Shell脚本从给定路径

Question

我的完整路径是/data7/stmt_data16/pdf/RL/20170202/INLAND/641/K_EDGE1_641 我想从pdf目录开始仅提取和打印路径， 即/pdf/RL/20170202/INLAND/641/K_EDGE1_641。

那么如何用shell脚本中的sed或awk命令实现呢？

Answer 1

如果你的shell是bash，你既不需要sed也不需要awk：

path='/data7/stmt_data16/pdf/RL/20170202/INLAND/641/K_EDGE1_641'
echo "${path#*/pdf/}"

RL/20170202/INLAND/641/K_EDGE1_641

这也会切断/pdf/，但可以手动添加：

echo "/pdf/${path#*/pdf/}"

Answer 2

也可以这样做：

sed -r 's#(.*)(/pdf/.*)#\2#' <<<"$yourpathvvar"

测试：

mypath="/data7/stmt_data16/pdf/RL/20170202/INLAND/641/K_EDGE1_641"
sed -r 's#(.*)(/pdf/.*)#\2#' <<<"$mypath"
/pdf/RL/20170202/INLAND/641/K_EDGE1_641

Answer 3

grep

$ path='/data7/stmt_data16/pdf/RL/20170202/INLAND/641/K_EDGE1_641'
$ echo $path | grep -o '/pdf.*'
$ /pdf/RL/20170202/INLAND/641/K_EDGE1_641

-o -print only match

Answer 4

使用字符串AWK和match函数

substr方法：

path='/data7/stmt_data16/pdf/RL/20170202/INLAND/641/K_EDGE1_641'
echo "$path" | awk 'match($0, /\/pdf.*/) {print substr($0, RSTART, RLENGTH) }'

输出：

/pdf/RL/20170202/INLAND/641/K_EDGE1_641

SED方法：

path='/data7/stmt_data16/pdf/RL/20170202/INLAND/641/K_EDGE1_641'
echo "$path" | sed -n 's/.*\(\/pdf.*\)/\1/p'

Answer 5

这里有很多解决方案......用纯粹的bash再抓一个：

echo "${mypath/*pdf//pdf}"

Answer 6

path='/data7/stmt_data16/pdf/RL/20170202/INLAND/641/K_EDGE1_641'

echo "$path" |awk -F '_data16' '{print $2}'
/pdf/RL/20170202/INLAND/641/K_EDGE1_641