Python刷问卷只刷到1/4，不知道怎么修改
```python
import logging
import random
import re
import traceback
from threading import Thread
import time
import numpy
import requests
from selenium import webdriver
from selenium.webdriver.common.action_chains import ActionChains
from selenium.webdriver.common.by import By
def zanip():
# 这里放你的ip链接，选择你想要的地区，1分钟，ip池无所谓，数据格式txt，提取数量1，其余默认即可
api = "https://service.ipzan.com/core-extract?num=1&no=???&minute=1&area=all&pool=quality&secret=???"
ip = requests.get(api).text
return ip
# 问卷
url = 'https://www.wjx.cn/vm/Ok6idb2.aspx#'
"""
题号其实不重要，只是为了填写概率值时方便记录我才加上去的，这个字典在真正使用前会转化为一个列表；（这一行没看懂没关系，下面一行懂了就行）
最重要的其实是保证single_prob的第n个参数对应第n个单选题，比如在示例问卷中第5题是滑块题，但是我single_prob却有“第5题”，因为这个"5"其实对应的是第5个单选题，也就是问卷中的第6题
这个single_prob的"5"可以改成其他任何值，当然我不建议你这么干，因为问卷中只有5个单选题，所以第6个单选题的参数其实是没有用上的，参数只能多不能少！！！（这一点其他类型的概率参数也适用）
"""
single_prob = {"1": [25,25,30,20], "2": [5,15,30,50], "3":[60,29,1,2,10], "4": [11,56,24,9], "5": [30,10,13,9,8,9,15,0,1,1], "6": [34,40,20,6], "7": [24,40,20,16],
"8": [25,15,45,15], "9": [17,25,45,13], "10": [4,20,20,56], "11": [27,43,24,6], "12": [1,54,26,19], "13": [34,40,20,6], "14": [23,50,20,7],"15": [15,30,40,10,5], }
# 下拉框参数，具体含义参考单选题，如果没有下拉框题也不要删，就让他躺在这儿吧，其他题也是哦，没有就不动他，别删，只改你有的题型的参数就好啦
droplist_prob = {"1": [1, 1, 1]}
# 多选题概率参数,0不选该选项，100必选，[10, 50]表示1:5,-1表示随机，
multiple_prob = {"1": [70,5,70,1,20], "2": [40,20,20,18,2], "3": [35,35,19,10,1], "4": [70,5,70,50], "5": [70,5,70,90], "6": [45,70,35,40,5],"7": [70,5,70,90,10],}
# 多选题选择的选项数量（去除必选后的数），这里填1与上面的multiple_prob表示在必选A后，会再从BCD中选1个选项
# 注意！！！如果选项数量比较少，建议多选的数量参数不要太大，因为数量参数值越大，最后刷出来的数据分布误差越大！！！4个选项建议选1-2个即可。
multiple_opts = {"9": 1, }
# 矩阵题概率参数,-1表示随机，其他含义参考单选题；同样的，题号不重要，保证第几个参数对应第几个矩阵小题就可以了；
# 在示例问卷中矩阵题是第10题，每个小题都要设置概率值才行！！以下参数表示第二题随机，其余题全选A
matrix_prob = {"1": [1,19,20,30,20], "2": [1,19,20,30,20], "3": [1,19,20,20,30], "4": [1,19,20,20,30],
"5": [1,19,20,20,30]}
# 量表题概率参数，参考单选题
scale_prob = {"7": [0, 1, 3, 4, 1], "12": [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]}
# 填空题参数，在题号后面按该格式填写需要填写的内容，
texts = {"8": ["内容1", "内容2", " 内容3"], }
# 每个内容对应的概率1:1:1,
texts_prob = {"8": [1, 1, 1]}
# --------------到此为止，参数设置完毕，可以直接运行啦！-------------------
# 如果需要设置浏览器窗口数量，请转到最后一个函数(main函数)，注意看里面的注释喔！
# 参数归一化，把概率值按比例缩放到概率值和为1，比如某个单选题[1,2,3,4]会被转化成[0.1,0.2,0.3,0.4],[1,1]会转化成[0.5,0.5]
for prob in [single_prob, matrix_prob, droplist_prob, scale_prob, texts_prob]:
for key in prob:
if isinstance(prob[key], list) and prob[key] != -1:
prob_sum = sum(prob[key])
prob[key] = [x / prob_sum for x in prob[key]]
# 转化为列表,去除题号
single_prob = list(single_prob.values())
droplist_prob = list(droplist_prob.values())
multiple_prob = list(multiple_prob.values())
multiple_opts = list(multiple_opts.values())
matrix_prob = list(matrix_prob.values())
scale_prob = list(scale_prob.values())
texts_prob = list(texts_prob.values())
texts = list(texts.values())
print("单选题参数: ", single_prob)
print("下拉框参数: ", droplist_prob)
print("多选题参数: ", multiple_prob)
print("矩阵题参数: ", matrix_prob)
print("量表题参数: ", scale_prob)
# 校验IP地址合法性
def validate(ip):
pattern = r'^((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?):(\d{1,5})$'
if re.match(pattern, ip):
