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Using wildcards for -group or -user with find command


Find a file where name starts with a capital letterFind files that end with numberPrint file name extension using -exec in findBatch Renaming ScriptFind files with group permissions more restrictive than owner permissionsCommand find and tar in scriptRegex not matching file using `find` despite being validFind and Replace with Sed, not a regular fileFind command - file path with white spaceBash array with folder paths and wildcards






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








1















I'm trying to find owners of files on the system that all start with a specific name, for example



[unu@here findtest]$ find . -ls
17295583 0 drwxrwxr-x 2 unu unu 74 Mar 28 03:14 .
17295584 0 -rw-rw-r-- 1 br_asd1 br_asd1 0 Mar 28 03:13 ./test1
17295585 0 -rw-rw-r-- 1 br_asd2 br_asd2 0 Mar 28 03:13 ./test2
17295586 0 -rw-rw-r-- 1 unu unu 0 Mar 28 03:13 ./test33
17295587 0 -rw-rw-r-- 1 br_bfg1 br_bfg1 0 Mar 28 03:14 ./test11
17295588 0 -rw-rw-r-- 1 unu unu 0 Mar 28 03:14 ./test22


I need find to only get files owned by users whose name starts with "br_".



So what I tried was



[unu@here findtest]$ find . -user "br_*" -ls
find: ‘br_*’ is not the name of a known user


Now a method I found that works is using awk, but I have certain problems with this method and it's not really usable for what I'm trying:



[unu@here findtest]$ find . -ls|awk 'if ($5 ~ "br_"'
17295584 0 -rw-rw-r-- 1 br_asd1 br_asd1 0 Mar 28 03:13 ./test1
17295585 0 -rw-rw-r-- 1 br_asd2 br_asd2 0 Mar 28 03:13 ./test2
17295587 0 -rw-rw-r-- 1 br_bfg1 br_bfg1 0 Mar 28 03:14 ./test11


Is there a way to add wildcards to the -user or -group options of find?










share|improve this question
















migrated from stackoverflow.com Mar 28 at 3:29


This question came from our site for professional and enthusiast programmers.



















  • With getent passwd get a list of users which match br_.* and build a list like: --user br_asd1 --user br_asd2 --user br_bfg1 for find command.

    – Cyrus
    Mar 28 at 3:33












  • GNU find does not have options to pass wildcards for -user and -group fields. One way would be to use GNU coreutils stat and use the %U quantifier to get the owner name. Do something like for file in *; do [ -f "$file" ] || continue; owner=$(stat -c '%U' "$file"); if [[ "$owner" =~ ^br.* ]]; then echo "$file"; fi; done

    – Inian
    Mar 28 at 3:52

















1















I'm trying to find owners of files on the system that all start with a specific name, for example



[unu@here findtest]$ find . -ls
17295583 0 drwxrwxr-x 2 unu unu 74 Mar 28 03:14 .
17295584 0 -rw-rw-r-- 1 br_asd1 br_asd1 0 Mar 28 03:13 ./test1
17295585 0 -rw-rw-r-- 1 br_asd2 br_asd2 0 Mar 28 03:13 ./test2
17295586 0 -rw-rw-r-- 1 unu unu 0 Mar 28 03:13 ./test33
17295587 0 -rw-rw-r-- 1 br_bfg1 br_bfg1 0 Mar 28 03:14 ./test11
17295588 0 -rw-rw-r-- 1 unu unu 0 Mar 28 03:14 ./test22


I need find to only get files owned by users whose name starts with "br_".



So what I tried was



[unu@here findtest]$ find . -user "br_*" -ls
find: ‘br_*’ is not the name of a known user


Now a method I found that works is using awk, but I have certain problems with this method and it's not really usable for what I'm trying:



[unu@here findtest]$ find . -ls|awk 'if ($5 ~ "br_"'
17295584 0 -rw-rw-r-- 1 br_asd1 br_asd1 0 Mar 28 03:13 ./test1
17295585 0 -rw-rw-r-- 1 br_asd2 br_asd2 0 Mar 28 03:13 ./test2
17295587 0 -rw-rw-r-- 1 br_bfg1 br_bfg1 0 Mar 28 03:14 ./test11


Is there a way to add wildcards to the -user or -group options of find?










share|improve this question
















migrated from stackoverflow.com Mar 28 at 3:29


This question came from our site for professional and enthusiast programmers.



















