修复send_api中command_to_stream未使用display_name的问题
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@@ -284,7 +284,7 @@ async def image_to_stream(image_base64: str, stream_id: str, storage_message: bo
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return await _send_to_target("image", image_base64, stream_id, "", typing=False, storage_message=storage_message)
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return await _send_to_target("image", image_base64, stream_id, "", typing=False, storage_message=storage_message)
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async def command_to_stream(command: Union[str, dict], stream_id: str, storage_message: bool = True) -> bool:
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async def command_to_stream(command: Union[str, dict], stream_id: str, storage_message: bool = True, display_message: str = "") -> bool:
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"""向指定流发送命令
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"""向指定流发送命令
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Args:
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Args:
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@@ -295,7 +295,7 @@ async def command_to_stream(command: Union[str, dict], stream_id: str, storage_m
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Returns:
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Returns:
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bool: 是否发送成功
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bool: 是否发送成功
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"""
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"""
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return await _send_to_target("command", command, stream_id, "", typing=False, storage_message=storage_message)
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return await _send_to_target("command", command, stream_id, display_message, typing=False, storage_message=storage_message)
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async def custom_to_stream(
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async def custom_to_stream(
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