return True
return False
# 检测题量和页数的函数，返回一个列表，第一个数表示第一页的题量，第二个数表示第二页的题量；比如示例问卷会返回：[3, 2, 2, 7]
# 虽然但是，我见识过问卷星再没有跳题逻辑的情况下有题被隐藏，我当时就??????
# 这会导致detect返回包含被隐藏的题，数值可能偏高，比如可见题目[3, 2, 2, 7]被detect成[4, 2, 2, 7]。。
def detect(driver):
q_list = [] # 长度等于页数，数字代表该页的题数
xpath = '//*[@id="divQuestion"]/fieldset'
page_num = len(driver.find_elements(By.XPATH, xpath)) # 页数
qs = driver.find_elements(By.XPATH, f'//*[@id="fieldset1"]/div') # 每一页的题
invalid_item = 0 # 无效问题数量
for qs_item in qs:
# 判断其topic属性值是否值包含数字
if qs_item.get_attribute("topic").isdigit() is False:
invalid_item += 1
# 如果只有1页
q_list.append(len(qs) - invalid_item)
if page_num >= 2:
for i in range(2, page_num + 1):
qs = driver.find_elements(By.XPATH, f'//*[@id="fieldset{i}"]/div')
invalid_item = 0 # 每一页的无效问题初始值为0
# 遍历每一个div，判断其是否可以回答
for qs_item in qs:
# 判断其topic属性值是否值包含数字，因为只有题的div的topic属性才是纯数字
if qs_item.get_attribute("topic").isdigit() is False:
invalid_item += 1
# [3, 2, 2, 7]
q_list.append(len(qs) - invalid_item)
return q_list
# 填空题处理函数
def vacant(driver, current, index):
content = texts[index]
# 对应填空题概率参数
p = texts_prob[index]
text_index = numpy.random.choice(a=numpy.arange(0, len(p)), p=p)
driver.find_element(By.CSS_SELECTOR, f'#q{current}').send_keys(content[text_index])
# 单选题处理函数
def single(driver, current, index):
xpath = f'//*[@id="div{current}"]/div[2]/div'
a = driver.find_elements(By.XPATH, xpath)
p = single_prob[index]
if p == -1:
r = random.randint(1, len(a))
else:
r = numpy.random.choice(a=numpy.arange(1, len(a) + 1), p=p)
driver.find_element(By.CSS_SELECTOR,
f'#div{current} > div.ui-controlgroup > div:nth-child({r})').click()
# 下拉框处理函数
def droplist(driver, current, index):
# 先点击“请选择”
driver.find_element(By.CSS_SELECTOR, f"#select2-q{current}-container").click()
time.sleep(0.5)
# 选项数量
options = driver.find_elements(By.XPATH, f"//*[@id='select2-q{current}-results']/li")
p = droplist_prob[index] # 对应概率
r = numpy.random.choice(a=numpy.arange(1, len(options)), p=p)
driver.find_element(By.XPATH, f"//*[@id='select2-q{current}-results']/li[{r + 1}]").click()
# 多选题处理函数：这个超复杂，要不是chatgpt我一辈子都写不出这代码
def multiple(driver, current, index):
xpath = f'//*[@id="div{current}"]/div[2]/div'
options = driver.find_elements(By.XPATH, xpath)
# 第current题对应的概率值
probabilities = multiple_prob[index]
if probabilities == 0: # 不选
return
elif probabilities == -1: # 随机
r = random.randint(1, len(options))
driver.find_element(By.CSS_SELECTOR,
f'#div{current} > div.ui-controlgroup > div:nth-child({r})').click()
else:
prob_copy = probabilities.copy()
opts_num = multiple_opts[index] # 第current题对应的选项数量参数
for i in prob_copy: # 如果存在列表中概率为100的项，则直接选择该项
if i == 100:
# 找到100元素位置
sure = prob_copy.index(i)
driver.find_element(By.CSS_SELECTOR,
f'#div{current} > div.ui-controlgroup > div:nth-child({sure + 1})').click()
# 将已选的概率修改为0，以便在后面按概率选择其他选项
prob_copy[sure] = 0
# 计算不为0的数值总和
total = sum([num for num in prob_copy])
if total == 0: return
# 将不为0的数值归一化
probabilities_norm = [num / total if num != 0 else 0 for num in prob_copy]
