根据unix中的文件名Timestamp查找目录中最旧的文件

Question

我希望首先列出基于文件名日期和时间戳的目录中最旧的文件.

例:

输入文件 :

AAAG11020709581.txt
AAAG13020709581.txt
AACL11020709581.txt
AACL13020709581.txt
AAFU11020709581.txt
AAFU13020709581.txt
AAHO11020709581.txt
AAHO13020709581.txt
AAPC11020709581.txt
AAPC13020709581.txt
AAPO11020709581.txt
AAPO13020709581.txt
AATR11020709581.txt
AATR13020709581.txt
AARC11020709581.txt
AARC13020709581.txt

预期产量:

AAAG11020709581.txt
AACL11020709581.txt
AAFU11020709581.txt
AAHO11020709581.txt
AAPC11020709581.txt
AAPO11020709581.txt
AARC11020709581.txt
AATR11020709581.txt
AAAG13020709581.txt
AACL13020709581.txt
AAFU13020709581.txt
AAHO13020709581.txt
AAPC13020709581.txt
AAPO13020709581.txt
AARC13020709581.txt
AATR13020709581.txt

任何人都可以建议吗？

Answer 1

默认情况下,排序将以行的开头作为键排序.您可以使用-k FIELD.OFFSET符号告诉它从不同的地方开始,例如,如果所有文件名以4个字母开头,您可以跳过这些:

sort -k1.5

输出:

AAAG11020709581.txt
AACL11020709581.txt
AAFU11020709581.txt
AAHO11020709581.txt
AAPC11020709581.txt
AAPO11020709581.txt
AARC11020709581.txt
AATR11020709581.txt
AAAG13020709581.txt
AACL13020709581.txt
AAFU13020709581.txt
AAHO13020709581.txt
AAPC13020709581.txt
AAPO13020709581.txt
AARC13020709581.txt
AATR13020709581.txt