从第n次出现的模式替换到sed的行尾

Question

For example:

/some/long/path/we/need/to/shorten

Need to delete after the 6th occurrence of '/', including itself:

/some/long/path/we/need

Using sed I came up with this solution, but it's kind of workaround-ish:

path=/some/long/path/we/need/to/shorten
slashesToKeep=5
n=2+slashesToKeep
echo $path | sed "s/[^/]*//$n;s/\/\/.*//g"

Cleaner solution much appreciated!

Answer 1

Input

/some/long/path/we/need/to/shorten

Code

Cut Solution

echo '/some/long/path/we/need/to/shorten' | cut -d '/' -f 1-6

AWK Solution

echo '/some/long/path/we/need/to/shorten' | awk -F '/'  '{ for(i=1; i<=6; i++) {print $i} }' | tr '\n' '/'|sed 's/.$//'

Output

/some/long/path/we/need

Answer 2

这可能适合你（GNU sed）：

sed 's/\/[^\/]*//6g' file

Answer 3

Awk:

awk -F'/' 'BEGIN{OFS=FS}{NF=6}1'

In action:

$ echo /some/long/path/we/need/to/shorten | awk -F'/' 'BEGIN{OFS=FS}{NF=6}1' 
/some/long/path/we/need