说我有路径gui/site/junior/profile.py
我如何得到这个?:
gui
gui/site
gui/site/junior
如果你告诉我如何遍历每条路径,可以获得奖金:D
答案 0 :(得分:2)
你可以用awk循环:
awk 'BEGIN{FS=OFS="/"}
{ for (i=1; i<=NF; i++) {
for (j=1; j<i; j++)
printf "%s/", $j
printf "%s\n", $i
}
}' <<< "gui/site/junior/profile.py"
看作一个班轮:
$ awk 'BEGIN{FS=OFS="/"}{for (i=1; i<=NF; i++) { for (j=1; j<i; j++) printf "%s%s", $j, OFS; printf "%s\n", $i}}' <<< "gui/site/junior/profile.py"
gui
gui/site
gui/site/junior
gui/site/junior/profile.py
这利用了NF
,它计算当前记录的字段数。基于此,它从第一个循环到最后一个循环,每次首先打印到该值。
答案 1 :(得分:2)
您可以使用shell的内置拆分工具。 IFS
指定要拆分的内容。
oldIFS=$IFS
IFS=/
set -f
set -- $path
set +f
IFS=$oldIFS
for component in "$@"; do
echo "$component"
done
这可以通过多种方式进行重构,但我希望对IFS
进行更改以仅控制实际的拆分。
使用set
将字符串拆分为位置参数有点模糊,但值得深入了解。
如果最初没有设置IFS
,你应该小心谨慎,但我对此不屑一顾。
答案 2 :(得分:2)
简短的回答:
path="gui/site with spaces/junior/profile.py"
oldIFS=$IFS
IFS=/
set -- $(dirname "$path")
IFS=$oldIFS
accumulated=""
for dir in "$@"; do
accumulated="${accumulated}${dir}/"
echo "$accumulated"
done
gui/
gui/site with spaces/
gui/site with spaces/junior/
答案 3 :(得分:1)
Perl变体:
perl -F/ -nlE 'say join("/",@F[0..$_])||"/"for(0..$#F-1)' <<< "gui/site with spaces/junior/profile.py"
产生
gui
gui/site with spaces
gui/site with spaces/junior
如果你有NULL分隔路径名,你可以将0
添加到参数:
perl -F/ -0nlE 'say join("/",@F[0..$_])||"/"for(0..$#F-1)'
^
例如
printf "/some/path/name/here/file.py\0" | perl -F/ -nlE 'say join("/",@F[0..$_])||"/"for(0..$#F-1)'
# ^^
产生
/
/some
/some/path
/some/path/name
/some/path/name/here
对于迭代路径,您可以使用下一个:
origpath="some/long/path/here/py.py"
do_something() {
local path="$1"
#add here what you need to do with the partial path
echo "Do something here with the ==$path=="
}
while read -r part
do
do_something "$part"
done < <(perl -F/ -nlE 'say join("/",@F[0..$_])||"/"for(0..$#F-1)' <<< "$origpath")
它产生:
Do something here with the ==some==
Do something here with the ==some/long==
Do something here with the ==some/long/path==
Do something here with the ==some/long/path/here==