Python运算符筛选器不工作（Selenium）

Question

嘿，伙计们，我正在尝试用selenium&；过滤一些instagram用户；python的目标是跟踪所有用户，但我不知道为什么他会返回相同的结果：

例如，此帐户：

通常情况下，他必须告诉我“有更多的追随者而不是追随者”

我的代码：

followers = []    #list of url users
for follower in followers:
    #Iterate into the list
    scrap_follower = browser.find_element_by_xpath('/html/body/div[1]/section/main/div/header/section/ul/li[2]/a/span').get_attribute("title")
    
    scrap_following = browser.find_element_by_xpath('/html/body/div[1]/section/main/div/header/section/ul/li[3]/a/span').text
    
    
    follower_count = int(scrap_follower)
    following_count = int(scrap_following)
    
    browser.get(follower)
    sleep(2)
    
    followButton = browser.find_element_by_css_selector('button')
    
    print(follower_count)
    print(following_count)     
    
    if follower_count == 0:
        pass
        print("0 followers")
        
        if follower_count > following_count :
            print("have more followers than following")
            pass
    else:
        print("eligible")
            #Go to follow

我的控制台日志返回：

谢谢你的帮助：）

Answer 1

第二个if语句需要与第一个语句对齐，否则它将永远不会执行，除非满足if follower_count == 0:条件

您的代码：

follower_count = 1500
following_count = 1000

print(follower_count)
print(following_count)     

if follower_count == 0:
    pass
    print("0 followers")
    
    if follower_count > following_count :
        print("have more followers than following")
        pass
else:
    print("eligible")

结果（不正确）：

1500
1000
eligible

正确的代码（第二个if语句与第一个if语句对齐）

follower_count = 1500
following_count = 1000

print(follower_count)
print(following_count)     

if follower_count == 0:
    pass
    print("0 followers")
    
if follower_count > following_count :
    print("have more followers than following")
    pass
else:
    print("eligible")

结果:

1500
1000
have more followers than following