How to print only txt files on linux terminal? - linux

Question

On my Linux directory I have 6 files. 5 files are txt files and 1 file a .tar.gz type file. How can I print to the terminal only the name of the txt files?
directory :dir
content:
ex1, ex2, ex3, ex4, ex5, ex6.tar.gz

Answer 1

Because you do not have a file extension (.txt) I would try to do it with exclusion.
ls | grep -v tar.gz
If you have multiple types then use extensions.

Answer 2

The command 'file', followed by the name of a file, will return the type of the file.
You can loop over the files in your directory, use each filename as input to the 'file' command, and if it is a text file, print that filename.
The following includes some extra output from the file command, which I'm not sure how to remove yet, but it does give you the filenames you want:
#!/bin/bash
for f in *
do
file $f | grep text
done
You can put this into a shell script in the directory you want to get the filenames from, and run it from the command line.

Answer 3

The suggestions of using the file command are correct. The problem here is parsing the output of this command, because (1) file names can contain pretty any character, and (2) the concrete output of the file command is a bit unpredictable, because it depends on how the so called magic files are present.
If we rely on the fact that the explanation text of the output of the file command - i.e. that part which explains what file it is - always contains the word text if it is a text file, and that it never contains a colon, we can process it as follows:
The last colon in the output must separated the filename from the explanation. Everything to the left is the filename, and if the word text (note the leading space before text!) occurs in the right part, we have a text file.
This still leaves us with those (hopefully rare) cases where a file name contains a non-printable character, they would be translated to their octal equivalent, which might or might not be what you want to see. You can suppress this by passing the -r option to the file command. This is useful if you want to process this filename further instead of just displaying it to the user, but it might corrupt your parsing logic, especially if the filename contains a newline.
Finally, don't forget that in any case, you see what the system considers a text file. This is not necessarily the same what you define to be a text file.

Answer 4

Updated Answer
As #hek2mgl points out in the comments, a more robust solution is to separate filenames using nul characters (which may not occur in filenames) and that will deal with filenames containing newlines, and colons:
file -0 * | awk -F'\0' '$2 ~ /text/{print $1}'
Original Answer
I would do this:
file * | awk -F: '$2~/text/{print $1}'
That runs file to see the type of each file and passes the names and types to awk separated by a colon. awk then looks for the word text in the second field and if it finds it, prints the first field - which is the filename.
Try running the following simpler command on its own to see how it works:
file *

Answer 5

Given this directory of files:
$ file *
1.txt: UTF-8 Unicode (with BOM) text, with CRLF line terminators
2.pdf: PDF document, version 1.5
3.pdf: PDF document, version 1.5
4.dat: data
5.txt: ASCII text
6.jpg: JPEG image data, JFIF standard 1.02, aspect ratio, density 100x100, segment length 16, baseline, precision 8, 2833x972, frames 3
7.html: HTML document text, UTF-8 Unicode text, with very long lines, with no line terminators
8.js: UTF-8 Unicode text
9.xml: XML 1.0 document text
A.pl: a /opt/local/bin/perl script text executable, ASCII text
B.Makefile: makefile script text, ASCII text
C.c: c program text, ASCII text
D.docx: Microsoft Word 2007+
You can see the only files that are pure ascii are 5.txt, 9.xml, and A-C. The rest are either binary or UTF according to file.
You can use a Bash glob to loop through files and use file to test each file. This save having to parse the output of file for the file names but relies on file to accurate identify what you consider to be 'text':
for fn in *; do
[ -f "$fn" ] || continue
fo=$(file "$fn")
[[ $fo =~ ^"$fn":.*text ]] || continue
echo "$fn"
done
If you cannot use file, which is certainly the easiest way, you can open the file and look for binary characters. Use Perl for that:
for fn in *; do
[ -f "$fn" ] || continue
head -c 2000 "$fn" | perl -lne '$tot+=length; $cnt+=s/[^[:ascii:]]//g; END{exit 1 if($cnt/$tot>0.03);}'
[ $? -eq 0 ] || continue
echo "$fn"
done
In this case, I am looking for a percentage of ascii vs non ascii in the first 2000 bytes of a file. YMMV but that allows finding a file that file would report as UTF (since it has a binary BOM) but most of the file is ascii.
For that directory, the two Bash scripts report (with my comments on each file):
1.txt # UTF file with a binary BOM but no UTF characters -- all ascii
4.dat # text based configuration file for a router. file does not report this
5.txt # Pure ascii file
7.html # html file
8.js # Javascript sourcecode
9.xml # xml file all text
A.pl # Perl file
B.Makefile # Unix make file
C.c # C source file
Since file does not consider the all ascii file 4.dat to be text, it is not reported by the first Bash script but is by the second. Otherwise -- same output.