Python shows No such file or directory Error though the file exists

I want to search a string from multiple files

What I tried:

import os

path= 'sample1/nvram2/logs' 

all_files=os.listdir(path) 

for my_file1 in all_files:

    print(my_file1)

    with open(my_file1, 'r') as my_file2:

        print(my_file2)

        for line in my_file2:

            if 'string' in line:

                print(my_file2)

Output:

C:Usersuser1scripts>python search_string_3.py

abcd.txt

Traceback (most recent call last):

  File "search_string_3.py", line 6, in <module>

    with open(my_file1, 'r') as my_file2:

FileNotFoundError: [Errno 2] No such file or directory: 'abcd.txt'

But file abcd.txt is present in C:Usersuser1scriptssample1nvram2logs

Why the Error shows that No such file or directory?

Using glob:

The following error was displayed when I use all_files=glob.glob(path) instead of all_files=os.listdir(path)

C:Usersuser1scripts>python search_string_3.py

sample1/nvram2/logs

Traceback (most recent call last):

  File "search_string_3.py", line 7, in <module>

    with open(my_file1, 'r') as my_file2:

PermissionError: [Errno 13] Permission denied: 'sample1/nvram2/logs'

asked Nov 14 '18 at 9:11

Dipankar Nalui

2321517

add a comment |

I want to search a string from multiple files

What I tried:

import os

path= 'sample1/nvram2/logs' 

all_files=os.listdir(path) 

for my_file1 in all_files:

    print(my_file1)

    with open(my_file1, 'r') as my_file2:

        print(my_file2)

        for line in my_file2:

            if 'string' in line:

                print(my_file2)

Output:

C:Usersuser1scripts>python search_string_3.py

abcd.txt

Traceback (most recent call last):

  File "search_string_3.py", line 6, in <module>

    with open(my_file1, 'r') as my_file2:

FileNotFoundError: [Errno 2] No such file or directory: 'abcd.txt'

But file abcd.txt is present in C:Usersuser1scriptssample1nvram2logs

Why the Error shows that No such file or directory?

Using glob:

The following error was displayed when I use all_files=glob.glob(path) instead of all_files=os.listdir(path)

C:Usersuser1scripts>python search_string_3.py

sample1/nvram2/logs

Traceback (most recent call last):

  File "search_string_3.py", line 7, in <module>

    with open(my_file1, 'r') as my_file2:

PermissionError: [Errno 13] Permission denied: 'sample1/nvram2/logs'

asked Nov 14 '18 at 9:11

Dipankar Nalui

2321517

add a comment |

I want to search a string from multiple files

What I tried:

import os

path= 'sample1/nvram2/logs' 

all_files=os.listdir(path) 

for my_file1 in all_files:

    print(my_file1)

    with open(my_file1, 'r') as my_file2:

        print(my_file2)

        for line in my_file2:

            if 'string' in line:

                print(my_file2)

Output:

C:Usersuser1scripts>python search_string_3.py

abcd.txt

Traceback (most recent call last):

  File "search_string_3.py", line 6, in <module>

    with open(my_file1, 'r') as my_file2:

FileNotFoundError: [Errno 2] No such file or directory: 'abcd.txt'

But file abcd.txt is present in C:Usersuser1scriptssample1nvram2logs

Why the Error shows that No such file or directory?

Using glob:

The following error was displayed when I use all_files=glob.glob(path) instead of all_files=os.listdir(path)

C:Usersuser1scripts>python search_string_3.py

sample1/nvram2/logs

Traceback (most recent call last):

  File "search_string_3.py", line 7, in <module>

    with open(my_file1, 'r') as my_file2:

PermissionError: [Errno 13] Permission denied: 'sample1/nvram2/logs'

asked Nov 14 '18 at 9:11

Dipankar Nalui

2321517

I want to search a string from multiple files

What I tried:

import os

path= 'sample1/nvram2/logs' 

all_files=os.listdir(path) 

for my_file1 in all_files:

    print(my_file1)

    with open(my_file1, 'r') as my_file2:

        print(my_file2)

        for line in my_file2:

            if 'string' in line:

                print(my_file2)

Output:

C:Usersuser1scripts>python search_string_3.py

abcd.txt

Traceback (most recent call last):

  File "search_string_3.py", line 6, in <module>

    with open(my_file1, 'r') as my_file2:

FileNotFoundError: [Errno 2] No such file or directory: 'abcd.txt'

But file abcd.txt is present in C:Usersuser1scriptssample1nvram2logs

Why the Error shows that No such file or directory?

