filename and line number of Python script

How can I get the file name and line number in a Python script?

Exactly the file information we get from an exception traceback. In this case without raising an exception.

Solution 1:

Thanks to mcandre, the answer is:

#python3
from inspect import currentframe, getframeinfo

frameinfo = getframeinfo(currentframe())

print(frameinfo.filename, frameinfo.lineno)

Solution 2:

Whether you use currentframe().f_back depends on whether you are using a function or not.

Calling inspect directly:

from inspect import currentframe, getframeinfo

cf = currentframe()
filename = getframeinfo(cf).filename

print "This is line 5, python says line ", cf.f_lineno 
print "The filename is ", filename

Calling a function that does it for you:

from inspect import currentframe

def get_linenumber():
    cf = currentframe()
    return cf.f_back.f_lineno

print "This is line 7, python says line ", get_linenumber()

Solution 3:

Handy if used in a common file - prints file name, line number and function of the caller:

import inspect
def getLineInfo():
    print(inspect.stack()[1][1],":",inspect.stack()[1][2],":",
          inspect.stack()[1][3])

Solution 4:

Filename:

__file__
# or
sys.argv[0]

Line:

inspect.currentframe().f_lineno

(not inspect.currentframe().f_back.f_lineno as mentioned above)