awk从输出中删除空间

Question

I am using @"df -lH | grep \\"/Volumes/*\\" | awk '{$1=$2=$3=$4=$5=$6=$7=$8\\"\\"; print $0 }'" for getting locally-mounted volumes path. but if volume name contain two or more space (Leopard 1). its(awk) removing space from output.

Output:

$df -lH

Filesystem     Size   Used  Avail Capacity  Mounted on
/dev/disk0s3    81G    61G    19G    77%    /
/dev/disk0s2    81G    72G   8.2G    90%    /Volumes/Leopard  1  2     3
/dev/disk0s4   158G    47G   111G    30%    /Volumes/Backup  

$ df -lH | grep "/Volumes/*"
/dev/disk0s2    81G    72G   8.2G    90%    /Volumes/Leopard  1  2     3
/dev/disk0s4   158G    47G   111G    30%    /Volumes/Backup  

$ df -lH | grep "/Volumes/*" | awk '{$1=$2=$3=$4=$5=""; print $0}'
     /Volumes/Leopard 1 2 3
     /Volumes/Backup

can anyone please help me out?

Answer 1

By default awk 's output field separator is a single space. So the output of your awk command is completely expected. To get the result you want, you'll need to use some sort of regex, so try this instead with GNU awk :

df -lH | awk '/Volumes/ { sub(/^(\S+\s+){5}/, ""); print }'

Or if you have GNU sed :

df -lH | sed -nr '/Volumes/s/^(\S+\s+){5}//p'

Results:

/Volumes/Leopard  1  2     3
/Volumes/Backup  

EDIT:

I see that BSD/OSX awk doesn't support interval expressions unfortunately. Therefore the safest way would be to do this with awk :

df -lH | awk '/Volumes/ { sub(/^[^ ]+ +[^ ]+ +[^ ]+ +[^ ]+ +[^ ]+ +/, ""); print }'

or with sed :

df -lH | sed -n '/Volumes/s/^[^ ]\+ \+[^ ]\+ \+[^ ]\+ \+[^ ]\+ \+[^ ]\+ \+[^ ]\+ \+[^ ]\+ \+//p'

This should also be highly portable! HTH.

EDIT:

For Mac OSX 10.8 with 8 columns, simply extend the regex:

df -lH | awk '/Volumes/ { sub(/^[^ ]+ +[^ ]+ +[^ ]+ +[^ ]+ +[^ ]+ +[^ ]+ +[^ ]+ +/, ""); print }'

or with sed :

df -lH | sed -n '/Volumes/s/^[^ ]\+ \+[^ ]\+ \+[^ ]\+ \+[^ ]\+ \+[^ ]\+ \+//p'

Answer 2

This is easier using the mount command, which has simpler output than df , and perl , which supports non-greedy matching:

:; mount | perl -pe 's/.*? on //;s/ \([^\)]*\)$//'
/
/dev
/Volumes/pro Time Machine 2
/Volumes/b
/Volumes/p
/net
/home

Notice that I do have a volume mounted with spaces in the mount point.

Answer 3

df -lH | grep "/Volumes/*" | perl -pe 's/[^\%]*\%//g'

如果要在awk中执行此操作：

df -lH | grep "/Volumes/*"|awk '{gsub(/[^\%]*\%/,"");print}'

Answer 4

检查此命令的输出，可能有助于您解析并获取信息

mount | column -t