当我捕获异常时，我如何获得类型、文件和行号?

import sys, os


try:
raise NotImplementedError("No error")
except Exception as e:
exc_type, exc_obj, exc_tb = sys.exc_info()
fname = os.path.split(exc_tb.tb_frame.f_code.co_filename)[1]
print(exc_type, fname, exc_tb.tb_lineno)

源 (Py v2.7.3)用于traceback.format_exception ()和被调用的/相关函数有很大帮助。令人尴尬的是，我总是忘记阅读原文。我只是在徒劳地搜索了类似的细节后才这样做的。一个简单的问题，“如何为异常重新创建与Python相同的输出，并具有所有相同的细节?”这将使任何人得到90%以上的他们想要的东西。沮丧的我想出了这个例子。我希望它能帮助其他人。(它确实帮助了我!: -)

import sys, traceback


traceback_template = '''Traceback (most recent call last):
File "%(filename)s", line %(lineno)s, in %(name)s
%(type)s: %(message)s\n''' # Skipping the "actual line" item


# Also note: we don't walk all the way through the frame stack in this example
# see hg.python.org/cpython/file/8dffb76faacc/Lib/traceback.py#l280
# (Imagine if the 1/0, below, were replaced by a call to test() which did 1/0.)


try:
1/0
except:
# http://docs.python.org/2/library/sys.html#sys.exc_info
exc_type, exc_value, exc_traceback = sys.exc_info() # most recent (if any) by default


'''
Reason this _can_ be bad: If an (unhandled) exception happens AFTER this,
or if we do not delete the labels on (not much) older versions of Py, the
reference we created can linger.


traceback.format_exc/print_exc do this very thing, BUT note this creates a
temp scope within the function.
'''


traceback_details = {
'filename': exc_traceback.tb_frame.f_code.co_filename,
'lineno'  : exc_traceback.tb_lineno,
'name'    : exc_traceback.tb_frame.f_code.co_name,
'type'    : exc_type.__name__,
'message' : exc_value.message, # or see traceback._some_str()
}


del(exc_type, exc_value, exc_traceback) # So we don't leave our local labels/objects dangling
# This still isn't "completely safe", though!
# "Best (recommended) practice: replace all exc_type, exc_value, exc_traceback
# with sys.exc_info()[0], sys.exc_info()[1], sys.exc_info()[2]


print
print traceback.format_exc()
print
print traceback_template % traceback_details
print

具体回答此问题:

sys.exc_info()[0].__name__, os.path.basename(sys.exc_info()[2].tb_frame.f_code.co_filename), sys.exc_info()[2].tb_lineno

下面是一个显示发生异常的行号的示例。

import sys
try:
print(5/0)
except Exception as e:
print('Error on line {}'.format(sys.exc_info()[-1].tb_lineno), type(e).__name__, e)


print('And the rest of program continues')

最简单的形式适合我。

import traceback


try:
print(4/0)
except ZeroDivisionError:
print(traceback.format_exc())

输出

Traceback (most recent call last):
File "/path/to/file.py", line 51, in <module>
print(4/0)
ZeroDivisionError: division by zero


Process finished with exit code 0

你可以在不导入traceback的情况下实现这一点:

try:
func1()
except Exception as ex:
trace = []
tb = ex.__traceback__
while tb is not None:
trace.append({
"filename": tb.tb_frame.f_code.co_filename,
"name": tb.tb_frame.f_code.co_name,
"lineno": tb.tb_lineno
})
tb = tb.tb_next
print(str({
'type': type(ex).__name__,
'message': str(ex),
'trace': trace
}))

输出:

{


'type': 'ZeroDivisionError',
'message': 'division by zero',
'trace': [
{
'filename': '/var/playground/main.py',
'name': '<module>',
'lineno': 16
},
{
'filename': '/var/playground/main.py',
'name': 'func1',
'lineno': 11
},
{
'filename': '/var/playground/main.py',
'name': 'func2',
'lineno': 7
},
{
'filename': '/var/playground/my.py',
'name': 'test',
'lineno': 2
}
]
}

没有任何导入，但也不兼容导入的模块:

try:
raise TypeError("Hello, World!")  # line 2
except Exception as e:
print(
type(e).__name__,          # TypeError
__file__,                  # /tmp/example.py
e.__traceback__.tb_lineno  # 2
)

重申一下，这在__abc0或模块上执行不，因此如果执行import X; try: X.example();，则文件名和行号将指向包含X.example()的行，而不是出错的行在 X.example()。如果有人知道如何从最后一个堆栈跟踪行中轻松获得文件名和行号(我期望类似e[-1].filename的东西，但没有这样的运气)，请改进这个答案。

这是我用来获取文件名的。

__file__.__str__

总之，我创建了一个页面来显示错误。返回此异常。

context={
'details':'Type of error:{}:Function name:{}:Line number:{}'.format(exc_type, fname, exc_tb.tb_lineno),
'error_details':str(e),
'filename':__file__.__str__,
})

发生异常时的情况: