我想在不退出的情况下捕获和记录异常,例如,
try:
do_stuff()
except Exception as err:
print(Exception, err)
# I want to print the entire traceback here,
# not just the exception name and details
我想打印与抛出异常时打印的完全相同的输出,而不使用try/,只是拦截异常,并且我不希望它退出程序。
我想在不退出的情况下捕获和记录异常,例如,
try:
do_stuff()
except Exception as err:
print(Exception, err)
# I want to print the entire traceback here,
# not just the exception name and details
我想打印与抛出异常时打印的完全相同的输出,而不使用try/,只是拦截异常,并且我不希望它退出程序。
当前回答
Traceback.format_exc()或sys.exc_info()将产生更多信息,如果这是你想要的。
import traceback
import sys
try:
do_stuff()
except Exception:
print(traceback.format_exc())
# or
print(sys.exc_info()[2])
其他回答
traceback.format_exception (exception_object)
如果你只有异常对象,你可以从Python 3中的任何代码点获得字符串形式的回溯:
import traceback
''.join(traceback.format_exception(None, exc_obj, exc_obj.__traceback__))
完整的例子:
#!/usr/bin/env python3
import traceback
def f():
g()
def g():
raise Exception('asdf')
try:
g()
except Exception as e:
exc_obj = e
tb_str = ''.join(traceback.format_exception(None, exc_obj, exc_obj.__traceback__))
print(tb_str)
输出:
Traceback (most recent call last):
File "./main.py", line 12, in <module>
g()
File "./main.py", line 9, in g
raise Exception('asdf')
Exception: asdf
文档:https://docs.python.org/3.9/library/traceback.html traceback.format_exception
请参见:从异常对象中提取回溯信息
在Python 3.9中测试
关于这个答案的注释:print(traceback.format_exc())对我来说比traceback.print_exc()更好。对于后者,hello有时会奇怪地与回溯文本“混合”在一起,比如两者都想同时写入stdout或stderr,产生奇怪的输出(至少在从文本编辑器内部构建并在“构建结果”面板中查看输出时)。
回溯(最近一次调用): 文件“C:\Users\User\Desktop\test.py”,第7行,在 地狱do_stuff () 文件“C:\Users\User\Desktop\test.py”,第4行,do_stuff 1/0 ZeroDivisionError:整数除法或对0取模 o [0.1s完成]
所以我用:
import traceback, sys
def do_stuff():
1/0
try:
do_stuff()
except Exception:
print(traceback.format_exc())
print('hello')
要获得精确的堆栈跟踪(作为字符串),如果没有try/except进行跨步处理,则会引发该字符串,只需将其放置在捕获违规异常的except块中。
desired_trace = traceback.format_exc(sys.exc_info())
下面是如何使用它(假设定义了flaky_func,并且log调用您最喜欢的日志系统):
import traceback
import sys
try:
flaky_func()
except KeyboardInterrupt:
raise
except Exception:
desired_trace = traceback.format_exc(sys.exc_info())
log(desired_trace)
捕获并重新引发KeyboardInterrupts是个好主意,这样您仍然可以使用Ctrl-C终止程序。日志记录不在这个问题的范围之内,但是日志记录是一个很好的选择。sys和traceback模块的文档。
这是我把错误写在日志文件和控制台的解决方案:
import logging, sys
import traceback
logging.basicConfig(filename='error.log', level=logging.DEBUG)
def handle_exception(exc_type, exc_value, exc_traceback):
if issubclass(exc_type, KeyboardInterrupt):
sys.__excepthook__(exc_type, exc_value, exc_traceback)
return
exc_info=(exc_type, exc_value, exc_traceback)
logging.critical("\nDate:" + str(datetime.datetime.now()), exc_info=(exc_type, exc_value, exc_traceback))
print("An error occured, check error.log to see the error details")
traceback.print_exception(*exc_info)
sys.excepthook = handle_exception
Python 3解决方案
stacktrace_helper.py:
from linecache import getline
import sys
import traceback
def get_stack_trace():
exc_type, exc_value, exc_tb = sys.exc_info()
trace = traceback.format_stack()
trace = list(filter(lambda x: ("\\lib\\" not in x and "/lib/" not in x and "stacktrace_helper.py" not in x), trace))
ex_type = exc_type.__name__
ex_line = exc_tb.tb_lineno
ex_file = exc_tb.tb_frame.f_code.co_filename
ex_message = str(exc_value)
line_code = ""
try:
line_code = getline(ex_file, ex_line).strip()
except:
pass
trace.insert(
0, f'File "{ex_file}", line {ex_line}, line_code: {line_code} , ex: {ex_type} {ex_message}',
)
return trace
def get_stack_trace_str(msg: str = ""):
trace = list(get_stack_trace())
trace_str = "\n".join(list(map(str, trace)))
trace_str = msg + "\n" + trace_str
return trace_str