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 ```

Accepted Answer

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`.

Add command arguments using inline if-statement in bash

Solution

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