Using glob:

The following error was displayed when I use all_files=glob.glob(path) instead of all_files=os.listdir(path)

C:Usersuser1scripts>python search_string_3.py

sample1/nvram2/logs

Traceback (most recent call last):

  File "search_string_3.py", line 7, in <module>

    with open(my_file1, 'r') as my_file2:

PermissionError: [Errno 13] Permission denied: 'sample1/nvram2/logs'

python python-3.x

asked Nov 14 '18 at 9:11

Dipankar Nalui

2321517

asked Nov 14 '18 at 9:11

Dipankar Nalui

2321517

asked Nov 14 '18 at 9:11

Dipankar Nalui

2321517

asked Nov 14 '18 at 9:11

Dipankar Nalui

2321517

asked Nov 14 '18 at 9:11

Dipankar Nalui

2321517

add a comment |

2 Answers
2

active

oldest

votes

you figured out/guessed the first issue. Joining the directory with the filename solves it. A classic:

with open(os.path.join(path,my_file1), 'r') as my_file2:

I wouldn't have cared to answer if you didn't attempted something with glob. Now:

for x in glob.glob(path):

since path is a directory, glob evaluates it as itself (you get a list with one element: [path]). You need to add a wildcard:

for x in glob.glob(os.path.join(path,"*")):

The other issue with glob is that if the directory (or the pattern) doesn't match anything you're not getting any error. It just does nothing... The os.listdir version crashes at least.

and also test if it's a file before opening (in both cases) because attempting to open a directory results in an I/O exception:

if os.path.isfile(x):

  with open ...

In a nutshell os.path package is your friend when manipulating files.

edited Nov 14 '18 at 9:46

answered Nov 14 '18 at 9:18

Jean-François Fabre

104k955112

This solution solved the problem.

– Dipankar Nalui
Nov 14 '18 at 10:54

add a comment |

-1

Since the file abcd.txt is present in C:Usersuser1scriptssample1nvram2logs, and the said path is not your working directory, you have to add the same to sys.path

import os, sys

path= 'sample1/nvram2/logs'

sys.path.append(path)





all_files=os.listdir(path) 

for my_file1 in all_files:

    print(my_file1)

    with open(my_file1, 'r') as my_file2:

        print(my_file2)

        for line in my_file2:

            if 'string' in line:

                print(my_file2)

answered Nov 14 '18 at 9:23

Pradip Gupta

6910

no, sys.path is for module loading.

– Jean-François Fabre
Nov 14 '18 at 9:25

sys.path is not only for module loading. Even when you have to import files without giving their complete path, you can append their path using sys.path and then call it directly. It only makes your files searchable by the current path.

– Pradip Gupta
Nov 14 '18 at 9:54

open doesn't use sys.path at all

– Jean-François Fabre
Nov 14 '18 at 10:17

I agree with that. However, in general, python would search for the file in all the paths found in sys.path. For example, when you have the file in your cwd, you do not give the absolute path with open as the file is already present in the path for you. Same is the logic with sys.path

– Pradip Gupta
Nov 14 '18 at 10:20

modules aren't data files.

– Jean-François Fabre
Nov 14 '18 at 10:21

|
show 1 more comment

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

you figured out/guessed the first issue. Joining the directory with the filename solves it. A classic:

with open(os.path.join(path,my_file1), 'r') as my_file2:

I wouldn't have cared to answer if you didn't attempted something with glob. Now:

for x in glob.glob(path):

since path is a directory, glob evaluates it as itself (you get a list with one element: [path]). You need to add a wildcard:

for x in glob.glob(os.path.join(path,"*")):

The other issue with glob is that if the directory (or the pattern) doesn't match anything you're not getting any error. It just does nothing... The os.listdir version crashes at least.

and also test if it's a file before opening (in both cases) because attempting to open a directory results in an I/O exception:

if os.path.isfile(x):

  with open ...

In a nutshell os.path package is your friend when manipulating files.

edited Nov 14 '18 at 9:46

answered Nov 14 '18 at 9:18

Jean-François Fabre

104k955112

This solution solved the problem.

– Dipankar Nalui
Nov 14 '18 at 10:54

add a comment |

you figured out/guessed the first issue. Joining the directory with the filename solves it. A classic:

with open(os.path.join(path,my_file1), 'r') as my_file2:

I wouldn't have cared to answer if you didn't attempted something with glob. Now:

for x in glob.glob(path):

since path is a directory, glob evaluates it as itself (you get a list with one element: [path]). You need to add a wildcard:

for x in glob.glob(os.path.join(path,"*")):

The other issue with glob is that if the directory (or the pattern) doesn't match anything you're not getting any error. It just does nothing... The os.listdir version crashes at least.

and also test if it's a file before opening (in both cases) because attempting to open a directory results in an I/O exception:

if os.path.isfile(x):

  with open ...

