LinuxQuestions.org - [SOLVED] Bash command substitution but with bash variables

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Bash command substitution but with bash variables

I have a simple bash script that prompts the user for a command and a filename and it works with this input:

Enter command: ls
Enter filename: datafile

but if I enter:
Enter command: ls -l
Enter filename: datafile

Then the program fails with:
line 10: ls -l: command not found

Code:

#! /usr/bin/env bash                                                                                                                    

set -euo pipefail                                                                                                                      

IFS=$'\n\t'                                                                                                                            

                                                                                                                                        

read -r -p 'Enter command: ' cmd                                                                                                        

read -r -p 'Enter filename: ' fName                                                                                                    

                                                                                                                                        

#IFS=$' \n\t'                                                                                                                          

                                                                                                                                        

ans=$(${cmd} "${fName}")                                                                                                                

                                                                                                                                                                                                                                                                        

echo "${ans}"

If I remove the comment on the second IFS=$' \n\t' then the command substitution will work with both ls and ls -l.

What's going on here?

I wanted to write, it is quite simple, it is what you have requested. But probably not that obvious:
IFS is the Input Field Separator, you set it to \n\t. Therefore ls -l is a single word (does not contain either \n or \t).
Actually there is no such command, "ls -l" is not a command itself.
BUT
if IFS is set to $' \n\t': ls -l will be split (because it contains a space, which is now a separator) and the result is: the command will be ls and -l will be an argument to it, which is the correct way to execute it.

Code:

- ans=$(${cmd} "${fName}")                                                                                                                

+ ans=$(eval ${cmd} "'${fName}'")

I tried a few experiments in my bash terminal..

If I type ls -l and hit return it produces the expected results.

Code:

ls -l

If I type 'ls' '-l' and hit return it produces the same results.

Code:

'ls' '-l'

And if I type '''ls''' '''-l''' and hit return(surprisingly) it produces the same results.

Code:

'''ls''' '''-l'''

But if I type 'ls -l' and hit return?

Code:

'ls -l'

ls -l: command not found

Regarding '''ls''' '''-l'''
Try to remove the empty strings:

Code:

$ echo "'''ls''' '''-l'''" | sed "s/''//g"

'ls' '-l'

Off:
As far as I know, there is no way to use quotes within quotes (duplication or escaping), you have to close the quote, use a standalone quote (\' or "'"), then again a staring quote:

Code:

$ echo 'she sometimes said '\''Supercalifragilisticexpialidocious'\'' I remembered'

she sometimes said 'Supercalifragilisticexpialidocious' I remembered

With double quotes escaping works:

Code:

$ echo "another version of this word is \"supercaliflawjalisticeexpialadoshus\" you might add"

another version of this word is "supercaliflawjalisticeexpialadoshus" you might add

Off/2 this is different in other languages, eg:

Code:

echo 'This works in \'PHP\' language';

SELECT 'This works in ''SQL'''

you can use set -x to see what's going on.
you type a string, press enter. First this string will be evaluated, and split into an array and finally will be executed

Code:

pan@host:/tmp/a$ ls -l

+ ls -l

total 0

pan@host:/tmp/a$ 'ls' '-l'

+ ls -l

total 0

pan@host:/tmp/a$ '''ls''' '''-l'''

+ ls -l

total 0

pan@host:/tmp/a$ 'ls -l'

+ 'ls -l'

bash: ls -l: command not found

What is probably not obvious, ls -l is always split into two parts, the only exception is the last case when you force it to keep in one (that is 'ls -l', with ', which is still not part of the string, just printed to show it)

Quote:

Originally Posted by pan64 (Post 6499210)

This was the revelation to me(but shouldn't have been one). All the early Bash chapters point out this behavior but somehow I never made the connection for this example. Thanks for pointing it out.

By quoting you force it to be one word, no word-splitting takes place.
Differerent quoting styles:

Code:

'ls -l'

"ls -l"

ls\ -l

The shell invokes a one-word command with an embedded space.
The same applies to the command arguments:

Code:

touch "two words"

ls -l 'two words'

rm two\ words

The shell recognizes the one-word argument, dequotes it, and hands it to the respective command (first word).

Code:

'''ls''' '''-l'''

is a concatenation of many '' strings, like ''+'ls'+''<space>''+'-l'+''

The word-splitting is according to $IFS (Input Field Separators).
With IFS=$'\n\t' the space is removed but the \t (tab) is still there. Then

Code:

ls -l

is one word, but

Code:

ls<tab>-l

is two words.
What happens with the \n (newline)? On the command line (actually readline, also read) a newline causes an end-of-input-line so it won't exist for a direct word-splitting.
But the following example demonstrates it:

Code:

twolines="line1

line2"

echo "with word-splitting"

echo $twolines

echo 'double quotes "" allow $-evaluation/substitution but not word-splitting:'

echo "$twolines"

Instead of
echo ...
you better use
printf "<%s>\n" ...
for clarity; the format is applied on each argument (wrapped in < > followed by a newline).