Python当我发现exception时，如何获取types，文件和行号？
捕获一个会打印这样的exception：
Traceback (most recent call last): File "c:/tmp.py", line 1, in <module> 4 / 0 ZeroDivisionError: integer division or modulo by zero 我想将其格式化为：
ZeroDivisonError, tmp.py, 1
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）为traceback.format_exception（）和调用/相关函数有很大的帮助。 令人尴尬的是，我总是忘记阅读资料来源 。 我只是在徒劳地寻找类似细节之后才这样做的。 一个简单的问题，“如何重新创build与Python相同的输出作为例外，具有相同的细节？” 无论他们在寻找什么，这都会得到90％以上的人。 沮丧，我想出了这个例子。 我希望它可以帮助别人。 （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
在这个查询的具体答案：
sys.exc_info()[0].__name__, os.path.basename(sys.exc_info()[2].tb_frame.f_code.co_filename), sys.exc_info()[2].tb_lineno
下面是显示发生exception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')