# 从位置1到列表长度之间随机选择 opts_num - 已选数 相同数量的选项
selection_indices = numpy.random.choice(
range(len(options)),
size=opts_num,
replace=False,
p=probabilities_norm)
# 选择随机选择的选项
for i in selection_indices:
driver.find_element(By.CSS_SELECTOR,
f'#div{current} > div.ui-controlgroup > div:nth-child({i + 1})').click()
# 矩阵题处理函数
def matrix(driver, current, index):
xpath1 = f'//*[@id="divRefTab{current}"]/tbody/tr'
a = driver.find_elements(By.XPATH, xpath1)
q_num = 0 # 矩阵的题数量
for tr in a:
if tr.get_attribute("rowindex") is not None:
q_num += 1
# 选项数量
xpath2 = f'//*[@id="drv{current}_1"]/td'
b = driver.find_elements(By.XPATH, xpath2) # 题的选项数量+1 = 6
# 遍历每一道小题
for i in range(1, q_num + 1):
p = matrix_prob[index]
index += 1
if p == -1:
opt = random.randint(2, len(b))
else:
opt = numpy.random.choice(a=numpy.arange(2, len(b) + 1), p=p)
driver.find_element(By.CSS_SELECTOR, f'#drv{current}_{i} > td:nth-child({opt})').click()
return index
# 排序题处理函数，排序暂时只能随机
def reorder(driver, current):
xpath = f'//*[@id="div{current}"]/ul/li'
a = driver.find_elements(By.XPATH, xpath)
for j in range(1, len(a) + 1):
b = random.randint(j, len(a))
driver.find_element(By.CSS_SELECTOR, f'#div{current} > ul > li:nth-child({b})').click()
time.sleep(0.4)
# 量表题处理函数
def scale(driver, current, index):
xpath = f'//*[@id="div{current}"]/div[2]/div/ul/li'
a = driver.find_elements(By.XPATH, xpath)
p = scale_prob[index]
if p == -1:
b = random.randint(1, len(a))
else:
b = numpy.random.choice(a=numpy.arange(1, len(a) + 1), p=p)
driver.find_element(By.CSS_SELECTOR,
f"#div{current} > div.scale-div > div > ul > li:nth-child({b})").click()
# 刷题逻辑函数
def brush(driver):
q_list = detect(driver) # 检测页数和每一页的题量
single_num = 0 # 第num个单选题
vacant_num = 0 # 第num个填空题
droplist_num = 0 # 第num个下拉框题
multiple_num = 0 # 第num个多选题
matrix_num = 0 # 第num个矩阵小题
scale_num = 0 # 第num个量表题
current = 0 # 题号
for j in q_list: # 遍历每一页
for k in range(1, j + 1): # 遍历该页的每一题
current += 1
# 判断题型
q_type = driver.find_element(By.CSS_SELECTOR, f'#div{current}').get_attribute("type")
if q_type == "1" or q_type == "2": # 填空题
vacant(driver, current, vacant_num)
vacant_num += 1 # 同时将vacant_num+1表示运行vacant函数时该使用texts参数的下一个值
elif q_type == "3": # 单选
single(driver, current, single_num)
single_num += 1 # single_num+1表示运行single函数时该使用single_prob参数的下一个值
elif q_type == "4": # 多选
multiple(driver, current, multiple_num)
multiple_num += 1
elif q_type == "5": # 量表题
scale(driver, current, scale_num)
scale_num += 1
elif q_type == "6": # 矩阵题
matrix_num = matrix(driver, current, matrix_num)
elif q_type == "7": # 下拉框
droplist(driver, current, droplist_num)
droplist_num += 1
elif q_type == "8": # 滑块题
score = random.randint(1, 100)
driver.find_element(By.CSS_SELECTOR, f'#q{current}').send_keys(score)
elif q_type == "11": # 排序题
reorder(driver, current)
else:
print(f"第{k}题为不支持题型！")
time.sleep(0.5)
# 一页结束过后要么点击下一页，要么点击提交
try:
driver.find_element(By.CSS_SELECTOR, '#divNext').click() # 点击下一页
time.sleep(0.5)
except:
# 点击提交
driver.find_element(By.XPATH, '//*[@id="ctlNext"]').click()
submit(driver)
# 提交函数
def submit(driver):
time.sleep(1)
# 点击对话框的确认按钮
try:
