CODEWITHSUNDEEP
bash
WraithLux
WraithLux

Reputation: 699

bash basename syntax error inside for in loop

I have a simple script:

for photo in $full_path/*.jpg; do
    $b_photo=$(basename $photo)
    echo $b_photo
done

it's simplified form of what I'm trying to do, but this gives me error like this:

./foo.sh: line 334: =foobar.jpg: command not found

Upvotes: 0

Views: 1002

Answers (2)

Gareth Rees
Gareth Rees

Reputation: 65854

In addition to the problem pointed out by fedorqui, basename $photo will go wrong if $photo contains a space:

 $ photo="directory/file name"
 $ basename $photo
 file

It will also go wrong if $photo starts with a -:

$ photo="-directory/file"
$ basename $photo
basename: illegal option -- d
usage: basename string [suffix]
       basename [-a] [-s suffix] string [...]

You could write:

for photo in "$full_path"/*.jpg; do
    basename -- "$photo"
done

but a simpler way to carry out this task (printing file names in a directory that match a pattern to standard output, one per line) would be to switch to the directory and use printf:

{ cd -- "$full_path" && printf "%s\n" *.jpg }

Upvotes: 3

fedorqui
fedorqui

Reputation: 290145

Assign variables with var=$(command) and use them with $var (or, better, ${var}):

for photo in $full_path/*.jpg; do
    b_photo=$(basename $photo)       <------ b_photo=$() instead of $b_photo=$()
    echo $b_photo                                                   ^
done

Note you can directly write:

for photo in $full_path/*.jpg; do
    basename "$photo"
done

Upvotes: 5

Related Questions