bash - 提取变量的星期几

too*_*oop 1 unix linux bash shell scripting

从存储的日期变量中提取星期几的语法是什么?该dateinfile格式始终[alphanum]_YYYYMMDD.

在这个伪代码示例中,试图在dayofweek星期六存储:

#! /bin/bash 

dateinfile="P_20090530"
dayofweek="$dateinfile -u +%A"
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Sha*_*hin 7

[me@home]$ date --date=${dateinfile#?_} "+%A"
Saturday
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或者,按照您的要求提出:

[me@home]$ dayofweek=$(date --date=${dateinfile#?_} "+%A")
[me@home]$ echo $dayofweek
Saturday
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