  • With getent passwd get a list of users which match br_.* and build a list like: --user br_asd1 --user br_asd2 --user br_bfg1 for find command.

    – Cyrus
    Mar 28 at 3:33












  • GNU find does not have options to pass wildcards for -user and -group fields. One way would be to use GNU coreutils stat and use the %U quantifier to get the owner name. Do something like for file in *; do [ -f "$file" ] || continue; owner=$(stat -c '%U' "$file"); if [[ "$owner" =~ ^br.* ]]; then echo "$file"; fi; done

    – Inian
    Mar 28 at 3:52













1












1








1








I'm trying to find owners of files on the system that all start with a specific name, for example



[unu@here findtest]$ find . -ls
17295583 0 drwxrwxr-x 2 unu unu 74 Mar 28 03:14 .
17295584 0 -rw-rw-r-- 1 br_asd1 br_asd1 0 Mar 28 03:13 ./test1
17295585 0 -rw-rw-r-- 1 br_asd2 br_asd2 0 Mar 28 03:13 ./test2
17295586 0 -rw-rw-r-- 1 unu unu 0 Mar 28 03:13 ./test33
17295587 0 -rw-rw-r-- 1 br_bfg1 br_bfg1 0 Mar 28 03:14 ./test11
17295588 0 -rw-rw-r-- 1 unu unu 0 Mar 28 03:14 ./test22


I need find to only get files owned by users whose name starts with "br_".



So what I tried was



[unu@here findtest]$ find . -user "br_*" -ls
find: ‘br_*’ is not the name of a known user


Now a method I found that works is using awk, but I have certain problems with this method and it's not really usable for what I'm trying:



[unu@here findtest]$ find . -ls|awk 'if ($5 ~ "br_"'
17295584 0 -rw-rw-r-- 1 br_asd1 br_asd1 0 Mar 28 03:13 ./test1
17295585 0 -rw-rw-r-- 1 br_asd2 br_asd2 0 Mar 28 03:13 ./test2
17295587 0 -rw-rw-r-- 1 br_bfg1 br_bfg1 0 Mar 28 03:14 ./test11


Is there a way to add wildcards to the -user or -group options of find?










share|improve this question
















I'm trying to find owners of files on the system that all start with a specific name, for example



[unu@here findtest]$ find . -ls
17295583 0 drwxrwxr-x 2 unu unu 74 Mar 28 03:14 .
17295584 0 -rw-rw-r-- 1 br_asd1 br_asd1 0 Mar 28 03:13 ./test1
17295585 0 -rw-rw-r-- 1 br_asd2 br_asd2 0 Mar 28 03:13 ./test2
17295586 0 -rw-rw-r-- 1 unu unu 0 Mar 28 03:13 ./test33
17295587 0 -rw-rw-r-- 1 br_bfg1 br_bfg1 0 Mar 28 03:14 ./test11
17295588 0 -rw-rw-r-- 1 unu unu 0 Mar 28 03:14 ./test22


I need find to only get files owned by users whose name starts with "br_".



So what I tried was



[unu@here findtest]$ find . -user "br_*" -ls
find: ‘br_*’ is not the name of a known user


Now a method I found that works is using awk, but I have certain problems with this method and it's not really usable for what I'm trying:



[unu@here findtest]$ find . -ls|awk 'if ($5 ~ "br_"'
17295584 0 -rw-rw-r-- 1 br_asd1 br_asd1 0 Mar 28 03:13 ./test1
17295585 0 -rw-rw-r-- 1 br_asd2 br_asd2 0 Mar 28 03:13 ./test2
17295587 0 -rw-rw-r-- 1 br_bfg1 br_bfg1 0 Mar 28 03:14 ./test11


Is there a way to add wildcards to the -user or -group options of find?







bash find wildcards






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 28 at 4:07









Prvt_Yadv

3,9113 gold badges17 silver badges34 bronze badges




3,9113 gold badges17 silver badges34 bronze badges










asked Mar 28 at 3:21







Lhakryma DL












migrated from stackoverflow.com Mar 28 at 3:29


This question came from our site for professional and enthusiast programmers.











migrated from stackoverflow.com Mar 28 at 3:29


This question came from our site for professional and enthusiast programmers.









migrated from stackoverflow.com Mar 28 at 3:29


This question came from our site for professional and enthusiast programmers.














