python try except返回异常的信息字符串代码实例
问题
https://docs.python.org/3/tutorial/errors.html
https://docs.python.org/3/library/exceptions.html
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，也希望大家多多支持【听图阁-专注于Python设计】。
