当程序在Selenium Python中遇到错误时，有没有办法让它保持运行

Question

我正在使用Selenium从网站中提取数据。所有节目的HTML通常是相同的，但某些节目的HTML是不同的。大多数应该有27种元素，但有些元素的含量较低，所以会出现错误

所以，我在问，是否有一种方法可以在错误停止程序之前中断代码部分

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

Anime = input("Enter Anime:")
driver = webdriver.Chrome(executable_path=r"C:\Users\amete\Documents\chromedriver.exe")

driver.get("https://myanimelist.net/search/all?q=one%20piece&cat=all")
print(driver.title)

search = driver.find_element_by_xpath('//input[@name="q"]')
wait = WebDriverWait(driver, 20)
wait.until(EC.element_to_be_clickable((By.XPATH, '//input[@name="q"]')))
# Clears the field
search.send_keys(Keys.CONTROL, 'a')
search.send_keys(Keys.DELETE)

# The field is now cleared and the program can type whatever it wants
search.send_keys(Anime)
search.send_keys(Keys.RETURN)

#Accept the cookies
accept = driver.find_element_by_xpath('//*[@id="qc-cmp2-ui"]/div[2]/div/button[3]').click()

#  Added this wait
wait.until(EC.element_to_be_clickable((By.XPATH, '//h2[@id="anime"]//ancestor::div[@class="content-left"]//article[1]/div[contains(@class, "list")][1]/div[contains(@class, "information")]/a[1]')))
link = driver.find_element_by_xpath('//h2[@id="anime"]//ancestor::div[@class="content-left"]//article[1]/div[contains(@class, "list")][1]/div[contains(@class, "information")]/a[1]').click()

#Extracting information

#English Title + Jap Title + Display Score
English_Title = driver.find_element_by_xpath('//*[@id="contentWrapper"]/div[1]').text
#Jap_Title = driver.find_element_by_xpath('//*[@id="contentWrapper"]/div[1]/div/div[1]/div/h1/strong').text
Score = driver.find_element_by_xpath('//*[@id="content"]/table/tbody/tr/td[2]/div[1]/table/tbody/tr[1]/td/div[1]/div[1]/div[1]/div[1]/div[1]/div').text
Episodes = driver.find_element_by_xpath('//*[@id="content"]/table/tbody/tr/td[1]/div/div[10]').text
print ("The English title :"+English_Title)
#print ("The Japanese title:"+Jap_Title)
print ("The Score of the Anime is:"+str(Score))
print (Episodes)

for i in range (7,28):
    Info = driver.find_element_by_xpath('//*[@id="content"]/table/tbody/tr/td[1]/div/div['+str(i)+']').text
    print (Info)

这是我的密码 我正在使用的网站是MyAnimeList

Answer 1

是，请尝试使用，但以下情况除外：

假设您认为这几行代码会有一些问题（风险代码），请尝试将其放入try子句中，如下所示：

try:
 for i in range (7,28):
    Info = driver.find_element_by_xpath('//*[@id="content"]/table/tbody/tr/td[1]/div/div['+str(i)+']').text
    print (Info)
except:
    pass

即使它在for循环中失败，它也会转到，只是说pass