python try except返回异常的信息字符串代码实例

系统 1532 0

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 "
            
            ", line 1, in 
            
              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 "
            
            ", line 1, in 
            
              
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,也希望大家多多支持脚本之家。


更多文章、技术交流、商务合作、联系博主

微信扫码或搜索:z360901061

微信扫一扫加我为好友

QQ号联系: 360901061

您的支持是博主写作最大的动力,如果您喜欢我的文章,感觉我的文章对您有帮助,请用微信扫描下面二维码支持博主2元、5元、10元、20元等您想捐的金额吧,狠狠点击下面给点支持吧,站长非常感激您!手机微信长按不能支付解决办法:请将微信支付二维码保存到相册,切换到微信,然后点击微信右上角扫一扫功能,选择支付二维码完成支付。

【本文对您有帮助就好】

您的支持是博主写作最大的动力,如果您喜欢我的文章,感觉我的文章对您有帮助,请用微信扫描上面二维码支持博主2元、5元、10元、自定义金额等您想捐的金额吧,站长会非常 感谢您的哦!!!

发表我的评论
最新评论 总共0条评论