unix - 根据unix中的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旧的文件。

例子：

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。你可以用符号告诉它从不同的地方-k FIELD.OFFSET开始，例如，如果所有文件名都以 4 个字母开头，你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