如何在 Bash 脚本中解码希捷的硬盘驱动器日期代码
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How to decode Seagate's hard drive date code in a Bash script
提问by starfry
Seagatehard drives display a code instead of the manufacturing date. The code is described hereand an online decoder is available here.
希捷硬盘显示代码而不是制造日期。该代码是描述在这里和在线译码器可用在这里。
In short, it's a 4 or 5 digit number of the form YYWWDor YYWD, where:
简而言之，它是形式YYWWDor的 4 位或 5 位数字YYWD，其中：
YYis the year, 00 is year 1999WorWWis the week number beginning 1Dis day of week beginning 1- Week 1 begins on the first saturday of July in the stated year
YY是年份，00 是 1999 年W或者WW是从 1 开始的周数D是星期几从 1 开始- 第 1 周从指定年份的 7 月的第一个星期六开始
Examples
例子
- 06212 means Sunday 20 November 2005
- 0051 means Saturday 31 July 1999
- 06212 表示 2005 年 11 月 20 日星期日
- 0051 表示 1999 年 7 月 31 日星期六
How can this be decoded in a bashscript ?
这如何在bash脚本中解码？
采纳答案by thriqon
This is what I did, it should work:
这就是我所做的，它应该可以工作：
#!/bin/bash
DATE=
REGEX="^(..)(..?)(.)$"
[[ $DATE =~ $REGEX ]]
YEAR=$(( ${BASH_REMATCH[1]} + 1999 ))
WEEK=$(( ${BASH_REMATCH[2]} - 1))
DAYOFWEEK=$(( ${BASH_REMATCH[3]} - 1))
OFFSET=$(( 6 - $(date -d "$YEAR-07-01" +%u) ))
DATEOFFIRSTSATURDAY=$(date -d "$YEAR-7-01 $OFFSET days" +%d)
FINALDATE=`date -d "$YEAR-07-$DATEOFFIRSTSATURDAY $WEEK weeks $DAYOFWEEK days"`
echo $FINALDATE
It worked for the two dates given above... If you want to customize the date output, add a format string at the end of the FINALDATe assignment.
它适用于上面给出的两个日期...如果您想自定义日期输出，请在 FINALDATe 分配的末尾添加一个格式字符串。
回答by starfry
Here is a short script, it takes two arguments: $1is the code to convert and $2is an optional format (see man date), otherwise defaulted (see code).
这是一个简短的脚本，它有两个参数：$1是要转换的代码，$2是可选格式（请参阅man date），否则为默认格式（请参阅代码）。
It uses the last Saturday in June instead of the first one in July because I found it easer to locate and it allowed me to just add the relevant number of weeks and days to it.
它使用 6 月的最后一个星期六而不是 7 月的第一个星期六，因为我发现它更容易定位，并且它允许我只添加相关的周数和天数。
#!/bin/bash
date_format=${2:-%A %B %-d %Y}
code=
[[ ${#code} =~ ^[4-5]$ ]] || { echo "bad code"; exit 1; }
let year=1999+${code:0:2}
[[ ${#code} == 4 ]] && week=${code:2:1} || week=${code:2:2}
day=${code: -1}
june_last_saturday=$(cal 06 ${year} | awk '{ && X= } END { print X }')
date -d "${year}-06-${june_last_saturday} + ${week} weeks + $((${day}-1)) days" "+${date_format}"
Examples:
例子：
$ seadate 06212
Sunday November 20 2005
$ seadate 0051
Saturday July 31 1999
回答by Jared Palmer
I created a Seagate Date Code Calculator that actually works with pretty good accuracy. I've posted it here on this forum for anyone to use: https://www.data-medics.com/forum/seagate-date-code-conversion-translation-tool-t1035.html#p3261
我创建了一个 Seagate 日期代码计算器，它实际上可以非常准确地工作。我已将其发布在此论坛上供任何人使用：https: //www.data-medics.com/forum/seagate-date-code-conversion-translation-tool-t1035.html#p3261
It's far more accurate than the other ones online which often point to the entirely wrong year. I know it's not a bash script, but will still get the job done for anyone else who's searching how to do this.
它比网上的其他那些经常指向完全错误的年份要准确得多。我知道它不是 bash 脚本，但仍然可以为正在搜索如何执行此操作的其他任何人完成工作。
Enjoy!
享受！
