在bash中转换日期格式

Question

I have similar different file of the format backup_2016-26-10_16-30-00 is it possible to rename using bash script to backup_26-10-2016_16:30:00 for all files. Kindly suggest some method to fix this.

Original file:

backup_2016-30-10_12-00-00

Expected output:

backup_30-10-2016_12:00:00

Answer 1

To perform only the name transformation, you can use awk :

echo 'backup_2016-30-10_12-00-00' |
  awk -F'[_-]' '{ print $1 "_" $3 "-" $4 "-" $2 "_" $5 ":" $6 ":" $7 }'

As fedorqui points out in a comment, awk 's printf function may be tidier in this case:

echo 'backup_2016-30-10_12-00-00' |
  awk -F'[_-]' '{ printf "%s_%s-%s-%s_%s:%s:%s\n", $1,$3,$4,$2,$5,$6,$7 }'

---

That said, your specific Linux distro may come with a rename tool that allows you to do the same while performing actual file renaming.

Answer 2

with perl based rename command:

$ touch backup_2016-30-10_12-00-00 backup_2016-26-10_16-30-00

$ rename -n 's/(\d{4})-([^_]+)_(\d+)-(\d+)-/$2-$1_$3:$4:/' backup*
rename(backup_2016-26-10_16-30-00, backup_26-10-2016_16:30:00)
rename(backup_2016-30-10_12-00-00, backup_30-10-2016_12:00:00)

remove the -n option for actual renaming

Answer 3

rename is for this task

$ rename 's/_(\d{4})-(\d\d-\d\d)_(\d\d)-(\d\d)-(\d\d)$/_$2-$1_$3:$4:$5/' backup_2016-30-10_12-00-00

but not sure will be simpler

Answer 4

you can also this script;

#!/bin/bash
fileName=$1
prefix=$(echo ${fileName} | cut -d _ -f1)
date=$(echo ${fileName} | cut -d _ -f2)
time=$(echo ${fileName} | cut -d _ -f3)

year=$(echo ${date} | cut -d - -f1)
day=$(echo ${date} | cut -d '-' -f2)
month=$(echo ${date} | cut -d '-' -f3)

formatedTime=$(echo $time |  sed 's/-/:/g')

formatedDate=$day"-"$month"-"$year

formatedFileName=$prefix"_"$formatedDate"_"$formatedTime

echo $formatedFileName

Eg;

user@host:/tmp$ ./test.sh backup_2016-30-10_12-00-00
backup_30-10-2016_12:00:00