将获得弹幕所需要的函数更仔细的划分

main
p6fxi93qh 2 months ago
parent 6a161daaec
commit d06df68b5b

@ -12,47 +12,56 @@ dic_cookies = {}
for i in cookies.split("; "):
dic_cookies[i.split("=")[0]] = i.split("=")[1]
#获取整页视频的url
def get_videos_url(keyword,pages=1):
def get_web_url(keyword, pages):#页面url
# 对关键词进行URL编码
keyword = quote(keyword) if keyword else "" # 确保keyword不为None
res = []
pages = int(pages)
# 当请求页数为0时直接返回空列表
if pages <= 0:
return res
else:
# 构建基础URL
url = f"https://search.bilibili.com/video?keyword={keyword}&from_source=webtop_search&spm_id_from=333.1007&search_source=5"
result = []
for i in range(0, pages):
if i > 0 :
# 根据页数构建完整的URL列表
for i in range(pages):
if i > 0:
url1 = f"{url}&page={i + 1}&o={i * 30}"
else:
url1 = url
response = requests.get(url1, headers=headers,cookies=dic_cookies)
response.encoding = 'utf-8'
e = etree.HTML(response.text)
# xpath定位
a = e.xpath('//div[@class="video-list row"]/div[not(contains(@class, "to_hide_md"))]/div[1]/div[2]/a[1]')
for element in a:
href = element.get('href') # 获取href属性
href1 = f"https:{href}"
result.append(href1)
print(href1)
time.sleep(5)
return result
res.append(url1)
return res
def get_videos_url(web_url):#获取当前页面的所有视频的url
videos_url = []
response = requests.get(web_url,headers=headers,cookies=dic_cookies)
e = etree.HTML(response.text)
# xpath定位
a = e.xpath('//div[@class="video-list row"]/div[not(contains(@class, "to_hide_md"))]/div[1]/div[2]/a[1]')
for element in a:
href = element.get('href') # 获取href属性
href1 = f"https:{href}"
videos_url.append(href1)
return videos_url
def get_danmu(video_id):
#打开视频页面获取视频cid
response = requests.get(video_id, headers=headers)
video_cid = re.search(r'"cid":(\d*),', response.text).group(1)
#用现有链接获取弹幕
url_dm = f"https://comment.bilibili.com/{video_cid}.xml"
response_dm = requests.get(url_dm,headers=headers,cookies=dic_cookies)
if response.status_code == 200:
if response_dm.status_code == 200:
response_dm.encoding = 'utf-8'
content = re.findall('<d p=".*?">(.*?)</d>', response_dm.text)
print("请求弹幕成功")
return content
else:
print("请求弹幕失败")
return
return []
def writeintxt (list,file_path):
with open(file_path, 'w', encoding='utf-8') as file:
for item in list:
file.write("%s\n" % item)
file.write(f"{item}\n ")

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