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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}

1

1

I try to create an array/list which stores file names inside a folder. The below command creates for unknown reasons a :. How is possible to remove :?

> a=$(ls split*)

> echo $a

split_sam.o4433568 split_sam.o4433616 split_sam.o4441795 split-data-1:

bash files array

share|improve this question

edited 2 hours ago

user977828

asked 3 hours ago

user977828user977828

3561617

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  What operating system and terminal are you running? I've tried your snippet both in Fedora (bash) and Alpine (ash) and it works properly. Isn't it actually a part of the file name?
  
  – danieldeveloper001
  2 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If that is the case, you could replace the colon by using echo ${a/:/} to output the names.
  
  – danieldeveloper001
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  We are using SUSE Linux Enterprise Server 12 SP2 and I use bash.
  
  – user977828
  24 mins ago
  
  
  

  
  
  
  

add a comment | 

1

1

I try to create an array/list which stores file names inside a folder. The below command creates for unknown reasons a :. How is possible to remove :?

> a=$(ls split*)

> echo $a

split_sam.o4433568 split_sam.o4433616 split_sam.o4441795 split-data-1:

bash files array

share|improve this question

edited 2 hours ago

user977828

asked 3 hours ago

user977828user977828

3561617

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  What operating system and terminal are you running? I've tried your snippet both in Fedora (bash) and Alpine (ash) and it works properly. Isn't it actually a part of the file name?
  
  – danieldeveloper001
  2 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If that is the case, you could replace the colon by using echo ${a/:/} to output the names.
  
  – danieldeveloper001
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  We are using SUSE Linux Enterprise Server 12 SP2 and I use bash.
  
  – user977828
  24 mins ago
  
  
  

  
  
  
  

add a comment | 

I try to create an array/list which stores file names inside a folder. The below command creates for unknown reasons a :. How is possible to remove :?

> a=$(ls split*)

> echo $a

split_sam.o4433568 split_sam.o4433616 split_sam.o4441795 split-data-1:

bash files array

share|improve this question

edited 2 hours ago

user977828

asked 3 hours ago

user977828user977828

3561617

> a=$(ls split*)

> echo $a

split_sam.o4433568 split_sam.o4433616 split_sam.o4441795 split-data-1:

share|improve this question

edited 2 hours ago

user977828

asked 3 hours ago

user977828user977828

3561617

share|improve this question

edited 2 hours ago

user977828

asked 3 hours ago

user977828user977828

3561617

asked 3 hours ago

user977828user977828

3561617

asked 3 hours ago

user977828user977828

3561617

1 Answer
1

active

oldest

votes

1

The $(ls split*) command didn't create any filenames -- luckily! It also didn't create an array. What it did do was call ls with the split* wildcard, then paste all of the resulting output together (removing newlines that ls added between each filename and at the very end) and assign that conglomerate to your variable. When you then called echo $a, an extra bit of processing happened that you don't want: the contents of your variable were further subjected to splitting on whitespace ($IFS) and wildcard expansions. You lucked out by not having any files named, for example, split space file or split more*!

The root of your confusion is that you do, indeed, have a file named split-data-1: in that directory, and ls is happily handing it back to you.

What you want to do, instead, is to find a wildcard that suits the files that you really want, and use a simple array assignment:

a=(split_sam*)

or

a=(split*)

as you originally wrote, which will still include that split-data-1: file.

You can then iterate over the files or operate on them as a whole:

for file in "${a[@]}"

do

  ls -d -- "$file"

done

or

ls -d -- "${a[@]}"

share|improve this answer

answered 2 hours ago

Jeff Schaller♦Jeff Schaller

45.6k1165149

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I tried a=(split_sam*) but I got only one file instead of 3. What did I miss?
  
  – user977828
  26 mins ago
  

  
  
  
  

add a comment | 

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1

The $(ls split*) command didn't create any filenames -- luckily! It also didn't create an array. What it did do was call ls with the split* wildcard, then paste all of the resulting output together (removing newlines that ls added between each filename and at the very end) and assign that conglomerate to your variable. When you then called echo $a, an extra bit of processing happened that you don't want: the contents of your variable were further subjected to splitting on whitespace ($IFS) and wildcard expansions. You lucked out by not having any files named, for example, split space file or split more*!

The root of your confusion is that you do, indeed, have a file named split-data-1: in that directory, and ls is happily handing it back to you.

What you want to do, instead, is to find a wildcard that suits the files that you really want, and use a simple array assignment:

a=(split_sam*)

or

a=(split*)

as you originally wrote, which will still include that split-data-1: file.

You can then iterate over the files or operate on them as a whole:

for file in "${a[@]}"

do

  ls -d -- "$file"

done

or

ls -d -- "${a[@]}"

share|improve this answer

answered 2 hours ago

Jeff Schaller♦Jeff Schaller

45.6k1165149

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I tried a=(split_sam*) but I got only one file instead of 3. What did I miss?
  
  – user977828
  26 mins ago
  

  
  
  
  

add a comment | 

1

The $(ls split*) command didn't create any filenames -- luckily! It also didn't create an array. What it did do was call ls with the split* wildcard, then paste all of the resulting output together (removing newlines that ls added between each filename and at the very end) and assign that conglomerate to your variable. When you then called echo $a, an extra bit of processing happened that you don't want: the contents of your variable were further subjected to splitting on whitespace ($IFS) and wildcard expansions. You lucked out by not having any files named, for example, split space file or split more*!

The root of your confusion is that you do, indeed, have a file named split-data-1: in that directory, and ls is happily handing it back to you.

What you want to do, instead, is to find a wildcard that suits the files that you really want, and use a simple array assignment:

a=(split_sam*)

or

a=(split*)

as you originally wrote, which will still include that split-data-1: file.

You can then iterate over the files or operate on them as a whole:

for file in "${a[@]}"

do

  ls -d -- "$file"

done

or

ls -d -- "${a[@]}"

share|improve this answer

answered 2 hours ago

Jeff Schaller♦Jeff Schaller

45.6k1165149

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I tried a=(split_sam*) but I got only one file instead of 3. What did I miss?
  
  – user977828
  26 mins ago
  

  
  
  
  

add a comment | 

The $(ls split*) command didn't create any filenames -- luckily! It also didn't create an array. What it did do was call ls with the split* wildcard, then paste all of the resulting output together (removing newlines that ls added between each filename and at the very end) and assign that conglomerate to your variable. When you then called echo $a, an extra bit of processing happened that you don't want: the contents of your variable were further subjected to splitting on whitespace ($IFS) and wildcard expansions. You lucked out by not having any files named, for example, split space file or split more*!

The root of your confusion is that you do, indeed, have a file named split-data-1: in that directory, and ls is happily handing it back to you.

What you want to do, instead, is to find a wildcard that suits the files that you really want, and use a simple array assignment:

a=(split_sam*)

or

a=(split*)

as you originally wrote, which will still include that split-data-1: file.

You can then iterate over the files or operate on them as a whole:

for file in "${a[@]}"

do

  ls -d -- "$file"

done

or

ls -d -- "${a[@]}"

share|improve this answer

answered 2 hours ago

Jeff Schaller♦Jeff Schaller

45.6k1165149

a=(split_sam*)

a=(split*)

for file in "${a[@]}"

do

  ls -d -- "$file"

done

ls -d -- "${a[@]}"

share|improve this answer

answered 2 hours ago

Jeff Schaller♦Jeff Schaller

45.6k1165149

answered 2 hours ago

Jeff Schaller♦Jeff Schaller

45.6k1165149

answered 2 hours ago

Jeff Schaller♦Jeff Schaller

45.6k1165149