用shell實現(xiàn)創(chuàng)建指定文件夾

linuxbrook

需求：用shell實現(xiàn)創(chuàng)建目錄，一年中12個月，創(chuàng)建12個文件夾，再在這12個文件中分別創(chuàng)建以每個月份多少的天的文件夾，即是：1的文件夾下有31個文件夾，2的文件夾下有29個文件夾。。。。。。。等大牛各位幫解答下，謝謝！

jomenxiao

for a  in $(seq 365)
do
       mkdir  -p   $(date -d "${a} days ago" +%m)/$(date  -d "${a} days ago" +%m%d)
done

回復(fù) 1# linuxbrook


   

Shell_HAT

2的文件夾下有29個文件夾

不管是否閏年都是29個？

ljwd1000

1. #!/bin/bash
   
2. 
   
3. BigMonth=($(seq 1 31))
   
4. SmallMonth=($(seq 1 30))
   
5. 
   
6. TargetDir="/tmp/test"
   
7. 
   
8. echo "Please Input Year:"
   
9. read year
   
10. mode=$(($year%4))
    
11. if [ $mode -eq 0 ];then
    
12.     Feb=($(seq 1 29))
    
13. else
    
14.     Feb=($(seq 1 28))
    
15. fi
    
16. 
    
17. for i in 1 3 5 7 8 10 12
    
18. do
    
19.     for j in ${BigMonth[@]}
    
20.     do
    
21.         mkdir -p $TargetDir/$i/$j
    
22.     done
    
23. done
    
24. 
    
25. for i in 4 6 9 11
    
26. do
    
27.     for j in ${SmallMonth[@]}
    
28.     do
    
29.         mkdir -p $TargetDir/$i/$j
    
30.     done
    
31. done
    
32. 
    
33. for i in ${Feb[@]}
    
34. do
    
35.     mkdir -p $TargetDir/2/$i
    
36. done

復(fù)制代碼

jason680

回復(fù) 1# linuxbrook

How about this

mkdir -p {1..12}/{1..31}
   
maybe you don't want to have 31 on 2,4,6,9,11 months and 2/30

rm -rf {2,4,6,9,11}/31 2/30

cuteorange

回復(fù) 1# linuxbrook

#!/bin/bash
year=${1:-2012}
for mon in {1..12}
do
  mkdir -p $(cal $mon $year | sed -r -e '/[a-zA-Z]/d' -e '/./{H;$!d;}' -e "{x;s/([0-9]{1,2})/$mon\/\1/g;s/\n+/ /g;}")
done

cuteorange

回復(fù) 6# cuteorange


    改進(jìn)一下:
#!/bin/bash
year=${1:-2012}
for mon in {1..12}
do
  mkdir -p $(cal $mon 2012 | sed -r "/[a-z]/d;s/([0-9]{1,2})/$mon\/\1/g")
done