python try except 出现异常时，except 中如何返回异常的信息字符串

Posted 2022-07-14 wpcnblog

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‘,)
201991633
(‘month must be in 1..12‘,)
20181633
(‘day is out of range for month‘,)
2018433
datetime.datetime(2018, 4, 20, 0, 0)