Permissions on Files:
up vote
7
down vote
favorite
I would like to give 755 permissions to a directory, so I use:
# chmod -R 755 /my/folder/
It works for all files inside my folder, but the problem is that I use scripts that create new files in this folder, and by default the permissions are 600.
How could I impose, 755 permissions on these 'future' files ?
--EDIT--
I use a script that gives me info about network traffic on my campus. I have a new file every 10 minutes, located in a folder called journey, located in the month folder like this :
ls -lrt /home/netmet/secure/2017-04/2017-04-27/
total 118548
-rwxr-sr-x 1 root root 85922 avril 27 00:10 zzaccounting.dmp-00-00
-rwxr-sr-x 1 root root 54874 avril 27 00:20 zzaccounting.dmp-00-10
-rwxr-sr-x 1 root root 33534 avril 27 00:30 zzaccounting.dmp-00-20
-rwxr-sr-x 1 root root 48890 avril 27 00:40 zzaccounting.dmp-00-30
-rwxr-sr-x 1 root root 36878 avril 27 00:50 zzaccounting.dmp-00-40
-rwxr-sr-x 1 root root 37034 avril 27 01:00 zzaccounting.dmp-00-50
-rwxr-sr-x 1 root root 38154 avril 27 01:10 zzaccounting.dmp-01-00
-rwxr-sr-x 1 root root 38318 avril 27 01:20 zzaccounting.dmp-01-10
-rwxr-sr-x 1 root root 26978 avril 27 01:30 zzaccounting.dmp-01-20
-rwxr-sr-x 1 root root 31558 avril 27 01:40 zzaccounting.dmp-01-30
-rwxr-sr-x 1 root root 23662 avril 27 01:50 zzaccounting.dmp-01-40
-rwxr-sr-x 1 root root 32298 avril 27 02:00 zzaccounting.dmp-01-50
-rwxr-sr-x 1 root root 30282 avril 27 02:10 zzaccounting.dmp-02-00
-rwxr-sr-x 1 root root 31110 avril 27 02:20 zzaccounting.dmp-02-10
-rwxr-sr-x 1 root root 25718 avril 27 02:30 zzaccounting.dmp-02-20
-rwxr-sr-x 1 root root 26306 avril 27 02:40 zzaccounting.dmp-02-30
-rwxr-sr-x 1 root root 23690 avril 27 02:50 zzaccounting.dmp-02-40
-rwxr-sr-x 1 root root 23002 avril 27 03:00 zzaccounting.dmp-02-50
-rwxr-sr-x 1 root root 21854 avril 27 03:10 zzaccounting.dmp-03-00
Here I changed permissions by hand but when new the file appears, I have this :
-rw------- 1 root root 3479106 avril 27 15:50 zzaccounting.dmp-15-40
The thing is that for every folder and file under /home/netmet/secure/ I want 755 permision by default.
I have already done this :
chmod -R g+s /home/netmet/secure
setfacl -d -m g::rwx /home/netmet/secure
setfacl -d -m o::rx /home/netmet/secure
debian permissions chmod umask
|
show 6 more comments
up vote
7
down vote
favorite
I would like to give 755 permissions to a directory, so I use:
# chmod -R 755 /my/folder/
It works for all files inside my folder, but the problem is that I use scripts that create new files in this folder, and by default the permissions are 600.
How could I impose, 755 permissions on these 'future' files ?
