如何使用检查从 Python 中的被调用者获取调用者的信息？
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How to use inspect to get the caller's info from callee in Python?
提问by prosseek
I need to get the caller info (what file/what line) from callee. I learned that I can use inpect module for that for purposes, but not exactly how.
我需要从被调用者那里获取调用者信息（什么文件/什么行）。我了解到我可以为此目的使用 inpect 模块，但不完全是如何使用。
How to get those info with inspect? Or is there any other way to get the info?
如何通过检查获取这些信息？或者有什么其他方法可以获取信息？
import inspect
print __file__
c=inspect.currentframe()
print c.f_lineno
def hello():
print inspect.stack
?? what file called me in what line?
hello()
采纳答案by unutbu
The caller's frame is one frame higher than the current frame. You can use inspect.currentframe().f_backto find the caller's frame.
Then use inspect.getframeinfoto get the caller's filename and line number.
调用者的帧比当前帧高一帧。您可以使用inspect.currentframe().f_back来查找调用者的框架。然后使用inspect.getframeinfo获取调用者的文件名和行号。
import inspect
def hello():
previous_frame = inspect.currentframe().f_back
(filename, line_number,
function_name, lines, index) = inspect.getframeinfo(previous_frame)
return (filename, line_number, function_name, lines, index)
print(hello())
# ('/home/unutbu/pybin/test.py', 10, '<module>', ['hello()\n'], 0)
回答by Dmitry K.
I would suggest to use inspect.stackinstead:
我建议inspect.stack改用：
import inspect
def hello():
frame,filename,line_number,function_name,lines,index = inspect.stack()[1]
print(frame,filename,line_number,function_name,lines,index)
hello()
回答by Jacob Cohen
If the caller is the main file, simply use sys.argv[0]
如果调用者是主文件，只需使用 sys.argv[0]
回答by acue
I published a wrapper for inspect with simple stackframe addressing covering the stack frame by a single parameter spos:
我发布了一个包装器，用于通过一个参数覆盖堆栈帧的简单堆栈帧寻址进行检查spos：
E.g. pysourceinfo.PySourceInfo.getCallerLinenumber(spos=1)
例如 pysourceinfo.PySourceInfo.getCallerLinenumber(spos=1)
where spos=0is the lib-function, spos=1is the caller, spos=2the caller-of-the-caller, etc.
spos=0lib函数在哪里，spos=1是调用者，spos=2调用者的调用者等。
