"find \ awk“命令(Linux)所需的解释

Question

我正在试图找出下面的命令；

find ./ -type f |
awk -F / -v OFS=/ '{$NF="";dir[$0]++} END {for (i in dir) print dir[i]i}'

输出如下：

6./Release_1_18_1_0_06_26/metadata/3_Control_Files/   
5./Release_1_18_1_0_06_26/metadata/7_SAS_Code/   
5./Release_1_18_1_0_06_26/others/1_content/   
1./.cache/pip/selfcheck/   
2./Release_1_18_1_0_06_26/metadata/5_Status/   
1./Release_1_18_1_0_06_26/compute/2_packages/   
1./sasuser.v94/   
4./metadata/FR1_Release_1.17.1.1/3_Control_Files/   
4./Release_1_18_1_0_06_26/metadata/6_Patches/   

此命令正在计算当前路径中的代码数。但是，我不理解{$NF="";dir[$0]++} END {for (i in dir) print dir[i]i}，特别是dir[$0]。有人能解释吗？

Answer 1

请您在这里详细解释OP的代码，好吗？

find ./ -type f |     ##Running find command to find files in current directory and passing output as input to awk command.
awk -F / -v OFS=/ '   ##Running awk command setting field separator and output field separator as /
{
  $NF=""              ##Nullifying last field of current line here.
  dir[$0]++           ##creating array dir with current line and keep increasing its value with one here.
}
END{                  ##starting END block of this code here.
  for(i in dir){      ##traversing through dir array here.
    print dir[i]i     ##printing index of dir array and it's index here.
  }
}'

当最终完成NOTE:END读取时，将执行任何awk代码的Input_file块。

Answer 2

-F /告诉awk用斜线分隔行。$0是整条线。

$NF=""将行中的最后一项(在本例中为文件名)替换为空。

然后，dir[$0]++取整行(在文件名被删除后)，并将其用作哈希索引，将该值递增1。有效地计算具有相同路径的项目数。

END块循环遍历dir[]哈希打印中的所有键，首先是计数，然后是目录名。