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添加删除订阅命令的取消功能
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parent
d0f56e76cd
commit
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@ -269,13 +269,16 @@ def do_del_sub(del_sub: Type[Matcher]):
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if platform.enable_tag:
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res += " {}".format(", ".join(sub["tags"]))
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res += "\n"
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res += "请输入要删除的订阅的序号"
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res += "请输入要删除的订阅的序号\n输入'取消'中止"
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await bot.send(event=event, message=Message(await parse_text(res)))
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@del_sub.receive()
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async def do_del(event: Event, state: T_State):
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user_msg = str(event.get_message()).strip()
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if user_msg == "取消":
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await del_sub.finish("删除中止")
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try:
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index = int(str(event.get_message()).strip())
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index = int(user_msg)
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config = Config()
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user_info = state["target_user_info"]
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assert isinstance(user_info, User)
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@ -387,8 +387,8 @@ async def test_add_with_get_id(app: App):
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True,
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)
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"""
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line 362:
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鬼知道为什么要在这里这样写,
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关于Message([MessageSegment(*BotReply.add_reply_on_id_input_search())]):
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异客知道为什么要在这里这样写,
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没有[]的话assert不了(should_call_send使用[MessageSegment(...)]的格式进行比较)
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不在这里MessageSegment()的话也assert不了(指不能让add_reply_on_id_input_search直接返回一个MessageSegment对象)
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amen
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@ -67,7 +67,7 @@ async def test_del_sub(app: App):
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ctx.should_call_send(
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event,
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Message(
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"订阅的帐号为:\n1 weibo 明日方舟Arknights 6279793937\n [图文] 明日方舟\n请输入要删除的订阅的序号"
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"订阅的帐号为:\n1 weibo 明日方舟Arknights 6279793937\n [图文] 明日方舟\n请输入要删除的订阅的序号\n输入'取消'中止"
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),
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True,
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)
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