根据年周,返回周一的具体日期（注：系统返回2012-1-1为52周）

plumdavid

根据年周数,如2013年32周,返回此周的周一日期.
注:周以星期一为第一天.

谢谢各位回复，大家多数的算法都是错误的。

代号：军刀

1. awk -vl=32 'BEGIN{t1=mktime("2013 01 01 00 00 00")+(l*7*24*3600);t2=(strftime("%u",tstamp)-1)*3600;print strftime("%Y%m%d",t1-t2)}'

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plumdavid

回复 2# 代号：军刀


    谢谢.
    刚研究了一下shell年周数,系统设定的有点复杂,比如:2012年1月1日,年周数是2011年52周.
    这个问题我还得再研究下.

plumdavid

1. [nms@localhost week]$ date -d '2012-01-01 00:00:01' '+%V'
   
2. 52

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jason680

回复 4# plumdavid

maybe you want to know there are many spec to define week date different

http://en.wikipedia.org/wiki/ISO_week_date
The ISO week date system is a leap week calendar system that is part of the ISO 8601 date and time standard. The system is used (mainly) in government and business for fiscal years, as well as in timekeeping.
...

plumdavid

jason680 发表于 2013-08-16 16:16
回复 4# plumdavid

maybe you want to know there are many spec to define week date different

多谢。

guogang225

wk=32;date -d "20121231 $wk week" +%Y-%m-%d

关阴月飞

回复 1# plumdavid

1. #!/bin/bash
   
2. Y=$1
   
3. W=$2
   
4. D=$[`date -d "${Y}0101" +%u`-1]
   
5. d=$[(W-1)*7-D]
   
6. 
   
7. date -d "${Y}0101 +${d} days" "+%G-%m-%d 第%V周 星期%u"

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1. [root@ awktest]# cat a.sh
   
2. #!/bin/bash
   
3. Y=$1
   
4. W=$2
   
5. D=$[`date -d "${Y}0101" +%u`-1]
   
6. d=$[(W-1)*7-D]
   
7. 
   
8. date -d "${Y}0101 +${d} days" "+%G-%m-%d 第%V周 星期%u"
   
9. [root@ awktest]# ./a.sh 2013 32
   
10. 2013-08-05 第32周 星期1
    
11. [root@ awktest]# ./a.sh 2013 34
    
12. 2013-08-19 第34周 星期1
    
13. [root@ awktest]# ./a.sh 2013 3
    
14. 2013-01-14 第03周 星期1
    
15. [root@ awktest]# ./a.sh 2013 25
    
16. 2013-06-17 第25周 星期1
    
17. [root@ awktest]#

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jason680

回复 6# plumdavid

which one do you like (%V,%U,%W) ...

# man date | grep '%[UVW]'
       %g     the 2-digit year corresponding to the %V week number
       %G     the 4-digit year corresponding to the %V week number
       %U     week number of year with Sunday as first day of week (00..53)
       %V     week number of year with Monday as first day of week (01..53)
       %W     week number of year with Monday as first day of week (00..53)

# seq 1 10 | xargs -i date -d "2012-1-{} 0:0:1" +"%V,%U,%W,%w"
52,01,00,0
01,01,01,1
01,01,01,2
01,01,01,3
01,01,01,4
01,01,01,5
01,01,01,6
01,02,01,0
02,02,02,1
02,02,02,2

dn833

1. #!/bin/bash
   
2. YEAR=2013
   
3. WEEK=32
   
4. DAY=$(echo $((WEEK*7-1)))
   
5. DATE=($(date -d "${YEAR}0101 +$DAY day" '+%Y %m %d'))
   
6. 
   
7. cal -m ${DATE[1]} ${DATE[0]}|awk -v year=${DATE[0]} -v month=${DATE[1]} -v day=${DATE[2]} '$0~day{print year,month,$1}'

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