python try except返回异常的信息字符串代码实例
2019-08-15 17:09
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问题
https://docs.python.org/3/tutorial/errors.html#handling-exceptions
https://docs.python.org/3/library/exceptions.html#ValueError
try: int("x") except Exception as e: '''异常的父类,可以捕获所有的异常''' print(e) # e变量是Exception类型的实例,支持__str__()方法,可以直接打印。 invalid literal for int() with base 10: 'x' try: int("x") except Exception as e: '''异常的父类,可以捕获所有的异常''' print(e.args)
# e变量有个属性是.args,它是错误信息的元组
("invalid literal for int() with base 10: 'x'",)try: datetime(2017,2,30)except ValueError as e: print(e) day is out of range for monthtry: datetime(22017,2,30)except ValueError as e: print(e) year 22017 is out of rangetry: datetime(2017,22,30)except ValueError as e: print(e) month must be in 1..12e = Nonetry: datetime(2017,22,30)except ValueError as e: print(e) month must be in 1..12e # e这个变量在异常过程结束后即被释放,再调用也无效 Traceback (most recent call last): File "<input>", line 1, in <module>NameError: name 'e' is not defined
errarg = None try: datetime(2017,22,30) except ValueError as errarg: print(errarg) month must be in 1..12 errarg Traceback (most recent call last): File "<input>", line 1, in <module> NameError: name 'errarg' is not defined try: datetime(2017,22,30) except ValueError as errarg: print(errarg.args) # ValueError.args 返回元组 ('month must be in 1..12',) message = None try: datetime(2017,22,30) except ValueError as errarg: print(errarg.args) message = errarg.args ('month must be in 1..12',) message ('month must be in 1..12',) try: datetime(2017,22,30) except ValueError as errarg: print(errarg.args) message = errarg ('month must be in 1..12',) message ValueError('month must be in 1..12',) str(message) 'month must be in 1..12'
分析异常信息,并根据异常信息的提示做出相应处理:
try: y = 2017 m = 22 d = 30 datetime(y,m,d) except ValueError as errarg: print(errarg.args) message = errarg m = re.search(u"month", str(message)) if m: dt = datetime(y,1,d) ('month must be in 1..12',) dt datetime.datetime(2017, 1, 30, 0, 0)
甚至可以再except中进行递归调用:
def validatedate(y, mo, d): dt = None try: dt = datetime(y, mo, d) except ValueError as e: print(e.args) print(str(y)+str(mo)+str(d)) message = e ma = re.search(u"^(year)|(month)|(day)", str(message)) ymd = ma.groups() if ymd[0]: dt = validatedate(datetime.now().year, mo, d) if ymd[1]: dt = validatedate(y, datetime.now().month, d) if ymd[2]: dt = validatedate(y, mo, datetime.now().day) finally: return dt validatedate(20199, 16, 33) ('year 20199 is out of range',) ('month must be in 1..12',) ('day is out of range for month',) datetime.datetime(2018, 4, 20, 0, 0)
以上就是本文的全部内容,希望对大家的学习有所帮助
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