--EDIT--
I use a script that gives me info about network traffic on my campus. I have a new file every 10 minutes, located in a folder called journey, located in the month folder like this :
ls -lrt /home/netmet/secure/2017-04/2017-04-27/
total 118548
-rwxr-sr-x 1 root root 85922 avril 27 00:10 zzaccounting.dmp-00-00
-rwxr-sr-x 1 root root 54874 avril 27 00:20 zzaccounting.dmp-00-10
-rwxr-sr-x 1 root root 33534 avril 27 00:30 zzaccounting.dmp-00-20
-rwxr-sr-x 1 root root 48890 avril 27 00:40 zzaccounting.dmp-00-30
-rwxr-sr-x 1 root root 36878 avril 27 00:50 zzaccounting.dmp-00-40
-rwxr-sr-x 1 root root 37034 avril 27 01:00 zzaccounting.dmp-00-50
-rwxr-sr-x 1 root root 38154 avril 27 01:10 zzaccounting.dmp-01-00
-rwxr-sr-x 1 root root 38318 avril 27 01:20 zzaccounting.dmp-01-10
-rwxr-sr-x 1 root root 26978 avril 27 01:30 zzaccounting.dmp-01-20
-rwxr-sr-x 1 root root 31558 avril 27 01:40 zzaccounting.dmp-01-30
-rwxr-sr-x 1 root root 23662 avril 27 01:50 zzaccounting.dmp-01-40
-rwxr-sr-x 1 root root 32298 avril 27 02:00 zzaccounting.dmp-01-50
-rwxr-sr-x 1 root root 30282 avril 27 02:10 zzaccounting.dmp-02-00
-rwxr-sr-x 1 root root 31110 avril 27 02:20 zzaccounting.dmp-02-10
-rwxr-sr-x 1 root root 25718 avril 27 02:30 zzaccounting.dmp-02-20
-rwxr-sr-x 1 root root 26306 avril 27 02:40 zzaccounting.dmp-02-30
-rwxr-sr-x 1 root root 23690 avril 27 02:50 zzaccounting.dmp-02-40
-rwxr-sr-x 1 root root 23002 avril 27 03:00 zzaccounting.dmp-02-50
-rwxr-sr-x 1 root root 21854 avril 27 03:10 zzaccounting.dmp-03-00
Here I changed permissions by hand but when new the file appears, I have this :
-rw------- 1 root root 3479106 avril 27 15:50 zzaccounting.dmp-15-40
The thing is that for every folder and file under /home/netmet/secure/ I want 755 permision by default.
I have already done this :
chmod -R g+s /home/netmet/secure
setfacl -d -m g::rwx /home/netmet/secure
setfacl -d -m o::rx /home/netmet/secure
debian permissions chmod umask
1
ACL is the answer. This has been discussed before.
– d3ag0s
Apr 27 '17 at 11:43
7
ACL is the answer for overriding theumask
, but not for overriding a program that creates files with mode0600
by design. See here for some references.
– ilkkachu
Apr 27 '17 at 12:06
1
And one more detail: by default you can't useumask
for adding execution bit to files, this need to be done by hand
– Romeo Ninov
Apr 27 '17 at 13:01
umask doesn't work guys
– klaypez
Apr 28 '17 at 9:40
Can you edit the script you use to make the files? How do the files get in to the final directory (eg.2017-04-27
) and how are the directories made in the first place? Please update the question with these details.
– Tigger
Apr 28 '17 at 10:46
|
show 6 more comments
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I would like to give 755 permissions to a directory, so I use:
# chmod -R 755 /my/folder/
It works for all files inside my folder, but the problem is that I use scripts that create new files in this folder, and by default the permissions are 600.
How could I impose, 755 permissions on these 'future' files ?
--EDIT--
I use a script that gives me info about network traffic on my campus. I have a new file every 10 minutes, located in a folder called journey, located in the month folder like this :
ls -lrt /home/netmet/secure/2017-04/2017-04-27/
total 118548
-rwxr-sr-x 1 root root 85922 avril 27 00:10 zzaccounting.dmp-00-00
-rwxr-sr-x 1 root root 54874 avril 27 00:20 zzaccounting.dmp-00-10
-rwxr-sr-x 1 root root 33534 avril 27 00:30 zzaccounting.dmp-00-20
-rwxr-sr-x 1 root root 48890 avril 27 00:40 zzaccounting.dmp-00-30
-rwxr-sr-x 1 root root 36878 avril 27 00:50 zzaccounting.dmp-00-40
-rwxr-sr-x 1 root root 37034 avril 27 01:00 zzaccounting.dmp-00-50
-rwxr-sr-x 1 root root 38154 avril 27 01:10 zzaccounting.dmp-01-00
-rwxr-sr-x 1 root root 38318 avril 27 01:20 zzaccounting.dmp-01-10
-rwxr-sr-x 1 root root 26978 avril 27 01:30 zzaccounting.dmp-01-20
-rwxr-sr-x 1 root root 31558 avril 27 01:40 zzaccounting.dmp-01-30
-rwxr-sr-x 1 root root 23662 avril 27 01:50 zzaccounting.dmp-01-40
-rwxr-sr-x 1 root root 32298 avril 27 02:00 zzaccounting.dmp-01-50
-rwxr-sr-x 1 root root 30282 avril 27 02:10 zzaccounting.dmp-02-00
-rwxr-sr-x 1 root root 31110 avril 27 02:20 zzaccounting.dmp-02-10
-rwxr-sr-x 1 root root 25718 avril 27 02:30 zzaccounting.dmp-02-20
-rwxr-sr-x 1 root root 26306 avril 27 02:40 zzaccounting.dmp-02-30
-rwxr-sr-x 1 root root 23690 avril 27 02:50 zzaccounting.dmp-02-40
-rwxr-sr-x 1 root root 23002 avril 27 03:00 zzaccounting.dmp-02-50
-rwxr-sr-x 1 root root 21854 avril 27 03:10 zzaccounting.dmp-03-00
Here I changed permissions by hand but when new the file appears, I have this :
-rw------- 1 root root 3479106 avril 27 15:50 zzaccounting.dmp-15-40
The thing is that for every folder and file under /home/netmet/secure/ I want 755 permision by default.
