python try except返回异常的信息字符串代码实例

时间:2021-05-22

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 definederrarg = Nonetry: datetime(2017,22,30)except ValueError as errarg: print(errarg) month must be in 1..12errargTraceback (most recent call last): File "<input>", line 1, in <module>NameError: name 'errarg' is not definedtry: datetime(2017,22,30)except ValueError as errarg: print(errarg.args)# ValueError.args 返回元组('month must be in 1..12',)message = Nonetry: 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',)messageValueError('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',)dtdatetime.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,也希望大家多多支持。

声明:本页内容来源网络,仅供用户参考;我单位不保证亦不表示资料全面及准确无误,也不保证亦不表示这些资料为最新信息,如因任何原因,本网内容或者用户因倚赖本网内容造成任何损失或损害,我单位将不会负任何法律责任。如涉及版权问题,请提交至online#300.cn邮箱联系删除。

相关文章