Change to TimeoutError #71
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sci = self.send('scientifica', timeout=0.2) (line 132) will lead to a TimeoutError, not RunetimeError. If want to try again need to catch TimeoutError.
campagnola
reviewed
Dec 30, 2017
| @@ -130,7 +130,7 @@ def __init__(self, port=None, name=None, baudrate=None, ctrl_version=2): | |||
| SerialDevice.__init__(self, port=self.port, baudrate=baudrate) | |||
| try: | |||
| sci = self.send('scientifica', timeout=0.2) | |||
| except RuntimeError: | |||
| except TimeoutError: | |||
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In some circumstances we might get a RuntimeError as well, so it should be
except (RuntimeError, TimeoutError):
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sci = self.send('scientifica', timeout=0.2) (line 132) will lead to a TimeoutError, not RunetimeError. If want to try again need to catch TimeoutError.