输入:
201103 1 /mnt/hdd/PUB/SOMETHING
201102 7 /mnt/hdd/PUB/SOMETH ING
201103 11 /mnt/hdd/PUB/SO METHING
201104 3 /mnt/hdd/PUB/SOMET HING
201106 1 /mnt/hdd/PUB/SOMETHI NG
期望的输出:
201103 01 /mnt/hdd/PUB/SOMETHING
201102 07 /mnt/hdd/PUB/SOMETH ING
201103 11 /mnt/hdd/PUB/SO METHING
201104 03 /mnt/hdd/PUB/SOMET HING
201106 01 /mnt/hdd/PUB/SOMETHI NG
0
如果只有一位数字,例如1
“日”部分,如何添加 a ?我需要这种日期格式:YYYYMM DD。
答案1
另一个解决方案:awk '{$2 = sprintf("%02d", $2); print}'
答案2
$ sed 's/\<[0-9]\>/0&/' ./infile
201103 01 /mnt/hdd/PUB/SOMETHING
201102 07 /mnt/hdd/PUB/SOMETH ING
201103 11 /mnt/hdd/PUB/SO METHING
201104 03 /mnt/hdd/PUB/SOMET HING
201106 01 /mnt/hdd/PUB/SOMETHI NG
答案3
这是一种(非 sed)使用方法巴什和扩展正则表达式..
此方法允许作用域对各个行进行更复杂的处理。 (即不仅仅是正则表达式替换)
while IFS= read -r line ; do
if [[ "$line" =~ ^(.+\ )([0-9]\ .+)$ ]]
then echo "${BASH_REMATCH[1]}0${BASH_REMATCH[2]}"
else echo "$line"
fi
done <<EOF
201103 1 /mnt/hdd/PUB/SOMETHING
201102 7 /mnt/hdd/PUB/SOMETH ING
201103 11 /mnt/hdd/PUB/SO METHING
201104 3 /mnt/hdd/PUB/SOMET HING
201106 1 /mnt/hdd/PUB/SOMETHI NG
EOF
输出:
201103 01 /mnt/hdd/PUB/SOMETHING
201102 07 /mnt/hdd/PUB/SOMETH ING
201103 11 /mnt/hdd/PUB/SO METHING
201104 03 /mnt/hdd/PUB/SOMET HING
201106 01 /mnt/hdd/PUB/SOMETHI NG
答案4
while read a b c
do
new_format=$(printf "%02d" $b)
echo "$a $new_format $c"
done </tmp/input