Add command arguments using inline if-statement in bash

Question

I'd like to add an argument to a command in bash only if a variable evaluates to a certain value. For example this works:

test=1 if [ "${test}" == 1 ]; then     ls -la -R else     ls -R    fi 

The problem with this approach is that I have to duplicate ls -R both when test is 1 or if it's something else. I'd prefer if I could write this in one line instead such as this (pseudo code that doesn't work):

ls (if ${test} == 1 then -la) -R 

I've tried the following but it doesn't work:

test=1 ls `if [ $test -eq 1 ]; then -la; fi` -R 

This gives me the following error:

./test.sh: line 3: -la: command not found 

Answer 1

A more idiomatic version of svlasov's answer:

ls $( (( test == 1 )) && printf %s '-la' ) -R 

^{Since echo understands a few options itself, it's safer to use printf %s to make sure that the text to print is not mistaken for an option.
Note that the command substitution must not be quoted here - which is fine in the case at hand, but calls for a more robust approach in general - see below.}

However, in general, the more robust approach is to build up arguments in an array and pass it as a whole:

# Build up array of arguments... args=() (( test == 1 )) && args+=( '-la' ) args+=( '-R' )  # ... and pass it to `ls`. ls "${args[@]}" 

Update: The OP asks how to conditionally add an additional, variable-based argument to yield ls -R -la "$PWD". In that case, the array approach is a must: each argument must become its own array element, which is crucial for supporting arguments that may have embedded whitespace:

(( test == 1 )) && args+= ( '-la' "$PWD" ) # Add each argument as its own array element. 

---

As for why your command,

 ls `if [ $test -eq 1 ]; then -la; fi` -R 

didn't work:

A command between backticks (or its modern, nestable equivalent, $(...)) - a so-called command substitution - is executed just like any other shell command (albeit in a sub-shell) and the whole construct is replaced with the command's stdout output.

Thus, your command tries to execute the string -la, which fails. To send it to stdout, as is needed here, you must use a command such as echo or printf.

Answer 2

Print the argument with echo:

test=1 ls `if [ $test -eq 1 ]; then echo "-la"; fi` -R