从当前目录获取文件并在同一行打印具有相同扩展名的文件

Question

脚本应该列出当前目录下除后缀的文件（并以不带后缀的形式结束列表）。

示例（也称为“.extension:”）

.c: first.c main.c var.c
.h: const.h first.h
.odt: relazione.odt
makefile README COPYING

我正在尝试使用 ls、sort、uniq，但我做不到。 有人可以帮助我吗？

我用python写了一个解决方案：

#!/usr/bin/env python3
import os

dictionary={}

for dirpath,_,files in os.walk("./"):
    for f in files:
        path = os.path.abspath(os.path.join(dirpath, f))
        _ , ext = os.path.splitext(path)
        if ext not in dictionary:
            dictionary[ext] = []
        dictionary[ext].append(f)

for k in dictionary:
    if k != "":
        print(k, end=": ")
        for i in dictionary.get(k):
            print(i, end= " ")
        print("")

for i in dictionary.get(""):
    print(i, end= " ")
print("")

Answer 1

给定：

$ ls -lndp *
-rw-r--r--  1 501  0   0 May  1 15:01 COPYING
-rw-r--r--  1 501  0   0 May  1 15:01 README
-rw-r--r--  1 501  0   0 May  2 06:36 a.b.c
drwxr-xr-x  2 501  0  68 May  2 06:15 adir.dir/
-rw-r--r--  1 501  0   0 May  1 15:00 const.h
-rw-r--r--  1 501  0   0 May  2 08:00 file name with a space.doc
-rw-r--r--  1 501  0   0 May  1 15:00 first.c
-rw-r--r--  1 501  0   0 May  1 15:00 first.h
-rw-r--r--  1 501  0   0 May  1 15:00 main.c
-rw-r--r--  1 501  0   0 May  1 15:01 makefile
-rw-r--r--  1 501  0   0 May  1 15:01 relazione.odt
-rw-r--r--  1 501  0   0 May  1 15:00 var.c

假设您不想包含目录，您可以过滤掉目录并将文件名提供给awk：

awk '{n=split($0, parts, /\./)
      ext = ( n>1 ? parts[n] : "none" )
      ext2files[ext] = ext2files[ext] OFS $0
      }
   END{
      for (ext in ext2files) 
         print ext ":" ext2files[ext]
      }' <(for fn in *; do  [ ! -d "$fn" ] && echo "$fn"; done)
h: const.h first.h
none: COPYING README makefile
odt: relazione.odt
doc: file name with a space.doc
c: a.b.c first.c main.c var.c

如果你想对输出进行排序，并且你有gawk，你可以编写一个比较函数：

gawk ' function cmp_idx(i1, v1, i2, v2) {
         if (i1=="" || i2=="")
            return (i1=="") ? 1 : -1 
         return (i1 < i2) ? -1 : (i1 != i2)
      }
      {
      n=split($0, parts, /\./)
      ext = ( n>1 ? parts[n] : "" )
      ext2files[ext] = (ext2files[ext] ? ext2files[ext] OFS $0 : $0)
      }
   END{
      PROCINFO["sorted_in"] = "cmp_idx"
      for (ext in ext2files) 
         print (ext ? ext ": " ext2files[ext] : ext2files[ext])
      }' <(for fn in *; do  [ ! -d "$fn" ] && echo "$fn"; done)
c: a.b.c first.c main.c var.c
doc: file name with a space.doc
h: const.h first.h
odt: relazione.odt
COPYING README makefile

Answer 2

#!/bin/bash
ls -d *.* 2>/dev/null | cut -d. -f2- | sort -u | while read i ; do
    echo -n ."$i":' '
    for j in *."$i" ; do
         echo -n "$j"' '
    done
    echo
done
ls | grep -v '\.' | tr '\n' ' '

Answer 3

awk '
BEGIN {
    for (i=1; i<ARGC; i++) {
       fname = ARGV[i]
       n = split(fname,parts,/\./)
       ext = ( n>1 ? parts[n] : "none" )
       ext2files[ext] = ext2files[ext] OFS fname
    }
    for (ext in ext2files) {
        print ext ":" ext2files[ext]
    }
    exit
}' *

Answer 4

可以使用 Perl。

use warnings;
use strict;

die "Too many args. Please supply one directory\n" if @ARGV > 1;
die "Too few args. Please supply one directory\n" if @ARGV < 1;

opendir (my $dir, "$ARGV[0]") || die "$ARGV[0]: $!\n" ;
my %extfiles;
my @Others;
while (my $file = readdir $dir){
        next if (-d "./$file");
        if($file =~ /^..*(\.[^\.]*)$/){
                $extfiles{$1}=$extfiles{$1}?"$extfiles{$1} $file":"$file"
        }
        else{
                push @Others,$file
        }
}
closedir $dir;

#print "Files with extensions\n\n";
for my $extension (keys %extfiles){
        print "$extension: $extfiles{$extension}\n";
}

#print "\nFiles without extensions\n\n";

print join (" ",@Others),"\n";

保存在文件中，例如

ListExt.pl

运行方式：

perl ListExt.pl $Dir

Answer 5

您可以使用这个find + awk 脚本：

find . -maxdepth 1 -type f -print0 |
awk -v RS='\0' -F. '{$0 = substr($0, 3)} NF>1{ext[$NF] = ext[$NF] OFS $0; next}
{noext = noext $0 OFS} END{for (e in ext) print "." e ":" ext[e]; print noext}'