bash - 提取变量的星期几
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bash - extract day of week of variable
提问by toop
What is the syntax to extract the day of the week from a stored date variable?
The dateinfileformat is always [alphanum]_YYYYMMDD.
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。
In this pseudocode example, trying to get dayofweekto store Saturday:
在这个伪代码示例中，尝试dayofweek存储星期六：
#! /bin/bash
dateinfile="P_20090530"
dayofweek="$dateinfile -u +%A"
回答by Shawn Chin
[me@home]$ date --date=${dateinfile#?_} "+%A"
Saturday
Or, to put it as you've requested:
或者，按照您的要求说：
[me@home]$ dayofweek=$(date --date=${dateinfile#?_} "+%A")
[me@home]$ echo $dayofweek
Saturday
回答by golimar
date -d $(echo $dateinfile | cut -f2 -d_) -u +%A
The inner expression separates the 20090530from P_20090530, and the outer one extracts the day of week from that date
内部表达式将20090530from分开P_20090530，外部表达式从该日期中提取星期几
回答by Jason Bullough
I needed the same output recently and googled upon this, but I had the same problem stating Bad Substition error.
我最近需要相同的输出并对此进行了搜索，但是我遇到了同样的问题，指出 Bad Substition 错误。
I then read the date manual and made my version up as follows:
然后我阅读了日期手册并按如下方式制作了我的版本：
#! /bin/sh
dateinfile="P_20090530"
dayofweek=`date --reference $dateinfile +%A`
echo $dayofweek
