Cut in CLI doesn't work as expected - returns full lines every time
I'm running MacOS Mojave and I'm trying to get a list of owners of files in given path. I'm trying to do it like this but it doesn't work.
$ ls -l /usr | cut -f3
total 0
drwxr-xr-x 971 root wheel 31072 23 sty 20:05 bin
drwxr-xr-x 304 root wheel 9728 23 sty 20:05 lib
drwxr-xr-x 248 root wheel 7936 23 sty 20:05 libexec
drwxr-xr-x 16 root wheel 512 3 lis 10:50 local
drwxr-xr-x 239 root wheel 7648 23 sty 20:05 sbin
drwxr-xr-x 46 root wheel 1472 3 lis 10:41 share
drwxr-xr-x 5 root wheel 160 21 wrz 06:06 standalone
Specifying delimiters seems to work but not for the TAB character (which should be default).
$ ls -l /usr | cut -f3 -d' '
971
304
248
239
I'm using ZSH with Oh my zsh and iTerm 2 if it matters.
command-line zsh
edited Jan 27 at 14:23
nohillside♦
52.4k13112155
asked Jan 27 at 13:55
NajkiNajki
1284
add a comment |
I'm running MacOS Mojave and I'm trying to get a list of owners of files in given path. I'm trying to do it like this but it doesn't work.
$ ls -l /usr | cut -f3
total 0
drwxr-xr-x 971 root wheel 31072 23 sty 20:05 bin
drwxr-xr-x 304 root wheel 9728 23 sty 20:05 lib
drwxr-xr-x 248 root wheel 7936 23 sty 20:05 libexec
drwxr-xr-x 16 root wheel 512 3 lis 10:50 local
drwxr-xr-x 239 root wheel 7648 23 sty 20:05 sbin
drwxr-xr-x 46 root wheel 1472 3 lis 10:41 share
drwxr-xr-x 5 root wheel 160 21 wrz 06:06 standalone
Specifying delimiters seems to work but not for the TAB character (which should be default).
$ ls -l /usr | cut -f3 -d' '
971
304
248
239
I'm using ZSH with Oh my zsh and iTerm 2 if it matters.
command-line zsh
edited Jan 27 at 14:23
nohillside♦
52.4k13112155
asked Jan 27 at 13:55
NajkiNajki
1284
2
Can you add some details about what you intend to do with the list of owners? Depending on this there might be better solutions than parsing the output ofls.
– nohillside♦
Jan 27 at 14:22
1
Tab is, indeed, the default forcut, but the output oflsisn't tab-delimited; it just uses calculated numbers of space characters to produce tabular output.
– chepner
Jan 28 at 16:54
add a comment |
I'm running MacOS Mojave and I'm trying to get a list of owners of files in given path. I'm trying to do it like this but it doesn't work.
$ ls -l /usr | cut -f3
total 0
drwxr-xr-x 971 root wheel 31072 23 sty 20:05 bin
drwxr-xr-x 304 root wheel 9728 23 sty 20:05 lib
drwxr-xr-x 248 root wheel 7936 23 sty 20:05 libexec
drwxr-xr-x 16 root wheel 512 3 lis 10:50 local
drwxr-xr-x 239 root wheel 7648 23 sty 20:05 sbin
drwxr-xr-x 46 root wheel 1472 3 lis 10:41 share
drwxr-xr-x 5 root wheel 160 21 wrz 06:06 standalone
Specifying delimiters seems to work but not for the TAB character (which should be default).
$ ls -l /usr | cut -f3 -d' '
971
304
248
239
I'm using ZSH with Oh my zsh and iTerm 2 if it matters.
command-line zsh
edited Jan 27 at 14:23
nohillside♦
52.4k13112155
asked Jan 27 at 13:55
NajkiNajki
1284
I'm running MacOS Mojave and I'm trying to get a list of owners of files in given path. I'm trying to do it like this but it doesn't work.
$ ls -l /usr | cut -f3
total 0
drwxr-xr-x 971 root wheel 31072 23 sty 20:05 bin
drwxr-xr-x 304 root wheel 9728 23 sty 20:05 lib
drwxr-xr-x 248 root wheel 7936 23 sty 20:05 libexec
drwxr-xr-x 16 root wheel 512 3 lis 10:50 local
drwxr-xr-x 239 root wheel 7648 23 sty 20:05 sbin
drwxr-xr-x 46 root wheel 1472 3 lis 10:41 share
drwxr-xr-x 5 root wheel 160 21 wrz 06:06 standalone
Specifying delimiters seems to work but not for the TAB character (which should be default).
