How to store NUL output of a program in bash script?

Question

Suppose there is a directory 'foo' which contains several files:

ls foo:
1.aa 2.bb 3.aa 4.cc

Now in a bash script, I want to count the number of files with specific suffix in 'foo', and display them, e.g.:

SUFF='aa'
FILES=`ls -1 *."$SUFF" foo`
COUNT=`echo $FILES | wc -l`
echo "$COUNT files have suffix $SUFF, they are: $FILES"

The problem is: if SUFF='dd', $COUNT also equal to 1. After google, the reason I found is when SUFF='dd', $FILES is an empty string, not really the null output of a program, which will be considered to have one line by wc. NUL output can only be passed through pipes. So one solution is:

COUNT=`ls -1 *."$SUFF" foo | wc -l`

but this will lead to the ls command being executed twice. So my question is: is there any more elegant way to achieve this?

the file-names end with foo? Can you give one such sample name? Do you want to recursively search in sub-directories also? — Inian
– Inian, Commented Sep 28, 2016 at 10:12
foo is the directory. file-names end with aa,bb,cc. Not need to search sub-directories. — xlwang
– xlwang, Commented Sep 28, 2016 at 10:15

Ignacio Vazquez-Abrams · Accepted Answer · 2016-09-28 10:16:44Z

2

$ shopt -s nullglob
$ FILES=(*)
$ echo "${#FILES[@]}"
4
$ FILES=(*aa)
$ echo "${#FILES[@]}"
2
$ FILES=(*dd)
$ echo "${#FILES[@]}"
0
$ SUFFIX=aa
$ FILES=(*"$SUFFIX")
$ echo "${#FILES[@]}"
2
$ SUFFIX=dd
$ FILES=(*"$SUFFIX")
$ echo "${#FILES[@]}"
0

answered Sep 28, 2016 at 10:16

Ignacio Vazquez-Abrams

804k160 gold badges1.4k silver badges1.4k bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

123 Over a year ago

Wow great explanation!

Ignacio Vazquez-Abrams Over a year ago

@123: That's funny, I didn't write one. Have you visited an ophthalmologist recently?

xlwang Over a year ago

Is there a simple way to output contents of FILES one by one?

Ignacio Vazquez-Abrams Over a year ago

stackoverflow.com/questions/8880603/…

xlwang Over a year ago

This is the earliest and best answer till now. So the answer is essentially not using programs like ls or find, but using pathname expansion built in bash.

Mustafa DOGRU · Accepted Answer · 2016-09-28 10:31:04Z

0

you can also try this;

#!/bin/bash
SUFF='aa'
FILES=`ls -1 *."$SUFF" foo`
FILENAMES=`echo $FILES | awk -F ':' '{print $2}'`
COUNT=`echo $FILENAMES | wc -w`
echo "$COUNT files have suffix $SUFF, they are: $FILENAMES"

if inserted echo $FILES in your script, output is foo: 1.aa 2.aa 3.aa so

awk -F ':' '{print $2}' gets 1.aa 2.aa 3.aa from $FILES variable

wc -w prints the word counts

edited Sep 28, 2016 at 10:31

answered Sep 28, 2016 at 10:24

Mustafa DOGRU

4,1121 gold badge18 silver badges24 bronze badges

Comments

redneb · Accepted Answer · 2016-09-28 10:33:19Z

0

If you only need the file count, I would actually use find for that:

find '/path/to/directory' -mindepth 1 -maxdepth 1 -name '*.aa' -printf '\n' | wc -l

This is more reliable as it handles correctly filenames with line breaks. The way this works is that find outputs one empty line for each matching file.

Edit: If you want to keep the file list in an array, you can use a glob:

GLOBIGNORE=".:.."
shopt -s nullglob
FILES=(*aa)
COUNT=${#arr[@]}
echo "$COUNT"

edited Sep 28, 2016 at 10:33

answered Sep 28, 2016 at 10:13

redneb

24.4k6 gold badges50 silver badges57 bronze badges

6 Comments

xlwang Over a year ago

But I also want to output the names of these files later in the script.

redneb Over a year ago

@Inian What do you mean?

123 Over a year ago

@Inian It does because they only output a newline for each filename, not the actual filename

123 Over a year ago

@Inian yes but print0 is only on gnu as well.

Inian Over a year ago

@123: My bad! thought it was a POSIX standard, guess only -print is portable

|

score 0 · Accepted Answer · 2016-09-28 11:34:51Z

0

The reason is that the option nullglob is unset by default in bash:

If no matching file names are found, and the shell option nullglob is not enabled, the word is left unchanged. If the nullglob option is set, and no matches are found, the word is removed.

So, just set the nullglob option, and run you code again:

shopt -s nullglob
SUFF='aa'
FILES="$(printf '%s\n' foo/*."$SUFF")"
COUNT="$(printf '%.0s\n' foo/*."$SUFF" | wc -l)"
echo "$COUNT files have suffix $SUFF, they are: $FILES"

Or better yet:

shopt -s nullglob
suff='aa'
files=( foo/*."$suff" )
count=${#file[@]}
echo "$count files have suffix $suff, they are: ${files[@]}"

edited Sep 28, 2016 at 11:34

answered Sep 28, 2016 at 10:32

user8017719

2 Comments

xlwang Over a year ago

Thx. But if file names contain \n, it may fail.

user8017719 Over a year ago

@xlwang Ok, solved. The %.0s\n prints only a new line per each file.

Collectives™ on Stack Overflow

How to store NUL output of a program in bash script?

4 Answers 4

5 Comments

Comments

6 Comments

2 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

5 Comments

Comments

6 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related