[求助] AIX中如何实现在一个目录下建立连续数字的子目录

liion631818

mkdir {1..31}
zooyo 发表于 2011-06-28 17:12

    这个牛叉。。。

lzx4456

#!/bin/sh
for((i=1;i<=31;i++)); do
   mkdir ${i}
done

zm_wl

回复 18# ly5066113


    谢谢，正解！ 呵呵

unixlinuxsys

1. [root@thinkpad ibm]# cd 2010/2011/
   
2. [root@thinkpad 2011]# dirlist=`seq -w -s " " 1 31`
   
3. [root@thinkpad 2011]# mkdir $dirlist
   
4. [root@thinkpad 2011]# ls
   
5. 01  03  05  07  09  11  13  15  17  19  21  23  25  27  29  31
   
6. 02  04  06  08  10  12  14  16  18  20  22  24  26  28  30
   

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unixlinuxsys

#!/bin/sh
for((i=1;i
lzx4456 发表于 2011-06-28 17:47

1. [root@thinkpad 01]# #!/bin/sh
   
2. [root@thinkpad 01]# for((i=1;i<=31;i++));do
   
3. > mkdir ${i}
   
4. > done
   
5. [root@thinkpad 01]# ls
   
6. 1   11  13  15  17  19  20  22  24  26  28  3   31  5  7  9
   
7. 10  12  14  16  18  2   21  23  25  27  29  30  4   6  8
   
8. [root@thinkpad 01]#
   

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为啥俩括号？

xiaopan3322

回复 25# unixlinuxsys


   

Similar to the let command, the ((...)) construct permits arithmetic expansion and evaluation. In its simplest form, a=$(( 5 + 3 )) would set "a" to "5 + 3", or 8. However, this double parentheses construct is also a mechanism for allowing C-type manipulation of variables in Bash.