$ ls -l /usr | cut -f3 -d' '
971
304
248
239
I'm using ZSH with Oh my zsh and iTerm 2 if it matters.
command-line zsh
command-line zsh
edited Jan 27 at 14:23
nohillside♦
52.4k13112155
asked Jan 27 at 13:55
NajkiNajki
1284
edited Jan 27 at 14:23
nohillside♦
52.4k13112155
asked Jan 27 at 13:55
NajkiNajki
1284
edited Jan 27 at 14:23
nohillside♦
52.4k13112155
edited Jan 27 at 14:23
nohillside♦
52.4k13112155
edited Jan 27 at 14:23
nohillside♦
52.4k13112155
52.4k13112155
asked Jan 27 at 13:55
NajkiNajki
1284
asked Jan 27 at 13:55
NajkiNajki
1284
asked Jan 27 at 13:55
NajkiNajki
1284
1284
2
Can you add some details about what you intend to do with the list of owners? Depending on this there might be better solutions than parsing the output ofls.
– nohillside♦
Jan 27 at 14:22
1
Tab is, indeed, the default forcut, but the output oflsisn't tab-delimited; it just uses calculated numbers of space characters to produce tabular output.
– chepner
Jan 28 at 16:54
add a comment |
2
Can you add some details about what you intend to do with the list of owners? Depending on this there might be better solutions than parsing the output ofls.
– nohillside♦
Jan 27 at 14:22
1
Tab is, indeed, the default forcut, but the output oflsisn't tab-delimited; it just uses calculated numbers of space characters to produce tabular output.
– chepner
Jan 28 at 16:54
2
2
Can you add some details about what you intend to do with the list of owners? Depending on this there might be better solutions than parsing the output of
ls.– nohillside♦
Jan 27 at 14:22
Can you add some details about what you intend to do with the list of owners? Depending on this there might be better solutions than parsing the output of
ls.– nohillside♦
Jan 27 at 14:22
1
1
Tab is, indeed, the default for
cut, but the output of ls isn't tab-delimited; it just uses calculated numbers of space characters to produce tabular output.– chepner
Jan 28 at 16:54
Tab is, indeed, the default for
cut, but the output of ls isn't tab-delimited; it just uses calculated numbers of space characters to produce tabular output.– chepner
Jan 28 at 16:54
add a comment |
2 Answers
2
active
oldest
votes
You can squeeze the white spaces into a single white space in ls 's output then use cut.
ls -l /usr | tr -s ' ' | cut -d ' ' -f3
but avoid parsing ls output. Here's an alternate solution.
stat -f'%Su' /usr/*
answered Jan 27 at 16:39
fd0fd0
6,35011430
6
+1 for not parsing the output from ls — see unix.stackexchange.com/questions/128985, mywiki.wooledge.org/ParsingLs,&c.
– gidds
Jan 27 at 18:13
statseems to be a good solution to my problem and lets me retrieve more info without all the hassle with extracting specific columns. Thanks!
– Najki
Jan 28 at 9:36
1
That second link should be just mywiki.wooledge.org/ParsingLs
– Chris
Jan 28 at 14:01
add a comment |
ls doesn't use tabs, cut doesn't work with a variable number of delimeters between fields.
ls -l /usr | awk '{print $3}'
will work, or
ls -l /usr | awk 'NR > 1 {print $3}'
if you want to skip the first line (total 0 in your example).
answered Jan 27 at 14:21
nohillside♦nohillside
52.4k13112155
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can squeeze the white spaces into a single white space in ls 's output then use cut.
ls -l /usr | tr -s ' ' | cut -d ' ' -f3
but avoid parsing ls output. Here's an alternate solution.
stat -f'%Su' /usr/*
answered Jan 27 at 16:39
fd0fd0
6,35011430
6
+1 for not parsing the output from ls — see unix.stackexchange.com/questions/128985, mywiki.wooledge.org/ParsingLs,&c.
– gidds
Jan 27 at 18:13
statseems to be a good solution to my problem and lets me retrieve more info without all the hassle with extracting specific columns. Thanks!
– Najki
Jan 28 at 9:36
1
That second link should be just mywiki.wooledge.org/ParsingLs
– Chris
Jan 28 at 14:01
add a comment |
You can squeeze the white spaces into a single white space in ls 's output then use cut.
ls -l /usr | tr -s ' ' | cut -d ' ' -f3
but avoid parsing ls output. Here's an alternate solution.
stat -f'%Su' /usr/*
answered Jan 27 at 16:39
fd0fd0
6,35011430
6
+1 for not parsing the output from ls — see unix.stackexchange.com/questions/128985, mywiki.wooledge.org/ParsingLs,&c.
– gidds
Jan 27 at 18:13
statseems to be a good solution to my problem and lets me retrieve more info without all the hassle with extracting specific columns. Thanks!
