I want awk etc to show all fields after a certain field

LANG=Cls-laN$1|awk'{
    re="^([^]+){7}[^]+"
    match ($0,re)
    num = RLENGTH+1
    print substr ($0,num)
}'

When you want to cut out a file that is the result of ls in awk's script, it's troublesome if the file name contains spaces.
The default delimiter for awk is blank, but the number of consecutive delimiters is interpreted as one, so once you break it down into fields, you cannot restore the number of blanks.

When I wanted to use the results of ls, the only solution I could think of was to delete the fields up to the filename.
In this environment, we have removed the first 50 characters from $0, since the 51st character was the file name.
If you run ls without the N option, the N option suppresses the file because it contains quotation marks before and after the file containing the blank space.

Version of ls used for verification.
ls(GNU coreutils) 8.28

Additional
"You pointed out that ""the starting position of the file name changes"" depending on the situation, so I modified the code."
Supports file names that begin with a blank space.

@cubick, if you can specify the information to display with the arguments of the command, you can use it.

If you want to retrieve information from information or commands that you cannot do, or if you want to extract it from a file that contains all the information you have already obtained and has been texted, can you use index or match and substr?

How to use the match() function in awk
cut strings with match() and substr() functions

How to specify the start position of substr() with index() in awk

How to use substr() to cut strings in awk

File name, folder name like this:
(If there are no 9 fields, another field was displayed, so I checked with the minimum digit position.)

#!/bin/sh
ls-la | awk'
{
  match ($0,$9)
  if(RSTART>=17){
    datext = substr ($0, RSTART)
    print datext
  }
}
'

If you want to cut out of the file date field, simply shift the number:

#!/bin/sh
ls-la | awk'
{
  match ($0, $6)
  if(RSTART>=11){
    datext = substr ($0, RSTART)
    print datext
  }
}
'

Use index instead of match here:

#!/bin/sh
ls-la | awk'
{
  column=index($0,6)
  if(column>=11){
    datext = substr ($0, column)
    print datext
  }
}
'

As Cubick pointed out, it would be better to use ls-1a (or ls-1A).

It is surprisingly troublesome to extract a file name that reproduces a blank space from the "ls-lA" output.
I tried the following just in case.
The split() function divides the column and separator into segments, and then outputs the necessary parts.
For ease of viewing, we place "|" at the beginning and end of the output.

$touch "000" "aaa" "bbbb" "cccccccccccc" "dddddddddd"
$ ls-lA
Total 0
-rw-rw-r -- 1 hidezz hidezz 0 June 22 11:21 '000'
-rw-rw-r -- 1 hidezzz hidezzz 0 June 22 11:21 aaa
-rw-rw-r -- 1 hidezzz hidezzz 0 June 22 11:21 'bbbbbb'
-rw-rw-r -- 1 hidezzz hidezzz 0 June 22 11:21 'cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc
-rw-rw-r--1 hidezzz hidezzz 0 June 22 11:21 'dddddddd'
$ls-lA | awk'NR>1 {split($0, a, "", sep); printf" | "substr(sep[8], 2);for(i=9;i<=length(a);++i){printfa[i]sep[i]}print"|"}}}'
|   000|
|aaa|
| bbbbbb|
| ccc ccc ccc |
|dddddddddd|