unix - 在unix中根据文件名时间戳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早的文件。

示例:

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

有人可以建议吗？

Answer 1

默认情况下，排序将以行的开头作为键进行排序。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都以 4 个字母开头，您可以像这样跳过这些:

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