– Najki
Jan 28 at 9:36
1
That second link should be just mywiki.wooledge.org/ParsingLs
– Chris
Jan 28 at 14:01
add a comment |
You can squeeze the white spaces into a single white space in ls 's output then use cut.
ls -l /usr | tr -s ' ' | cut -d ' ' -f3
but avoid parsing ls output. Here's an alternate solution.
stat -f'%Su' /usr/*
answered Jan 27 at 16:39
fd0fd0
6,35011430
You can squeeze the white spaces into a single white space in ls 's output then use cut.
ls -l /usr | tr -s ' ' | cut -d ' ' -f3
but avoid parsing ls output. Here's an alternate solution.
stat -f'%Su' /usr/*
answered Jan 27 at 16:39
fd0fd0
6,35011430
answered Jan 27 at 16:39
fd0fd0
6,35011430
answered Jan 27 at 16:39
fd0fd0
6,35011430
answered Jan 27 at 16:39
fd0fd0
6,35011430
6,35011430
6
+1 for not parsing the output from ls — see unix.stackexchange.com/questions/128985, mywiki.wooledge.org/ParsingLs,&c.
– gidds
Jan 27 at 18:13
statseems to be a good solution to my problem and lets me retrieve more info without all the hassle with extracting specific columns. Thanks!
– Najki
Jan 28 at 9:36
1
That second link should be just mywiki.wooledge.org/ParsingLs
– Chris
Jan 28 at 14:01
add a comment |
6
+1 for not parsing the output from ls — see unix.stackexchange.com/questions/128985, mywiki.wooledge.org/ParsingLs,&c.
– gidds
Jan 27 at 18:13
statseems to be a good solution to my problem and lets me retrieve more info without all the hassle with extracting specific columns. Thanks!
– Najki
Jan 28 at 9:36
1
That second link should be just mywiki.wooledge.org/ParsingLs
– Chris
Jan 28 at 14:01
6
6
+1 for not parsing the output from ls — see unix.stackexchange.com/questions/128985, mywiki.wooledge.org/ParsingLs,&c.
– gidds
Jan 27 at 18:13
+1 for not parsing the output from ls — see unix.stackexchange.com/questions/128985, mywiki.wooledge.org/ParsingLs,&c.
– gidds
Jan 27 at 18:13
stat seems to be a good solution to my problem and lets me retrieve more info without all the hassle with extracting specific columns. Thanks!– Najki
Jan 28 at 9:36
stat seems to be a good solution to my problem and lets me retrieve more info without all the hassle with extracting specific columns. Thanks!– Najki
Jan 28 at 9:36
1
1
That second link should be just mywiki.wooledge.org/ParsingLs
– Chris
Jan 28 at 14:01
That second link should be just mywiki.wooledge.org/ParsingLs
– Chris
Jan 28 at 14:01
add a comment |
ls doesn't use tabs, cut doesn't work with a variable number of delimeters between fields.
ls -l /usr | awk '{print $3}'
will work, or
ls -l /usr | awk 'NR > 1 {print $3}'
if you want to skip the first line (total 0 in your example).
answered Jan 27 at 14:21
nohillside♦nohillside
52.4k13112155
add a comment |
ls doesn't use tabs, cut doesn't work with a variable number of delimeters between fields.
ls -l /usr | awk '{print $3}'
will work, or
ls -l /usr | awk 'NR > 1 {print $3}'
if you want to skip the first line (total 0 in your example).
answered Jan 27 at 14:21
nohillside♦nohillside
52.4k13112155
add a comment |
ls doesn't use tabs, cut doesn't work with a variable number of delimeters between fields.
ls -l /usr | awk '{print $3}'
will work, or
ls -l /usr | awk 'NR > 1 {print $3}'
if you want to skip the first line (total 0 in your example).
answered Jan 27 at 14:21
nohillside♦nohillside
52.4k13112155
ls doesn't use tabs, cut doesn't work with a variable number of delimeters between fields.
ls -l /usr | awk '{print $3}'
will work, or
ls -l /usr | awk 'NR > 1 {print $3}'
if you want to skip the first line (total 0 in your example).
answered Jan 27 at 14:21
nohillside♦nohillside
52.4k13112155
answered Jan 27 at 14:21
nohillside♦nohillside
52.4k13112155
answered Jan 27 at 14:21
nohillside♦nohillside
52.4k13112155
answered Jan 27 at 14:21
nohillside♦nohillside
52.4k13112155
52.4k13112155
add a comment |
add a comment |
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2
Can you add some details about what you intend to do with the list of owners? Depending on this there might be better solutions than parsing the output of
ls.– nohillside♦
Jan 27 at 14:22
1
Tab is, indeed, the default for
cut, but the output oflsisn't tab-delimited; it just uses calculated numbers of space characters to produce tabular output.– chepner
Jan 28 at 16:54