  • With getent passwd get a list of users which match br_.* and build a list like: --user br_asd1 --user br_asd2 --user br_bfg1 for find command.

    – Cyrus
    Mar 28 at 3:33












  • GNU find does not have options to pass wildcards for -user and -group fields. One way would be to use GNU coreutils stat and use the %U quantifier to get the owner name. Do something like for file in *; do [ -f "$file" ] || continue; owner=$(stat -c '%U' "$file"); if [[ "$owner" =~ ^br.* ]]; then echo "$file"; fi; done

    – Inian
    Mar 28 at 3:52

















  • With getent passwd get a list of users which match br_.* and build a list like: --user br_asd1 --user br_asd2 --user br_bfg1 for find command.

    – Cyrus
    Mar 28 at 3:33












  • GNU find does not have options to pass wildcards for -user and -group fields. One way would be to use GNU coreutils stat and use the %U quantifier to get the owner name. Do something like for file in *; do [ -f "$file" ] || continue; owner=$(stat -c '%U' "$file"); if [[ "$owner" =~ ^br.* ]]; then echo "$file"; fi; done

    – Inian
    Mar 28 at 3:52
















With getent passwd get a list of users which match br_.* and build a list like: --user br_asd1 --user br_asd2 --user br_bfg1 for find command.

– Cyrus
Mar 28 at 3:33






With getent passwd get a list of users which match br_.* and build a list like: --user br_asd1 --user br_asd2 --user br_bfg1 for find command.

– Cyrus
Mar 28 at 3:33














GNU find does not have options to pass wildcards for -user and -group fields. One way would be to use GNU coreutils stat and use the %U quantifier to get the owner name. Do something like for file in *; do [ -f "$file" ] || continue; owner=$(stat -c '%U' "$file"); if [[ "$owner" =~ ^br.* ]]; then echo "$file"; fi; done

– Inian
Mar 28 at 3:52





GNU find does not have options to pass wildcards for -user and -group fields. One way would be to use GNU coreutils stat and use the %U quantifier to get the owner name. Do something like for file in *; do [ -f "$file" ] || continue; owner=$(stat -c '%U' "$file"); if [[ "$owner" =~ ^br.* ]]; then echo "$file"; fi; done

– Inian
Mar 28 at 3:52










2 Answers
2






active

oldest

votes


















1
















If you're writing for bash you can use an array to hold the set of users and another array to generate the corresponding syntax for find to match those users:



#!/bin/bash

# The beginning of the userids we need to match
match=br

# Find the matching set of users
users=($(
getent passwd |
awk -F: -vm="$match" 'BEGIN re = "^" m $1 ~ re print $1'
))

# Build the list of users ("find ( -user XX -o -user YY -o user ZZ ) ...")
finds=()
for user in "$users[@]"
do
finds+=('-o' '-user' "$user")
done
[[ $#finds[@] -gt 0 ]] && finds=('(' "$finds[@]:1" ')')

# Execute the find command with the set of users
find . "$finds[@]" -ls


As ever, you can prefix the find command with something like echo to see what would be executed. (Or you could run with bash -x to enable line-by-line debug reporting.)






share|improve this answer



























  • Instead of the test in each iteration of that loop, you could just use "$finds[@]:1" when calling find to ignore that first -o in the list.

    – Kusalananda
    Mar 28 at 8:46


















0
















If all you need is just a listing, I would go for a good old grep over find’s output. It would be slower, but much less typing.



At the very basic you might try this, for your example case:



find -printf '%u %pn' | egrep '^br_'


This gives you the list of files owned by users br_*, shown with username and filename on each line.



You might then expand the output by fiddling the -printf format string.