In a nutshell os.path package is your friend when manipulating files.

edited Nov 14 '18 at 9:46

answered Nov 14 '18 at 9:18

Jean-François Fabre

104k955112

This solution solved the problem.

– Dipankar Nalui
Nov 14 '18 at 10:54

add a comment |

you figured out/guessed the first issue. Joining the directory with the filename solves it. A classic:

with open(os.path.join(path,my_file1), 'r') as my_file2:

I wouldn't have cared to answer if you didn't attempted something with glob. Now:

for x in glob.glob(path):

since path is a directory, glob evaluates it as itself (you get a list with one element: [path]). You need to add a wildcard:

for x in glob.glob(os.path.join(path,"*")):

The other issue with glob is that if the directory (or the pattern) doesn't match anything you're not getting any error. It just does nothing... The os.listdir version crashes at least.

and also test if it's a file before opening (in both cases) because attempting to open a directory results in an I/O exception:

if os.path.isfile(x):

  with open ...

In a nutshell os.path package is your friend when manipulating files.

edited Nov 14 '18 at 9:46

answered Nov 14 '18 at 9:18

Jean-François Fabre

104k955112

you figured out/guessed the first issue. Joining the directory with the filename solves it. A classic:

with open(os.path.join(path,my_file1), 'r') as my_file2:

I wouldn't have cared to answer if you didn't attempted something with glob. Now:

for x in glob.glob(path):

since path is a directory, glob evaluates it as itself (you get a list with one element: [path]). You need to add a wildcard:

for x in glob.glob(os.path.join(path,"*")):

The other issue with glob is that if the directory (or the pattern) doesn't match anything you're not getting any error. It just does nothing... The os.listdir version crashes at least.

and also test if it's a file before opening (in both cases) because attempting to open a directory results in an I/O exception:

if os.path.isfile(x):

  with open ...

In a nutshell os.path package is your friend when manipulating files.

edited Nov 14 '18 at 9:46

answered Nov 14 '18 at 9:18

Jean-François Fabre

104k955112

edited Nov 14 '18 at 9:46

answered Nov 14 '18 at 9:18

Jean-François Fabre

104k955112

answered Nov 14 '18 at 9:18

Jean-François Fabre

104k955112

answered Nov 14 '18 at 9:18

Jean-François Fabre

104k955112

This solution solved the problem.

– Dipankar Nalui
Nov 14 '18 at 10:54

add a comment |

This solution solved the problem.

– Dipankar Nalui
Nov 14 '18 at 10:54

This solution solved the problem.

– Dipankar Nalui
Nov 14 '18 at 10:54

add a comment |

-1

Since the file abcd.txt is present in C:Usersuser1scriptssample1nvram2logs, and the said path is not your working directory, you have to add the same to sys.path

import os, sys

path= 'sample1/nvram2/logs'

sys.path.append(path)





all_files=os.listdir(path) 

for my_file1 in all_files:

    print(my_file1)

    with open(my_file1, 'r') as my_file2:

        print(my_file2)

        for line in my_file2:

            if 'string' in line:

                print(my_file2)

answered Nov 14 '18 at 9:23

Pradip Gupta

6910

no, sys.path is for module loading.

– Jean-François Fabre
Nov 14 '18 at 9:25

sys.path is not only for module loading. Even when you have to import files without giving their complete path, you can append their path using sys.path and then call it directly. It only makes your files searchable by the current path.

– Pradip Gupta
Nov 14 '18 at 9:54

open doesn't use sys.path at all

– Jean-François Fabre
Nov 14 '18 at 10:17

I agree with that. However, in general, python would search for the file in all the paths found in sys.path. For example, when you have the file in your cwd, you do not give the absolute path with open as the file is already present in the path for you. Same is the logic with sys.path

– Pradip Gupta
Nov 14 '18 at 10:20

modules aren't data files.

– Jean-François Fabre
Nov 14 '18 at 10:21

|
show 1 more comment

-1

Since the file abcd.txt is present in C:Usersuser1scriptssample1nvram2logs, and the said path is not your working directory, you have to add the same to sys.path

import os, sys

path= 'sample1/nvram2/logs'

sys.path.append(path)





all_files=os.listdir(path) 

for my_file1 in all_files:

    print(my_file1)

    with open(my_file1, 'r') as my_file2:

        print(my_file2)

        for line in my_file2:

            if 'string' in line:

                print(my_file2)

answered Nov 14 '18 at 9:23

Pradip Gupta

6910

no, sys.path is for module loading.

