?osq=Chinese+Food', 'https://www.yelp.com/biz/dumpling-baby-china-bistro-san-francisco-4?osq=Chinese+Food', 'https://www.yelp.com/search?find_desc=Dumplings&find_loc=San+Francisco%2C+CA']

Process finished with exit code 0