python try except返回异常的信息字符串代码实例

Python  /  管理员 发布于 7年前   230

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 = 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)

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