– Jean-François Fabre
Nov 14 '18 at 9:25

sys.path is not only for module loading. Even when you have to import files without giving their complete path, you can append their path using sys.path and then call it directly. It only makes your files searchable by the current path.

– Pradip Gupta
Nov 14 '18 at 9:54

open doesn't use sys.path at all

– Jean-François Fabre
Nov 14 '18 at 10:17

I agree with that. However, in general, python would search for the file in all the paths found in sys.path. For example, when you have the file in your cwd, you do not give the absolute path with open as the file is already present in the path for you. Same is the logic with sys.path

– Pradip Gupta
Nov 14 '18 at 10:20

modules aren't data files.

– Jean-François Fabre
Nov 14 '18 at 10:21

|
show 1 more comment

-1

Since the file abcd.txt is present in C:Usersuser1scriptssample1nvram2logs, and the said path is not your working directory, you have to add the same to sys.path

import os, sys

path= 'sample1/nvram2/logs'

sys.path.append(path)





all_files=os.listdir(path) 

for my_file1 in all_files:

    print(my_file1)

    with open(my_file1, 'r') as my_file2:

        print(my_file2)

        for line in my_file2:

            if 'string' in line:

                print(my_file2)

answered Nov 14 '18 at 9:23

Pradip Gupta

6910

Since the file abcd.txt is present in C:Usersuser1scriptssample1nvram2logs, and the said path is not your working directory, you have to add the same to sys.path

import os, sys

path= 'sample1/nvram2/logs'

sys.path.append(path)





all_files=os.listdir(path) 

for my_file1 in all_files:

    print(my_file1)

    with open(my_file1, 'r') as my_file2:

        print(my_file2)

        for line in my_file2:

            if 'string' in line:

                print(my_file2)

answered Nov 14 '18 at 9:23

Pradip Gupta

6910

answered Nov 14 '18 at 9:23

Pradip Gupta

6910

answered Nov 14 '18 at 9:23

Pradip Gupta

6910

answered Nov 14 '18 at 9:23

Pradip Gupta

6910

no, sys.path is for module loading.

– Jean-François Fabre
Nov 14 '18 at 9:25

sys.path is not only for module loading. Even when you have to import files without giving their complete path, you can append their path using sys.path and then call it directly. It only makes your files searchable by the current path.

– Pradip Gupta
Nov 14 '18 at 9:54

open doesn't use sys.path at all

– Jean-François Fabre
Nov 14 '18 at 10:17

I agree with that. However, in general, python would search for the file in all the paths found in sys.path. For example, when you have the file in your cwd, you do not give the absolute path with open as the file is already present in the path for you. Same is the logic with sys.path

– Pradip Gupta
Nov 14 '18 at 10:20

modules aren't data files.

– Jean-François Fabre
Nov 14 '18 at 10:21

|
show 1 more comment

no, sys.path is for module loading.

– Jean-François Fabre
Nov 14 '18 at 9:25

sys.path is not only for module loading. Even when you have to import files without giving their complete path, you can append their path using sys.path and then call it directly. It only makes your files searchable by the current path.

– Pradip Gupta
Nov 14 '18 at 9:54

open doesn't use sys.path at all

– Jean-François Fabre
Nov 14 '18 at 10:17

I agree with that. However, in general, python would search for the file in all the paths found in sys.path. For example, when you have the file in your cwd, you do not give the absolute path with open as the file is already present in the path for you. Same is the logic with sys.path

– Pradip Gupta
Nov 14 '18 at 10:20

modules aren't data files.

– Jean-François Fabre
Nov 14 '18 at 10:21

no, sys.path is for module loading.

– Jean-François Fabre
Nov 14 '18 at 9:25

sys.path is not only for module loading. Even when you have to import files without giving their complete path, you can append their path using sys.path and then call it directly. It only makes your files searchable by the current path.

– Pradip Gupta
Nov 14 '18 at 9:54

open doesn't use sys.path at all

– Jean-François Fabre
Nov 14 '18 at 10:17

I agree with that. However, in general, python would search for the file in all the paths found in sys.path. For example, when you have the file in your cwd, you do not give the absolute path with open as the file is already present in the path for you. Same is the logic with sys.path

– Pradip Gupta
Nov 14 '18 at 10:20

modules aren't data files.

– Jean-François Fabre
Nov 14 '18 at 10:21

|
show 1 more comment

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