如何为此进行模式匹配:4,7,9....n - 逗号分隔的数字字符串作为用户输入?我为此使用了 case 语句,我相信,case 宁愿使用模式匹配而不是正则表达式。这是用户得到的:
Do you want to delete any of these?
[ 1 ] launch-EOsgR4
[ 2 ] launch-SWZQdJ
[ 3 ] launch-tHAdIm
[ 4 ] launchd-235.z4KTVx
[ 5 ] launchd-257.nM2wOZ
[ 6 ] progress.log
[ 7 ] ssh-8pISGGnlZ5
----------------------------------------
Single: 4; Multiple: 2,3; Range: 4..7
a to delete all; n to cancel and exit
----------------------------------------
( [1][2][3][4][5][6][7] | a | n ):
而且,如上所示,用户可以选择单个数字(简单),范围:6..9(也不是那么难)或多个:3,5,6(使用“单个数字”有点难)选项)。这是我到目前为止所做的......
#!/usr/bin/env bash
NL=$(echo -e "\033[0;0m")
BD=$(echo -e "\033[0;1m")
ERR=$(printf "\n%-25s" "$(echo -e "\033[1;31m[ ERROR ]\033[0m")")
WRN=$(printf "\n%-25s" "$(echo -e "\033[1;33m[ WARN ]\033[0m")")
ls /tmp | tail -n9 > list_of_file
s=$(printf "%-40s" "-")
function lstFile()
{
local file=$1
if [[ -s $file ]]
then
LINES=( $(cat $file) ); echo ""
for ix in ${!LINES[@]}
do
printf "%-5s%-14s%s\n" "" "${BD}[ $(( ix+1 )) ]" "${LINES[$ix]}${NL}"
done
else
exit 0
fi
LST=$(echo ${!LINES[@]}|awk -v ORS=']' '{for (i=1; i<=NF; i++) print "["($i+1)}')
}
function delOpt()
{
echo "${s// /-}"
echo -e "Single: 4; Multiple: 2,3; Range: 4..7"
echo -e "${BD}a${NL} to delete all; ${BD}n${NL} to cancel and exit"
echo "${s// /-}"
echo -n "( $1 | a | n ): "
}
echo -e "\nDo you want to delete any of these?"
lstFile list_of_file
ANS=
until [[ "${ANS}" == "N" || "${ANS}" == "n" || "${ANS}" == "e" ]]
do
delOpt $LST
read ANS && echo ""
ANS=$( tr '[:upper:]' '[:lower:]' <<< "$ANS" )
[[ -n $(echo $ANS|grep -E -w "^[aen0-9,]{1,}") ]] && : || ANS="X"
case ${ANS} in
[0-9]..[0-9] )
for ix in $(eval echo \{$ANS\}); do
LINE=${LINES[(( $ix-1 ))]}
echo -e "Deleting: ${BD}${LINE}${NL}"
sed -i -c "/$LINE/d" list_of_file
done
unset LINE
lstFile list_of_file
;;
[0-9,]* )
ANS=$(echo $ANS | awk -F' |,' '{for (i=1; i<=NF; i++) print $i}')
for ix in $ANS; do
if [[ $ix -gt ${#LINES[@]} ]]
then
echo "${ERR}Out-of-range value: ${bd}$ix${NL}"
else
LINE=${LINES[(( $ix-1 ))]}
echo -e "Deleting: ${BD}${LINE}${NL}"
sed -i -c "/$LINE/d" list_of_file > /dev/null 2>&1
fi
done
unset LINE
lstFile list_of_file
;;
a )
for ix in ${LINES[@]}; do
echo "Deleting: ${BD}${ix}${NL}"
sed -i -c "/$ix/d" list_of_file
done
exit 0
;;
n|e ) exit 0
;;
* ) echo "${WRN}Invalid entry! Should be digit or ${BD}a${NL} for All."
printf "%-14s%s\n\n" "" "Otherwise, enter ${UL}n${NL}o or ${UL}e${NL}xit to quit"
;;
esac
done
这工作正常(有点),但有一些竞争条件。例如
2,d,7 - throws in: bad array subscript
6..10 - throws in: (( 6..11-1 )): syntax error: invalid arithmetic operator (error token is "..11-1 ))")
but,
6..9 - throws in: first RE may not be empty
有没有办法有单独的选项来捕获“单个”和“多个”数字输入?另外,对整体改进有什么建议吗?
非常感谢任何帮助。干杯!!
更新:31/10
现在正在工作。感谢 Alepac 的建议。
以防万一,如果其他人也在寻找类似的东西,我把它放在这里。根据我的原始代码,此函数将从文件中删除行:
list_of_file
根据用户输入。
function chkINPUT()
{
local I_PUT=( "$@" )
local MAX=${#LINES[@]}
#IFS=', ' read -a SPLITTED <<< "$I_PUT"
local SPLITTED=( $(echo "${I_PUT[@]}" | awk -F',| ' '{for (i=1; i<=NF; i++) print $i}') )
for idx in "${!SPLITTED[@]}"
do
SPLTD=${SPLITTED[idx]}
# Check if it's a range [4..7]
if [[ "${SPLTD}" =~ ^[0-9]{1,2}\.\.[0-9]{1,2}$ ]]
then
for ix in $(eval echo \{$SPLTD\}); do
if (( ${ix} <= $MAX )); then
LINE=${LINES[(( ix-1 ))]}
echo -e "Deleting: ${BD}${LINE}${NL}"
sed -i -e "/$LINE/d" list_of_file 2>&1 > /dev/null
unset LINE
else
echo -e "\t${ix} => Out of range"
break
fi
done
# Check if it's a single input
elif [[ "${SPLTD}" =~ ^[[:digit:]]+$ ]]
then
if (( ${SPLTD} <= $MAX )); then
LINE=${LINES[(( SPLTD-1 ))]}
echo -e "Deleting: ${BD}${LINE}${NL}"
sed -i -e "/$LINE/d" list_of_file 2>&1 > /dev/null
unset LINE
else
echo "${ERR}Out-of-range value: ${bd}$SPLTD${NL}"
fi
else
echo "${ERR}Invalid entry: ${bd}$SPLTD${NL}; must be an integer from the list!"
fi
done
}
然后,像这样使用它:
ANS=
until [[ "${ANS}" == "N" || "${ANS}" == "n" || "${ANS}" == "e" ]]
do
delOpt $LST
read ANS && echo ""
ANS=$( tr '[:upper:]' '[:lower:]' <<< "$ANS" )
case ${ANS} in
[0-9]* )
chkINPUT ${ANS}
;;
希望能帮助到你。干杯!!