[英]Python traceback with module names
Python中的堆棧跟蹤顯示文件路徑。 有沒有辦法讓他們顯示完全合格的功能名稱?
例:
class Foo(object):
def bar(self):
raise Exception, "Some error."
def inner():
return Foo().bar()
def outer():
return inner()
我希望我的輸出看起來像這樣:
In __main__.Foo.bar ("scratchpad.py", line 3)
__main__.inner ("scratchpad.py", line 6)
__main__.outer ("scratchpad.py", line 9)
Exception: Some error.
如果它改變了什么,我正在使用Python 2.7。
這是我到目前為止所擁有的:
import sys
class Foo(object):
def bar(self):
raise Exception, "Dummy value."
def inner():
"Inner function..."
return Foo().bar()
def outer():
return inner()
try:
outer()
except Exception, error:
traceback = sys.exc_info()[2]
while traceback is not None:
frame = traceback.tb_frame
print 'Name', frame.f_globals['__name__']+'.'+frame.f_code.co_name
docs = frame.f_code.co_consts[0]
if docs and docs != -1: # docs can be None or -1.
print 'Docs "%s"' % docs
print 'Args', frame.f_code.co_argcount
print 'File "%s", line %d' % (frame.f_code.co_filename, frame.f_lineno)
print
traceback = traceback.tb_next
當我運行它時,它會打印出來
$ python pretty-stack.py
Name __main__.<module>
Args 0
File "pretty-stack.py", line 28
Name __main__.outer
Args 0
File "pretty-stack.py", line 14
Name __main__.inner
Docs "Inner function..."
Args 0
File "pretty-stack.py", line 11
Name __main__.bar
Args 1
File "pretty-stack.py", line 7
它幾乎就在那里,但我遇到了重要的用例問題。 例如,我無法獲得Foo.bar()
的類名Foo
。
沒有直接的方法來從回溯中訪問符號,因為只有“代碼對象”是可訪問的,並且正如代碼對象上的Python文檔所說:
與函數對象不同,代碼對象是不可變的,並且不包含(直接或間接)可變對象的引用。
似乎要檢索回溯中涉及的模塊,函數和類,我們需要搜索它們。
我有一個似乎有效的實驗版本。 此實現基於遍歷代碼對象引用的模塊,以查找引用相關代碼對象的函數或方法。
from collections import namedtuple
import inspect
import sys
from nested.external import Foo
def inner(a, b='qux'):
"Inner function..."
return Foo().bar()
def outer(a, *args, **kwds):
return inner(a)
def resolve_signature(function):
"""Get a formatted string that looks like the function's signature."""
prgs, vrgs, kwds, defs = inspect.getargspec(function)
arguments = []
if defs:
for name in prgs[:len(defs)-1]:
arguments.append(name)
for i,name in enumerate(prgs[len(defs)-1]):
arguments.append('%s=%r'%(name,defs[i]))
else:
arguments.extend(prgs)
if vrgs:
arguments.append('*'+vrgs)
if kwds:
arguments.append('**'+kwds)
return '('+', '.join(arguments)+')'
def resolve_scope(module_name, code):
"""Resolve the scope name for a code object provided its module name."""
# Resolve module.
module = sys.modules.get(module_name, None)
if not module:
return '<hidden-module>' + '.' + code.co_name + '(?)'
# Check module's functions.
symbols = inspect.getmembers(module, inspect.isfunction)
for symbol_name,symbol_info in symbols:
if symbol_info.func_code is code:
scope = module_name + '.'
return scope + code.co_name + resolve_signature(symbol_info)
# Check module's classes.
symbols = inspect.getmembers(module, inspect.isclass)
for symbol_name,symbol_info in symbols:
# Check class' methods.
members = inspect.getmembers(symbol_info, inspect.ismethod)
for method_name,method_info in members:
if method_info.__func__.func_code is code:
scope = module_name + '.' + symbol_name + '.'
return scope + code.co_name + resolve_signature(method_info)
# Default to the thing's name. This usually happens
# when resolving the stack frame for module-level code.
return code.co_name
Frame = namedtuple('Frame', ['call', 'file', 'line', 'help'])
def pretty_stack(traceback=None):
"""Returns a simple stack frame."""
frames = []
if traceback is None:
traceback = sys.exc_info()[2]
while traceback is not None:
frame = traceback.tb_frame
call = resolve_scope(frame.f_globals['__name__'], frame.f_code)
path = frame.f_code.co_filename.replace('\\', '/')
line = frame.f_lineno
docs = frame.f_code.co_consts[0]
if docs == -1:
docs = None
frames.append(Frame(call, path, line, docs))
traceback = traceback.tb_next
return frames
try:
outer(1)
except Exception, error:
frames = pretty_stack()
for frame in frames:
print frame.call
print ' -> "%s", line %d.' % (frame.file, frame.line)
if frame.help:
print frame.help
print
當我運行這個時,我會得到類似的東西:
$ python pretty-stack.py
<module>
-> "pretty-stack.py", line 84.
__main__.outer(a, *args, **kwds)
-> "pretty-stack.py", line 14.
__main__.inner(a='qux')
-> "pretty-stack.py", line 11.
Inner function...
nested.external.Foo.bar(self)
-> "C:/Users/acaron/Desktop/nested/external.py", line 3.
請注意,這甚至會打印函數簽名和doc字符串,這在調試期間可能會有所幫助。
在描述符和諸如此類的情況下,它可能無法正常工作,我沒有嘗試過非常復雜的用例。 如果你能找到一個不起作用的案例,請告訴我,我會嘗試修補它。
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