将 tree 命令的输出转换为 json 格式

Question

有没有一种方便的方法可以将 *nix 命令“tree”的输出转换为 JSON 格式？

编辑： 我想我没有很好地描述我的问题。我的目标是转换如下内容：

.
|-- dir1
|   |-- dirA
|   |   |-- dirAA
|   |   `-- dirBB
|   `-- dirB
`-- dir2
    |-- dirA
    `-- dirB

进入：

{"dir1" : [{"dirA":["dirAA", "dirAB"]}, "dirB"], "dir2": ["dirA", "dirB"]}

Answer 1

1.7 版包括对 JSON 的支持：http :
//mama.indstate.edu/users/ice/tree/changes.html

每页man（在 下XML/JSON/HTML OPTIONS）：

-J     Turn on JSON output. Outputs the directory tree as an JSON formatted array.

例如

$ tree -J                                                                                                 

/home/me/trash/tree-1.7.0
[{"type":"directory","name": ".","contents":[
    {"type":"file","name":"CHANGES"},
    {"type":"file","name":"color.c"},
    {"type":"file","name":"color.o"},
    {"type":"directory","name":"doc","contents":[
      {"type":"file","name":"tree.1"},
      {"type":"file","name":"tree.1.fr"},
      {"type":"file","name":"xml.dtd"}
    ]},
    {"type":"file","name":"hash.c"},
    {"type":"file","name":"hash.o"},
    {"type":"file","name":"html.c"},
    {"type":"file","name":"html.o"},
    {"type":"file","name":"INSTALL"},
    {"type":"file","name":"json.c"},
    {"type":"file","name":"json.o"},
    {"type":"file","name":"LICENSE"},
    {"type":"file","name":"Makefile"},
    {"type":"file","name":"README"},
    {"type":"file","name":"strverscmp.c"},
    {"type":"file","name":"TODO"},
    {"type":"file","name":"tree"},
    {"type":"file","name":"tree.c"},
    {"type":"file","name":"tree.h"},
    {"type":"file","name":"tree.o"},
    {"type":"file","name":"unix.c"},
    {"type":"file","name":"unix.o"},
    {"type":"file","name":"xml.c"},
    {"type":"file","name":"xml.o"}
  ]},
  {"type":"report","directories":1,"files":26}
]

Answer 2

尝试 1

仅使用 perl 的解决方案，返回哈希结构的简单哈希。在 OP 阐明 JSON 的数据格式之前。

#! /usr/bin/perl

use File::Find;
use JSON;

use strict;
use warnings;

my $dirs={};
my $encoder = JSON->new->ascii->pretty;

find({wanted => \&process_dir, no_chdir => 1 }, ".");
print $encoder->encode($dirs);

sub process_dir {
    return if !-d $File::Find::name;
    my $ref=\%$dirs;
    for(split(/\//, $File::Find::name)) {
        $ref->{$_} = {} if(!exists $ref->{$_});
        $ref = $ref->{$_};
    }
}

File::Find模块的工作方式与 unixfind命令类似。该JSON模块采用 perl 变量并将它们转换为 JSON。

find({wanted => \&process_dir, no_chdir => 1 }, ".");

将从当前工作目录向下迭代文件结构，process_dir为“.”下的每个文件/目录调用子例程，并no_chdir告诉 perl 不要为chdir()它找到的每个目录发出 a 。

process_dir 如果当前检查的文件不是目录，则返回：

return if !-d $File::Find::name;

