排序问题请教

sunbw001

[root@localhost public]# cat data
b_cs_iter_1.am
a_c_cs_iter_1.am
b_cs_iter_10.am
c_b_cs_iter_4.am
b_cs_iter_11.am
a_c_cs_iter_12.am
b_cs_iter_2.am
c_b_cs_iter_11.am
a_c_cs_iter_13.am
b_cs_iter_6.am
c_b_cs_iter_1.am
a_c_cs_iter_5.am

想要的结果：
a_c_cs_iter_1.am
a_c_cs_iter_5.am
a_c_cs_iter_12.am
a_c_cs_iter_13.am
b_cs_iter_1.am
b_cs_iter_2.am
b_cs_iter_6.am
b_cs_iter_10.am
b_cs_iter_11.am
c_b_cs_iter_1.am
c_b_cs_iter_4.am
c_b_cs_iter_11.am

就是iter前面按字母顺序排序，前面相同的按iter后的数字按大小排序

我是DBA

[test@ ttt ] $ sort -t_ -k1,1 -k4,4n -k5,5n data
a_c_cs_iter_1.am
a_c_cs_iter_5.am
a_c_cs_iter_12.am
a_c_cs_iter_13.am
b_cs_iter_1.am
b_cs_iter_2.am
b_cs_iter_6.am
b_cs_iter_10.am
b_cs_iter_11.am
c_b_cs_iter_1.am
c_b_cs_iter_4.am
c_b_cs_iter_11.am

ywlscpl

通用点的，不管用_分为几个域

[root@Mylinux 1]# cat file
b_zzz_jjj_dddd_xxxx_cs_iter_1.am
a_c_cs_iter_1.am
b_cs_iter_10.am
c_b_cs_iter_4.am
b_cs_iter_11.am
a_c_cs_iter_12.am
b_cs_iter_2.am
c_b_cs_iter_11.am
a_c_cs_iter_13.am
b_cs_iter_6.am
c_b_cs_iter_1.am
a_xxx_yyyc_cs_iter_5.am
[root@Mylinux 1]# awk -F '[_.]' '{print $(NF-1)"_"$0}' file|sort -t_ -k2,2 -k1,1n |awk '{sub(/^[^_]*_/,"")}1'
a_c_cs_iter_1.am
a_xxx_yyyc_cs_iter_5.am
a_c_cs_iter_12.am
a_c_cs_iter_13.am
b_zzz_jjj_dddd_xxxx_cs_iter_1.am
b_cs_iter_2.am
b_cs_iter_6.am
b_cs_iter_10.am
b_cs_iter_11.am
c_b_cs_iter_1.am
c_b_cs_iter_4.am
c_b_cs_iter_11.am

[ 本帖最后由 ywlscpl 于 2009-4-23 20:00 编辑 ]

cssjtuer

sub(/^[^_]*_/,"")
请问这里为什么不是 正则表达式的 贪婪匹配原则， 而正好只把加在前面的：number_  给替换为空了？
我知道有个gsub是替换所有出现的，sub 是替换第一个 这里的第一个不能使用贪婪匹配原则进行扩展么？

greendays

sub 是替换第一个  就对了

blackold

try:

1. sed 's/iter_/ /' urfile|sort -t ' ' -k 1,1 -k 2n|sed 's/ /iter_/'

复制代码

我是DBA

因为前面有^开头。