bash 用 sed 替换从第 n 次出现的模式到行尾
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Replace from nth occurrence of pattern till the end of line with sed
提问by user1612686
For example:
例如：
/some/long/path/we/need/to/shorten
Need to delete after the 6th occurrence of '/', including itself:
需要在第 6 次出现 '/' 后删除，包括其自身：
/some/long/path/we/need
Using sed I came up with this solution, but it's kind of workaround-ish:
使用 sed 我想出了这个解决方案，但它是一种解决方法：
path=/some/long/path/we/need/to/shorten
slashesToKeep=5
n=2+slashesToKeep
echo $path | sed "s/[^/]*//$n;s/\/\/.*//g"
Cleaner solution much appreciated!
更清洁的解决方案非常感谢！
回答by Debaditya
Input
输入
/some/long/path/we/need/to/shorten
Code
代码
Cut Solution
切割解决方案
echo '/some/long/path/we/need/to/shorten' | cut -d '/' -f 1-6
AWK Solution
AWK解决方案
echo '/some/long/path/we/need/to/shorten' | awk -F '/' '{ for(i=1; i<=6; i++) {print $i} }' | tr '\n' '/'|sed 's/.$//'
Output
输出
/some/long/path/we/need
回答by potong
This might work for you (GNU sed):
这可能对你有用（GNU sed）：
sed 's/\/[^\/]*//6g' file
回答by Fredrik Pihl
Awk:
惊：
awk -F'/' 'BEGIN{OFS=FS}{NF=6}1'
In action:
在行动：
$ echo /some/long/path/we/need/to/shorten | awk -F'/' 'BEGIN{OFS=FS}{NF=6}1'
/some/long/path/we/need
