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修复某些情况下剧集命名预览界面异常提示获取文件夹列表失败的问题

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Merge pull request #12 from x1ao4/dev
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x1ao4 2025-04-28 17:51:35 +08:00 committed by GitHub
commit 5fdd5a14eb
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1 changed files with 6 additions and 2 deletions
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app/run.py
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@ -301,6 +301,9 @@ def get_share_detail():
shareurl = request.json.get("shareurl", "") shareurl = request.json.get("shareurl", "")
stoken = request.json.get("stoken", "") stoken = request.json.get("stoken", "")
account = Quark("", 0) account = Quark("", 0)
# 设置account的必要属性
account.episode_patterns = request.json.get("regex", {}).get("episode_patterns", [])
pwd_id, passcode, pdir_fid, paths = account.extract_url(shareurl) pwd_id, passcode, pdir_fid, paths = account.extract_url(shareurl)
if not stoken: if not stoken:
is_sharing, stoken = account.get_stoken(pwd_id, passcode) is_sharing, stoken = account.get_stoken(pwd_id, passcode)
@ -411,14 +414,15 @@ def get_share_detail():
# 实现高级排序算法 # 实现高级排序算法
def extract_sorting_value(file): def extract_sorting_value(file):
if file["dir"]: # 跳过文件夹 if file["dir"]: # 跳过文件夹
return float('inf') return (float('inf'), 0, 0, 0) # 返回元组以确保类型一致性
filename = file["file_name"] filename = file["file_name"]
# 尝试获取剧集序号 # 尝试获取剧集序号
episode_num = extract_episode_number_local(filename) episode_num = extract_episode_number_local(filename)
if episode_num is not None: if episode_num is not None:
return episode_num # 返回元组以确保类型一致性
return (0, episode_num, 0, 0)
# 如果无法提取剧集号，则使用通用的排序函数 # 如果无法提取剧集号，则使用通用的排序函数
return sort_file_by_name(file) return sort_file_by_name(file)