I have already done this :
chmod -R g+s /home/netmet/secure
setfacl -d -m g::rwx /home/netmet/secure
setfacl -d -m o::rx /home/netmet/secure
debian permissions chmod umask
I would like to give 755 permissions to a directory, so I use:
# chmod -R 755 /my/folder/
It works for all files inside my folder, but the problem is that I use scripts that create new files in this folder, and by default the permissions are 600.
How could I impose, 755 permissions on these 'future' files ?
--EDIT--
I use a script that gives me info about network traffic on my campus. I have a new file every 10 minutes, located in a folder called journey, located in the month folder like this :
ls -lrt /home/netmet/secure/2017-04/2017-04-27/
total 118548
-rwxr-sr-x 1 root root 85922 avril 27 00:10 zzaccounting.dmp-00-00
-rwxr-sr-x 1 root root 54874 avril 27 00:20 zzaccounting.dmp-00-10
-rwxr-sr-x 1 root root 33534 avril 27 00:30 zzaccounting.dmp-00-20
-rwxr-sr-x 1 root root 48890 avril 27 00:40 zzaccounting.dmp-00-30
-rwxr-sr-x 1 root root 36878 avril 27 00:50 zzaccounting.dmp-00-40
-rwxr-sr-x 1 root root 37034 avril 27 01:00 zzaccounting.dmp-00-50
-rwxr-sr-x 1 root root 38154 avril 27 01:10 zzaccounting.dmp-01-00
-rwxr-sr-x 1 root root 38318 avril 27 01:20 zzaccounting.dmp-01-10
-rwxr-sr-x 1 root root 26978 avril 27 01:30 zzaccounting.dmp-01-20
-rwxr-sr-x 1 root root 31558 avril 27 01:40 zzaccounting.dmp-01-30
-rwxr-sr-x 1 root root 23662 avril 27 01:50 zzaccounting.dmp-01-40
-rwxr-sr-x 1 root root 32298 avril 27 02:00 zzaccounting.dmp-01-50
-rwxr-sr-x 1 root root 30282 avril 27 02:10 zzaccounting.dmp-02-00
-rwxr-sr-x 1 root root 31110 avril 27 02:20 zzaccounting.dmp-02-10
-rwxr-sr-x 1 root root 25718 avril 27 02:30 zzaccounting.dmp-02-20
-rwxr-sr-x 1 root root 26306 avril 27 02:40 zzaccounting.dmp-02-30
-rwxr-sr-x 1 root root 23690 avril 27 02:50 zzaccounting.dmp-02-40
-rwxr-sr-x 1 root root 23002 avril 27 03:00 zzaccounting.dmp-02-50
-rwxr-sr-x 1 root root 21854 avril 27 03:10 zzaccounting.dmp-03-00
Here I changed permissions by hand but when new the file appears, I have this :
-rw------- 1 root root 3479106 avril 27 15:50 zzaccounting.dmp-15-40
The thing is that for every folder and file under /home/netmet/secure/ I want 755 permision by default.
I have already done this :
chmod -R g+s /home/netmet/secure
setfacl -d -m g::rwx /home/netmet/secure
setfacl -d -m o::rx /home/netmet/secure
debian permissions chmod umask
debian permissions chmod umask
edited Nov 26 at 15:12
Michael Prokopec
72116
72116
asked Apr 27 '17 at 11:41
klaypez
6819
6819
1
ACL is the answer. This has been discussed before.
– d3ag0s
Apr 27 '17 at 11:43
7
ACL is the answer for overriding theumask
, but not for overriding a program that creates files with mode0600
by design. See here for some references.
