如何使用bash和printf转换日期格式? [英] How to convert date format using bash and printf?

问题描述

我想使用awk和printf在bash中将2019-02-16转换为Feb 16 15:29.

I want to convert 2019-02-16 to Feb 16 15:29 in bash using awk and printf.

例如:

[root@localhost ~]# who | awk '{print $1, $3, $4}'
root 2019-02-16 15:29
root 2019-02-16 15:30
john 2019-02-01 10:34
emmett 2019-01-12 09:45

所需的输出:

root Feb 16 15:29
root Feb 16 15:30
john Feb 1  10:34
emmett Jan 12 09:45

请提供帮助并提供解决方案的说明.

Please help and provide an explanation with your solution.

推荐答案

在任何UNIX盒上的任何shell中使用任何awk:

With any awk in any shell on any UNIX box:

$ who | awk '{split($3,d,/-/); print $1, substr("JanFebMarAprMayJunJulAugSepOctNovDec",(d[2]*3)-2,3), d[3]+0, $4}'

例如:

$ cat file
root 2019-02-16 15:29
root 2019-02-16 15:30
john 2019-02-01 10:34
emmett 2019-01-12 09:45

$ awk '{split($2,d,/-/); print $1, substr("JanFebMarAprMayJunJulAugSepOctNovDec",(d[2]*3)-2,3), d[3]+0, $3}' file
root Feb 16 15:29
root Feb 16 15:30
john Feb 1 10:34
emmett Jan 12 09:45

如果对齐方式很重要，则有多种解决方案，包括使用printf代替print:

and if alignment matters there's various solutions, including using printf instead of print:

$ awk -v OFS='\t' '{split($2,d,/-/); printf "%s %s %-2d %s\n", $1, substr("JanFebMarAprMayJunJulAugSepOctNovDec",(d[2]*3)-2,3), d[3]+0, $3}' file
root Feb 16 15:29
root Feb 16 15:30
john Feb 1  10:34
emmett Jan 12 09:45

或使用制表符(而不是空格)分隔输出:

or separate the output with tabs instead of blanks:

$ awk -v OFS='\t' '{split($2,d,/-/); print $1, substr("JanFebMarAprMayJunJulAugSepOctNovDec",(d[2]*3)-2,3), d[3]+0, $3}' file
root    Feb     16      15:29
root    Feb     16      15:30
john    Feb     1       10:34
emmett  Jan     12      09:45

或将输出通过管道传输到column -t:

or pipe the output to column -t:

$ awk '{split($2,d,/-/); print $1, substr("JanFebMarAprMayJunJulAugSepOctNovDec",(d[2]*3)-2,3), d[3]+0, $3}' file | column -t
root    Feb  16  15:29
root    Feb  16  15:30
john    Feb  1   10:34
emmett  Jan  12  09:45

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