import os ,fnmatch
import os.path
import os


file_dir= '/home/deeghayu/Desktop/mec_sim/up_folder'
file_type = ['*.py']
        
for root, dirs,files in os.walk( file_dir ):
            for extension in ( tuple(file_type) ):
                for filename in fnmatch.filter(files, extension):
                    filepath = os.path.join(root, filename)
                    if os.path.isfile( filepath ):
                        print(filename , 'executing...');
                       
                        exec(open('/home/deeghayu/Desktop/mec_sim/up_folder/{}'.format(filename)).read())
                        
                    else:
                        print('Execution failure!!!')

Hello everyone I am working on this code which execute a python file using a python code. I need to save my output of the code as a text file. Here I have shown my code. Can any one give me a solution how do I save my output into a text file?

CodePudding user response：

Piggybacking off of the original answer since they are close but it isn't a best practice to open and close files that way.

It's better to use a context manager instead of saying f = open() since the context manager will handle closing the resource for you regardless of whether your code succeeds or not.

You use it like,

with open("file.txt","w ") as f:
    for i in range(10):
        f.write("This is line %d\r\n" % (i 1))

CodePudding user response：

try

Open file

f= open("file.txt","w ")

Insert data into file

for i in range(10):
     f.write("This is line %d\r\n" % (i 1))

Close the file

f.close()