[英]How to get the Directory name and file name of a bash script by bash?
跟隨是已知的。 可能有幫助:
獲取 filename.extension 包括。 完整路徑:
腳本: /path1/path2/path3/path4/path5/bashfile.sh
#!/bin/bash
echo $0
read -r
Output:
/path1/path2/path3/path4/path5/bashfile.sh
獲取文件名.擴展名:
腳本: /path/path/path/path/path/bashfile.sh
#!/bin/bash
echo ${0##*/}
read -r
Output:
bashfile.sh
Script: `/path1/path2/path3/path4/path5/bashfile.sh`
通緝output:
/path5/bashfile.sh
備注:也許是有可能的,如果你從右邊看,把“/*/”左邊的都去掉
比第一個擬合解決方案短一點:
腳本: /path1/path2/path3/path4/path5/bashfile.sh
#!/bin/bash
n=$(($(echo $0 | tr -dc "/" | wc -m)+1))
echo "/""$(echo "$0" | cut -d"/" -f$(($n-1)),$n)"
read -r
Output:
/path5/bashfile.sh
也許它們是一個更短的解決方案。
readlink -f $0 |awk -F"/" '{print "/"$(NF-1)"/"$NF}'
# or
awk -F"/" '{print "/"$(NF-1)"/"$NF}' <(readlink -f $0)
# or
awk -F"/" '{print "/"$(NF-1)"/"$NF}' <<<$(readlink -f $0)
# or
sed -E 's/^(.*)(\/\w+\/\w+\.\w+$)/\2/g' <(readlink -f $0)
output
/path5/bashfile.sh
#!/bin/bash
echo "/$(basename "$(dirname "$0")")/$(basename "$0")"
echo
echo
read -r
Output:
/Dirname/Filname.Extension
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