我想对find命令的输出进行唯一排序，在任何目录中都没有重复的文件名。
find /path/to/first_directory/* /path/to/second_directory/* /path/to/third_directory/* -mtime -1 -name "filename_pattern*"
示例输出：
/path/to/first_directory/sample_file1_2017Dec25.dat
/path/to/first_directory/sample_file2_2017Nov01.dat
/path/to/first_directory/sample_file3_2017Oct08.dat
/path/to/first_directory/archive/sample_file1_2017Dec25.dat.Z
/path/to/first_directory/archive/sample_file2_2017Nov01.dat.Z
/path/to/second_directory/sample_file4_2017Sep11.dat
/path/to/second_directory/sample_file5_2017Oct05.dat
/path/to/third_directory/sample_file1_2017Dec25.dat
/path/to/third_directory/sample_file2_2017Nov01.dat
/path/to/third_directory/sample_file3_2017Oct08.dat
/path/to/third_directory/sample_file4_2017Sep11.dat
/path/to/third_directory/sample_file5_2017Oct05.dat
/path/to/third_directory/sample_file6_2017July04.dat
/path/to/third_directory/sample_file6_2017June12.dat
/path/to/third_directory/sample_file7_2017May01.dat
从输出中，你可以看到，在重复的文件名/first_directory/，并/first_directory/archive/还从所有文件/first_directory/*和/second_directory/*也内/third_directory/*。这意味着/third_directory/*是所有文件的存档目录中发现的/first_directory/*和/second_directory/*，但也有只能在被找到的文件/third_directory/*（检查sample_file6和sample_file7）
所有我想要打印的文件从始发/first_directory/到/first_directory/archive/到/second_directory/到/third_directory/这个顺序不存在重复，也通过他们的日期排序。
所需的输出：
/path/to/first_directory/sample_file1_2017Dec25.dat
/path/to/first_directory/sample_file2_2017Nov01.dat
/path/to/first_directory/sample_file3_2017Oct08.dat
/path/to/second_directory/sample_file4_2017Sep11.dat
/path/to/second_directory/sample_file5_2017Oct05.dat
/path/to/third_directory/sample_file6_2017July04.dat
/path/to/third_directory/sample_file6_2017June12.dat
/path/to/third_directory/sample_file7_2017May01.dat
如果find命令的输出保存在名为的文件中filelist，请尝试：
$ awk -F/ '{f=$NF; sub(/\.Z$/,"",f)} !a[f]++' filelist
/path/to/first_directory/sample_file1_2017Dec25.dat
/path/to/first_directory/sample_file2_2017Nov01.dat
/path/to/first_directory/sample_file3_2017Oct08.dat
/path/to/second_directory/sample_file4_2017Sep11.dat
/path/to/second_directory/sample_file5_2017Oct05.dat
/path/to/third_directory/sample_file6_2017July04.dat
/path/to/third_directory/sample_file6_2017June12.dat
/path/to/third_directory/sample_file7_2017May01.dat
如果要执行相同的操作而不创建文件，请执行以下操作：
find /path/to/first_directory/* /path/to/second_directory/* /path/to/third_directory/* -mtime -1 -name "filename_pattern*" | awk -F/ '{f=$NF; sub(/\.Z$/,"",f)} !a[f]++'
或者，您更喜欢将命令分布在多行中，请使用：
find /path/to/first_directory/* /path/to/second_directory/* \
/path/to/third_directory/* -mtime -1 -name "filename_pattern*" |
awk -F/ '{f=$NF; sub(/\.Z$/,"",f)} !a[f]++'
我们将其添加\到第一行末尾的位置，因为那是bash的行连续字符。因为第二行以结尾|，所以不需要行继续符。
首先，find按照优先级顺序在命令中列出目录很重要。我知道您已经做到了。
-F/
这告诉awk/用作字段分隔符。这意味着文件名将是最后一个字段$NF。
f=$NF; sub(/\.Z$/,"",f)
这会将文件名分配给变量f，然后.Z从中删除最后一个（f如果存在）。
!a[f]++'
如果f以前没有看到过，请打印此行。
根据评论，.Z不是唯一需要删除的扩展名。可能有其他的扩展.dat.edi，并.dat.bak应与简单地更换.dat。在这种情况下：
awk -F/ '{f=$NF; sub(/\.dat.*/,".dat",f)} !a[f]++' filelist
awk -F/ '{f=$NF; sub(/\.dat.*/,".dat",f)} !a[f]++' filelist | xargs -d'\n' -r ls -t
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