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Check for file permission

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# 1  
Old 02-25-2014
Check for file permission
Hi,

I have an array that has several directory paths and file paths.

I use a for loop to traverse through the array and check for files and not directories using the if condition.

I wish to display all files [not directories] that have permissions lesser than what the user has specified.

For example:

Code:
-rwxrwx---   2 user1      sie   8273920 Feb 24 15:13 /opt/app/hello.txt
-rwxr-x---   2 user1      sie   8273920 Feb 24 15:13 /opt/app/yellow.txt
-rwxrwxr-x   2 user1      sie   8273920 Feb 24 15:13 /opt/app/fellow.txt

So, if the user gives 770 which is read in a variable $1 then.

Expected result should be:

Code:
-rwxrwx---   2 user1      sie   8273920 Feb 24 15:13 /opt/app/hello.txt
-rwxrwxr-x   2 user1      sie   8273920 Feb 24 15:13 /opt/app/fellow.txt

Because the second entry is with permission 750 which is lesser permission the 770.
# 2  
Old 02-25-2014
find with perm will do.
Below code will print all the files with permissions >= 750

Code:
find . -type f -perm -750

Below is the test case
Code:
-bash-4.1$ ls -lrt
total 12
-rwxr-x--- 1 username groupnm  55 Feb 25 05:43 table.dat
-rw-r--r-- 1 username groupnm 109 Feb 25 05:43 sample.dat
-rwxrwx--- 1 username groupnm 113 Feb 25 05:43 manipu.dat
-bash-4.1$ find . -type f -perm -750 | xargs ls -lrt
-rwxr-x--- 1 username groupnm  55 Feb 25 05:43 ./table.dat
-rwxrwx--- 1 username groupnm 113 Feb 25 05:43 ./manipu.dat
-bash-4.1$ find . -type f -perm -770 | xargs ls -lrt
-rwxrwx--- 1 username groupnm 113 Feb 25 05:43 ./manipu.dat
-bash-4.1$ find . -type f -perm -600 | xargs ls -lrt
-rwxr-x--- 1 username groupnm  55 Feb 25 05:43 ./table.dat
-rw-r--r-- 1 username groupnm 109 Feb 25 05:43 ./sample.dat
-rwxrwx--- 1 username groupnm 113 Feb 25 05:43 ./manipu.dat
-bash-4.1$

# 3  
Old 02-25-2014
What have you tried so far?
# 4  
Old 02-25-2014
Quote:
Originally Posted by chacko193
What have you tried so far?
I have the exact file name in an Array, so how could i use find ?

Code:
echo $arr[3]

-rwxrwx--- 2 user1 sie 8273920 Feb 24 15:13 /opt/app/hello.txt
Please let me know how can i use find with the array, if i can't, then what is the alternative ?
# 5  
Old 02-25-2014
If the array has 10 elements, i.e., $arr[0] to $arr[9], below is the code

Code:
for i in $(seq 0 9);
do
find $(dirname $(echo ${arr[${i}]} | awk '{print $9}')) -type f -name $(basename $(echo ${arr[${i}]} | awk '{print $9}')) -perm -750
done

This User Gave Thanks to SriniShoo For This Post:
mohtashims
# 6  
Old 02-25-2014
As per the requirement that you specified in your first post you can try this:
Code:
find /path/to/your/directory -type f -perm -$yourPerm -exec ls -l {} \;

# 7  
Old 02-25-2014
Quote:
Originally Posted by mohtashims
I have an array that has several directory paths and file paths.
I use a for loop to traverse through the array
You shouldn't do that. for-loops can break if you have too many elements. Use a while-loop instead:

Code:
(( counter = 0 ))
while [ $counter -lt ${#array[*]} ] ; do
     command "${array[$counter]}"
     (( counter += 1 ))
done

Quote:
I wish to display all files [not directories] that have permissions lesser than what the user has specified.
What do you mean by "lesser"? If i specify "660" and the file turns out to be "750", is that lesser, more or something entirely different? How about "755" or "555"?

Please clarify.

bakunin
This User Gave Thanks to bakunin For This Post:
mohtashims
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