driver.find_element(By.XPATH, '//*[@id="layui-layer1"]/div[3]/a').click()
time.sleep(1)
except:
pass
# 点击智能检测按钮，因为可能点击提交过后直接提交成功的情况，所以智能检测也要try
try:
driver.find_element(By.XPATH, '//*[@id="SM_BTN_1"]').click()
time.sleep(3)
except:
pass
# 滑块验证
try:
slider = driver.find_element(By.XPATH, '//*[@id="nc_1__scale_text"]/span')
if str(slider.text).startswith("请按住滑块"):
width = slider.size.get('width')
ActionChains(driver).drag_and_drop_by_offset(slider, width, 0).perform()
except:
pass
def run(xx, yy):
# 躲避智能检测，将webDriver设置为false
option = webdriver.ChromeOptions()
option.add_experimental_option('excludeSwitches', ['enable-automation'])
option.add_experimental_option('useAutomationExtension', False)
global count
global stop
global fail # 失败次数
while not stop:
ip = zanip()
if validate(ip):
option.add_argument(f'--proxy-server={ip}')
driver = webdriver.Chrome(options=option)
driver.set_window_size(550, 650)
driver.set_window_position(x=xx, y=yy)
driver.execute_cdp_cmd('Page.addScriptToEvaluateOnNewDocument',
{'source': 'Object.defineProperty(navigator, "webdriver", {get: () => undefined})'})
try:
driver.get(url)
url1 = driver.current_url # 表示问卷链接
brush(driver)
# 刷完后给一定时间让页面跳转
time.sleep(10)
url2 = driver.current_url # 表示问卷填写完成后跳转的链接，一旦跳转说明填写成功
if url1 != url2:
count += 1
print(f"已填写{count}份 - 失败{fail}次 - {time.strftime('%H:%M:%S', time.localtime(time.time()))} ")
driver.quit()
except:
traceback.print_exc()
fail += 1
logging.warning(f"已失败{fail}次,失败超过10次(左右)将强制停止------------------------------")
if fail >= 10: # 失败阈值
stop = True
logging.critical('失败次数过多，为防止耗尽ip余额，程序将强制停止，请检查代码是否正确')
quit()
driver.quit()
continue
# 多线程执行run函数
if __name__ == "__main__":
count = 0 # 记录已刷份数
fail = 0 # 失败次数
stop = False
if validate(zanip()):
print("IP设置成功, 将使用代理ip填写")
else:
print("IP设置失败, 将使用本机ip填写")
# 需要几个窗口同时刷就设置几个thread_?，默认两个，args里的数字表示设置浏览器窗口打开时的初始xy坐标
thread_1 = Thread(target=run, args=(50, 50))
thread_2 = Thread(target=run, args=(650, 50))
# thread_3 = Thread(target=run, args=(650, 280))
thread_1.start()
thread_2.start()
# thread_3.start()
thread_1.join()
thread_2.join()
# thread_3.join()
报错：Exception managing chrome: error sending request for url (https://googlechromelabs.github.io/chrome-for-testing/known-good-versions-with-downloads.json)
Traceback (most recent call last):
File "C:/Users/1/Desktop/8.py", line 337, in run
brush(driver)
File "C:/Users/1/Desktop/8.py", line 263, in brush
multiple(driver, current, multiple_num)
File "C:/Users/1/Desktop/8.py", line 169, in multiple
opts_num = multiple_opts[index] # 第current题对应的选项数量参数
IndexError: list index out of range
WARNING:root:已失败1次,失败超过10次(左右)将强制停止------------------------------
Traceback (most recent call last):
File "C:/Users/1/Desktop/8.py", line 337, in run
brush(driver)
File "C:/Users/1/Desktop/8.py", line 263, in brush
multiple(driver, current, multiple_num)
File "C:/Users/1/Desktop/8.py", line 169, in multiple
opts_num = multiple_opts[index] # 第current题对应的选项数量参数
IndexError: list index out of range
WARNING:root:已失败2次,失败超过10次(左右)将强制停止--
```