For instance, to make it resemble more an ls output:



find -printf '%-8.8u %-8.8g %M %8s %t %pn' | egrep '^br_'


The important thing to keep the command line as short as possible is to place the username at the very beginning, so that the egrep’s part can be kept short like that.



To see all that you can specify in the format string, go here, then search for -printf format.



Be careful though that if your files have newlines in their names then this solution might have problems in showing them.






share|improve this answer



























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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1
















    If you're writing for bash you can use an array to hold the set of users and another array to generate the corresponding syntax for find to match those users:



    #!/bin/bash

    # The beginning of the userids we need to match
    match=br

    # Find the matching set of users
    users=($(
    getent passwd |
    awk -F: -vm="$match" 'BEGIN re = "^" m $1 ~ re print $1'
    ))

    # Build the list of users ("find ( -user XX -o -user YY -o user ZZ ) ...")
    finds=()
    for user in "$users[@]"
    do
    finds+=('-o' '-user' "$user")
    done
    [[ $#finds[@] -gt 0 ]] && finds=('(' "$finds[@]:1" ')')

    # Execute the find command with the set of users
    find . "$finds[@]" -ls


    As ever, you can prefix the find command with something like echo to see what would be executed. (Or you could run with bash -x to enable line-by-line debug reporting.)






    share|improve this answer



























    • Instead of the test in each iteration of that loop, you could just use "$finds[@]:1" when calling find to ignore that first -o in the list.

      – Kusalananda
      Mar 28 at 8:46















    1
















    If you're writing for bash you can use an array to hold the set of users and another array to generate the corresponding syntax for find to match those users:



    #!/bin/bash

    # The beginning of the userids we need to match
    match=br

    # Find the matching set of users
    users=($(
    getent passwd |
    awk -F: -vm="$match" 'BEGIN re = "^" m $1 ~ re print $1'
    ))

    # Build the list of users ("find ( -user XX -o -user YY -o user ZZ ) ...")
    finds=()
    for user in "$users[@]"
    do
    finds+=('-o' '-user' "$user")
    done
    [[ $#finds[@] -gt 0 ]] && finds=('(' "$finds[@]:1" ')')

    # Execute the find command with the set of users
    find . "$finds[@]" -ls


    As ever, you can prefix the find command with something like echo to see what would be executed. (Or you could run with bash -x to enable line-by-line debug reporting.)






    share|improve this answer



























    • Instead of the test in each iteration of that loop, you could just use "$finds[@]:1" when calling find to ignore that first -o in the list.

      – Kusalananda
      Mar 28 at 8:46













    1














    1










    1









    If you're writing for bash you can use an array to hold the set of users and another array to generate the corresponding syntax for find to match those users:



    #!/bin/bash

    # The beginning of the userids we need to match
    match=br

    # Find the matching set of users
    users=($(
    getent passwd |
    awk -F: -vm="$match" 'BEGIN re = "^" m $1 ~ re print $1'
    ))

    # Build the list of users ("find ( -user XX -o -user YY -o user ZZ ) ...")
    finds=()
    for user in "$users[@]"
    do
    finds+=('-o' '-user' "$user")
    done
    [[ $#finds[@] -gt 0 ]] && finds=('(' "$finds[@]:1" ')')

    # Execute the find command with the set of users
    find . "$finds[@]" -ls


    As ever, you can prefix the find command with something like echo to see what would be executed. (Or you could run with bash -x to enable line-by-line debug reporting.)






    share|improve this answer















    If you're writing for bash you can use an array to hold the set of users and another array to generate the corresponding syntax for find to match those users:



    #!/bin/bash

    # The beginning of the userids we need to match
    match=br

    # Find the matching set of users
    users=($(
    getent passwd |
    awk -F: -vm="$match" 'BEGIN re = "^" m $1 ~ re print $1'
    ))

    # Build the list of users ("find ( -user XX -o -user YY -o user ZZ ) ...")
    finds=()
    for user in "$users[@]"
    do
    finds+=('-o' '-user' "$user")
    done
    [[ $#finds[@] -gt 0 ]] && finds=('(' "$finds[@]:1" ')')

    # Execute the find command with the set of users
    find . "$finds[@]" -ls


    As ever, you can prefix the find command with something like echo to see what would be executed. (Or you could run with bash -x to enable line-by-line debug reporting.)