然后我们将现有散列的引用抓取%$dirs到 中$ref，拆分文件路径/并循环，for为每个路径添加一个新的散列键。

制作像 slm 那样的目录结构：

mkdir -p dir{1..5}/dir{A,B}/subdir{1..3}

输出是：

{
   "." : {
      "dir3" : {
         "dirA" : {
            "subdir2" : {},
            "subdir3" : {},
            "subdir1" : {}
         },
         "dirB" : {
            "subdir2" : {},
            "subdir3" : {},
            "subdir1" : {}
         }
      },
      "dir2" : {
         "dirA" : {
            "subdir2" : {},
            "subdir3" : {},
            "subdir1" : {}
         },
         "dirB" : {
            "subdir2" : {},
            "subdir3" : {},
            "subdir1" : {}
         }
      },
      "dir5" : {
         "dirA" : {
            "subdir2" : {},
            "subdir3" : {},
            "subdir1" : {}
         },
         "dirB" : {
            "subdir2" : {},
            "subdir3" : {},
            "subdir1" : {}
         }
      },
      "dir1" : {
         "dirA" : {
            "subdir2" : {},
            "subdir3" : {},
            "subdir1" : {}
         },
         "dirB" : {
            "subdir2" : {},
            "subdir3" : {},
            "subdir1" : {}
         }
      },
      "dir4" : {
         "dirA" : {
            "subdir2" : {},
            "subdir3" : {},
            "subdir1" : {}
         },
         "dirB" : {
            "subdir2" : {},
            "subdir3" : {},
            "subdir1" : {}
         }
      }
   }
}

尝试 2

好的，现在使用不同的数据结构......

#! /usr/bin/perl

use warnings;
use strict;
use JSON;

my $encoder = JSON->new->ascii->pretty;   # ascii character set, pretty format
my $dirs;                                 # used to build the data structure

my $path=$ARGV[0] || '.';                 # use the command line arg or working dir

# Open the directory, read in the file list, grep out directories and skip '.' and '..'
# and assign to @dirs
opendir(my $dh, $path) or die "can't opendir $path: $!";
my @dirs = grep { ! /^[.]{1,2}/ && -d "$path/$_" } readdir($dh);
closedir($dh);

# recurse the top level sub directories with the parse_dir subroutine, returning
# a hash reference.
%$dirs = map { $_ => parse_dir("$path/$_") } @dirs;

# print out the JSON encoding of this data structure
print $encoder->encode($dirs);

sub parse_dir {
    my $path = shift;    # the dir we're working on

    # get all sub directories (similar to above opendir/readdir calls)
    opendir(my $dh, $path) or die "can't opendir $path: $!";
    my @dirs = grep { ! /^[.]{1,2}/ && -d "$path/$_" } readdir($dh);
    closedir($dh);

    return undef if !scalar @dirs; # nothing to do here, directory empty

    my $vals = [];                            # set our result to an empty array
    foreach my $dir (@dirs) {                 # loop the sub directories         
        my $res = parse_dir("$path/$dir");    # recurse down each path and get results

        # does the returned value have a result, and is that result an array of at 
        # least one element, then add these results to our $vals anonymous array 
        # wrapped in a anonymous hash
        # ELSE
        # push just the name of that directory our $vals anonymous array
        push(@$vals, (defined $res and scalar @$res) ? { $dir => $res } : $dir);
    }

    return $vals;  # return the recursed result
}

然后在建议的目录结构上运行脚本......

./tree2json2.pl .
{
   "dir2" : [
      "dirB",
      "dirA"
   ],
   "dir1" : [
      "dirB",
      {
         "dirA" : [
            "dirBB",
            "dirAA"
         ]
      }
   ]
}

我发现这非常棘手（特别是考虑到“如果子目录是哈希，如果不是数组，哦，除非顶级，那么无论如何都只是哈希”逻辑）。所以，如果这是你可以用sed/做的事情，我会感到惊讶awk……但我敢打赌，斯蒂芬还没有看过这个：)

Answer 3

这是使用 Perl 和 JSON perl 模块的一种方法。

$ tree | perl -e 'use JSON; @in=grep(s/\n$//, <>); \
     print encode_json(\@in)."\n";'

例子

创建一些示例数据。

$ mkdir -p dir{1..5}/dir{A,B}

这是它的样子：

$ tree 
.
|-- dir1
|   |-- dirA
|   `-- dirB
|-- dir2
|   |-- dirA
|   `-- dirB
|-- dir3
|   |-- dirA
|   `-- dirB
|-- dir4
|   |-- dirA
|   `-- dirB
`-- dir5
    |-- dirA
    `-- dirB

15 directories, 0 files

这是使用 Perl 命令的运行：

$ tree | perl -e 'use JSON; @in=grep(s/\n$//, <>); print encode_json(\@in)."\n";'

返回此输出：

[".","|-- dir1","|   |-- dirA","|   `-- dirB","|-- dir2","|   |-- dirA","|   `-- dirB","|-- dir3","|   |-- dirA","|   `-- dirB","|-- dir4","|   |-- dirA","|   `-- dirB","`-- dir5","    |-- dirA","    `-- dirB","","15 directories, 0 files"]

注意：这只是tree. 不是嵌套层次结构。在我提出这个建议后，OP 改变了这个问题！