– ilkkachu
Apr 27 '17 at 12:06
1
And one more detail: by default you can't useumask
for adding execution bit to files, this need to be done by hand
– Romeo Ninov
Apr 27 '17 at 13:01
umask doesn't work guys
– klaypez
Apr 28 '17 at 9:40
Can you edit the script you use to make the files? How do the files get in to the final directory (eg.2017-04-27
) and how are the directories made in the first place? Please update the question with these details.
– Tigger
Apr 28 '17 at 10:46
|
show 6 more comments
1
ACL is the answer. This has been discussed before.
– d3ag0s
Apr 27 '17 at 11:43
7
ACL is the answer for overriding theumask
, but not for overriding a program that creates files with mode0600
by design. See here for some references.
– ilkkachu
Apr 27 '17 at 12:06
1
And one more detail: by default you can't useumask
for adding execution bit to files, this need to be done by hand
– Romeo Ninov
Apr 27 '17 at 13:01
umask doesn't work guys
– klaypez
Apr 28 '17 at 9:40
Can you edit the script you use to make the files? How do the files get in to the final directory (eg.2017-04-27
) and how are the directories made in the first place? Please update the question with these details.
– Tigger
Apr 28 '17 at 10:46
1
1
ACL is the answer. This has been discussed before.
– d3ag0s
Apr 27 '17 at 11:43
ACL is the answer. This has been discussed before.
– d3ag0s
Apr 27 '17 at 11:43
7
7
ACL is the answer for overriding the
umask
, but not for overriding a program that creates files with mode 0600
by design. See here for some references.– ilkkachu
Apr 27 '17 at 12:06
ACL is the answer for overriding the
umask
, but not for overriding a program that creates files with mode 0600
by design. See here for some references.– ilkkachu
Apr 27 '17 at 12:06
1
1
And one more detail: by default you can't use
umask
for adding execution bit to files, this need to be done by hand– Romeo Ninov
Apr 27 '17 at 13:01
And one more detail: by default you can't use
umask
for adding execution bit to files, this need to be done by hand– Romeo Ninov
Apr 27 '17 at 13:01
umask doesn't work guys
– klaypez
Apr 28 '17 at 9:40
umask doesn't work guys
– klaypez
Apr 28 '17 at 9:40
Can you edit the script you use to make the files? How do the files get in to the final directory (eg.
2017-04-27
) and how are the directories made in the first place? Please update the question with these details.– Tigger
Apr 28 '17 at 10:46
Can you edit the script you use to make the files? How do the files get in to the final directory (eg.
2017-04-27
) and how are the directories made in the first place? Please update the question with these details.– Tigger
Apr 28 '17 at 10:46
|
show 6 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
Try to run umask
in your folder. If it returns anything other than '0022' then this is your problem. In your case it should initialy output '0177'. The permission system when creating directory is basically computed:
default - umask
0777 is the default mode for directories, and 0666 to ordinary files, but there are different umasks, if I understand these things right. Try to execute umask a=rx,uu+w
.
EDIT: You can use umask to give execute bit to directory to be able to cd into it, but not to files. These have to be given execute bit manually because of security. Simply add chmod +x <file>
to your script. And, execute flag set on file anything other than executable has no effect.
1
Theumask
shell command does not depend on the path, it only depends on the user running it, each user has a specific umask value, defined at login and that can be changed later. This defines how new file permissions are created.
– Patrick Mevzek
Nov 25 '17 at 22:21
That should bechmod a=rx,u+w
in your reply.
– Patrick Mevzek
Nov 25 '17 at 22:22
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Try to run umask
in your folder. If it returns anything other than '0022' then this is your problem. In your case it should initialy output '0177'. The permission system when creating directory is basically computed:
default - umask
0777 is the default mode for directories, and 0666 to ordinary files, but there are different umasks, if I understand these things right. Try to execute umask a=rx,uu+w
.
EDIT: You can use umask to give execute bit to directory to be able to cd into it, but not to files. These have to be given execute bit manually because of security. Simply add chmod +x <file>
to your script. And, execute flag set on file anything other than executable has no effect.
1
Theumask
shell command does not depend on the path, it only depends on the user running it, each user has a specific umask value, defined at login and that can be changed later. This defines how new file permissions are created.
– Patrick Mevzek
Nov 25 '17 at 22:21
That should bechmod a=rx,u+w
in your reply.