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Mar 28 at 9:03

























    answered Mar 28 at 8:29









    roaimaroaima

    49.4k7 gold badges66 silver badges133 bronze badges




    49.4k7 gold badges66 silver badges133 bronze badges















    • Instead of the test in each iteration of that loop, you could just use "$finds[@]:1" when calling find to ignore that first -o in the list.

      – Kusalananda
      Mar 28 at 8:46

















    • Instead of the test in each iteration of that loop, you could just use "$finds[@]:1" when calling find to ignore that first -o in the list.

      – Kusalananda
      Mar 28 at 8:46
















    Instead of the test in each iteration of that loop, you could just use "$finds[@]:1" when calling find to ignore that first -o in the list.

    – Kusalananda
    Mar 28 at 8:46





    Instead of the test in each iteration of that loop, you could just use "$finds[@]:1" when calling find to ignore that first -o in the list.

    – Kusalananda
    Mar 28 at 8:46













    0
















    If all you need is just a listing, I would go for a good old grep over find’s output. It would be slower, but much less typing.



    At the very basic you might try this, for your example case:



    find -printf '%u %pn' | egrep '^br_'


    This gives you the list of files owned by users br_*, shown with username and filename on each line.



    You might then expand the output by fiddling the -printf format string.



    For instance, to make it resemble more an ls output:



    find -printf '%-8.8u %-8.8g %M %8s %t %pn' | egrep '^br_'


    The important thing to keep the command line as short as possible is to place the username at the very beginning, so that the egrep’s part can be kept short like that.



    To see all that you can specify in the format string, go here, then search for -printf format.



    Be careful though that if your files have newlines in their names then this solution might have problems in showing them.






    share|improve this answer





























      0
















      If all you need is just a listing, I would go for a good old grep over find’s output. It would be slower, but much less typing.



      At the very basic you might try this, for your example case:



      find -printf '%u %pn' | egrep '^br_'


      This gives you the list of files owned by users br_*, shown with username and filename on each line.



      You might then expand the output by fiddling the -printf format string.



      For instance, to make it resemble more an ls output:



      find -printf '%-8.8u %-8.8g %M %8s %t %pn' | egrep '^br_'


      The important thing to keep the command line as short as possible is to place the username at the very beginning, so that the egrep’s part can be kept short like that.



      To see all that you can specify in the format string, go here, then search for -printf format.



      Be careful though that if your files have newlines in their names then this solution might have problems in showing them.






      share|improve this answer



























        0














        0










        0









        If all you need is just a listing, I would go for a good old grep over find’s output. It would be slower, but much less typing.



        At the very basic you might try this, for your example case:



        find -printf '%u %pn' | egrep '^br_'


        This gives you the list of files owned by users br_*, shown with username and filename on each line.



        You might then expand the output by fiddling the -printf format string.



        For instance, to make it resemble more an ls output:



        find -printf '%-8.8u %-8.8g %M %8s %t %pn' | egrep '^br_'


        The important thing to keep the command line as short as possible is to place the username at the very beginning, so that the egrep’s part can be kept short like that.



        To see all that you can specify in the format string, go here, then search for -printf format.



        Be careful though that if your files have newlines in their names then this solution might have problems in showing them.






        share|improve this answer













        If all you need is just a listing, I would go for a good old grep over find’s output. It would be slower, but much less typing.



        At the very basic you might try this, for your example case:



        find -printf '%u %pn' | egrep '^br_'


        This gives you the list of files owned by users br_*, shown with username and filename on each line.



        You might then expand the output by fiddling the -printf format string.



        For instance, to make it resemble more an ls output:



        find -printf '%-8.8u %-8.8g %M %8s %t %pn' | egrep '^br_'


        The important thing to keep the command line as short as possible is to place the username at the very beginning, so that the egrep’s part can be kept short like that.



        To see all that you can specify in the format string, go here, then search for -printf format.



        Be careful though that if your files have newlines in their names then this solution might have problems in showing them.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Mar 29 at 0:13









        LL3LL3

        2,1772 silver badges14 bronze badges




        2,1772 silver badges14 bronze badges






























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