How to use inspect to get the caller's info from callee in Python?
PythonInspectPython Problem Overview
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.
How to get those info with inspect? Or is there any other way to get the info?
import inspect
print __file__
c=inspect.currentframe()
print c.f_lineno
def hello():
print inspect.stack
?? what file called me in what line?
hello()
Python Solutions
Solution 1 - Python
The caller's frame is one frame higher than the current frame. You can use inspect.currentframe().f_back
to find the caller's frame.
Then use inspect.getframeinfo to get the caller's filename and line number.
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)
Solution 2 - Python
I would suggest to use inspect.stack
instead:
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()
Solution 3 - Python
I published a wrapper for inspect with simple stackframe addressing covering the stack frame by a single parameter spos
:
E.g. pysourceinfo.PySourceInfo.getCallerLinenumber(spos=1)
where spos=0
is the lib-function, spos=1
is the caller, spos=2
the caller-of-the-caller, etc.
Solution 4 - Python
If the caller is the main file, simply use sys.argv[0]