– Patrick Mevzek
Nov 25 '17 at 22:22
add a comment |
up vote
0
down vote
Try to run umask
in your folder. If it returns anything other than '0022' then this is your problem. In your case it should initialy output '0177'. The permission system when creating directory is basically computed:
default - umask
0777 is the default mode for directories, and 0666 to ordinary files, but there are different umasks, if I understand these things right. Try to execute umask a=rx,uu+w
.
EDIT: You can use umask to give execute bit to directory to be able to cd into it, but not to files. These have to be given execute bit manually because of security. Simply add chmod +x <file>
to your script. And, execute flag set on file anything other than executable has no effect.
1
Theumask
shell command does not depend on the path, it only depends on the user running it, each user has a specific umask value, defined at login and that can be changed later. This defines how new file permissions are created.
– Patrick Mevzek
Nov 25 '17 at 22:21
That should bechmod a=rx,u+w
in your reply.
– Patrick Mevzek
Nov 25 '17 at 22:22
add a comment |
up vote
0
down vote
up vote
0
down vote
Try to run umask
in your folder. If it returns anything other than '0022' then this is your problem. In your case it should initialy output '0177'. The permission system when creating directory is basically computed:
default - umask
0777 is the default mode for directories, and 0666 to ordinary files, but there are different umasks, if I understand these things right. Try to execute umask a=rx,uu+w
.
EDIT: You can use umask to give execute bit to directory to be able to cd into it, but not to files. These have to be given execute bit manually because of security. Simply add chmod +x <file>
to your script. And, execute flag set on file anything other than executable has no effect.
Try to run umask
in your folder. If it returns anything other than '0022' then this is your problem. In your case it should initialy output '0177'. The permission system when creating directory is basically computed:
default - umask
0777 is the default mode for directories, and 0666 to ordinary files, but there are different umasks, if I understand these things right. Try to execute umask a=rx,uu+w
.
EDIT: You can use umask to give execute bit to directory to be able to cd into it, but not to files. These have to be given execute bit manually because of security. Simply add chmod +x <file>
to your script. And, execute flag set on file anything other than executable has no effect.
edited Jul 15 '17 at 13:47
answered Jul 15 '17 at 12:44
M. H.
12029
12029
1
Theumask
shell command does not depend on the path, it only depends on the user running it, each user has a specific umask value, defined at login and that can be changed later. This defines how new file permissions are created.
– Patrick Mevzek
Nov 25 '17 at 22:21
That should bechmod a=rx,u+w
in your reply.
– Patrick Mevzek
Nov 25 '17 at 22:22
add a comment |
1
Theumask
shell command does not depend on the path, it only depends on the user running it, each user has a specific umask value, defined at login and that can be changed later. This defines how new file permissions are created.
– Patrick Mevzek
Nov 25 '17 at 22:21
That should bechmod a=rx,u+w
in your reply.
– Patrick Mevzek
Nov 25 '17 at 22:22
1
1
The
umask
shell command does not depend on the path, it only depends on the user running it, each user has a specific umask value, defined at login and that can be changed later. This defines how new file permissions are created.– Patrick Mevzek
Nov 25 '17 at 22:21
The
umask
shell command does not depend on the path, it only depends on the user running it, each user has a specific umask value, defined at login and that can be changed later. This defines how new file permissions are created.– Patrick Mevzek
Nov 25 '17 at 22:21
That should be
chmod a=rx,u+w
in your reply.– Patrick Mevzek
Nov 25 '17 at 22:22
That should be
chmod a=rx,u+w
in your reply.– Patrick Mevzek
Nov 25 '17 at 22:22
add a comment |
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1
ACL is the answer. This has been discussed before.
– d3ag0s
Apr 27 '17 at 11:43
7
ACL is the answer for overriding the
umask
, but not for overriding a program that creates files with mode0600
by design. See here for some references.– ilkkachu
Apr 27 '17 at 12:06
1
And one more detail: by default you can't use
umask
for adding execution bit to files, this need to be done by hand– Romeo Ninov
Apr 27 '17 at 13:01
umask doesn't work guys
– klaypez
Apr 28 '17 at 9:40
Can you edit the script you use to make the files? How do the files get in to the final directory (eg.
2017-04-27
) and how are the directories made in the first place? Please update the question with these details.– Tigger
Apr 28 '17 at 10:46