Embeded shell variable in command line

Shell Programming and Scripting
1

Hello,
I know this is a situation about the single quote and double literal, but I could not figure out after many search.

I need to loop through thousands of different BACs sequencing to optimize kmer individually.

IN=/PATH/TO/INPUT/FILES
for sample in S{01..1096}
do
     run_program in='$IN/${sample}_file1.fa $IN/${sample}_file2.fa' 
done

The package asks

in='input_file1 input_file2'

for the input files. Because of the single quote in the command option, the embedded shell variables cannot be interpreted literally. Is there a way to handle this situation?

Thanks a lot!

Yifang

2 
[user@host ~]$ x=2
[user@host ~]$ echo '$x'
$x
[user@host ~]$ echo "$x"
2
[user@host ~]$

HTH

3

Yes, I am aware of the difference. The problem I met is the $x should be changed each loop, but the single quote of the command parameter escaped the $ sign of $x. Maybe I should ask like this:

MYVAR=sometext 
echo 'single quotes gives you $MYVAR'

Is there away to get output from the single quotes expression as the double quoted one? i.e.

echo 'single quotes gives you $MYVAR'
Output: 
single quotes gives you sometext

I cannot change the single quote as it has been set by the package. Thanks again!

4
5

Thanks!
The awk examples seems a possible way to what I want, but not exactly. It seems there is no way to change the single quote literal in bash for my case. I thought there could be a way to get what I want. Waiting for more ideas.
Thanks a lot in advance!

6

Well, what can you change? The easiest solution is:

run_program in="$IN/${sample}_file1.fa $IN/${sample}_file2.fa"
1 Like
7

Ah, how simple it is!!! It works now.
Does this replacement apply to all the case IF there is embedded variable inside the single quote in general? I mean generally.
Thanks a lot again!

8

Enclosing a referenced value in double quotes (" ... ") does not interfere with variable substitution.

This is called partial quoting, sometimes referred to as weak quoting.

Using single quotes (' ... ') causes the variable name to be used literally, and no substitution will take place.

This is full quoting, sometimes referred to as strong quoting.

9

Yes, 100% of the time, '$var' is interpreted as the literal four characters, the dollar sign retained, and "$var" is interpreted as a shell variable. The following examples show this:

$ x=100
$ echo "$x"
100
$ echo '$x'
$x
$ echo "$y"
[ blank line ]

In the final example, the shell tries to interpret $y as a variable. Since there is no such variable, the shell substitutes "nothing", and a blank line results.

This distinction between "single quotes" and 'double quotes' is really an important concept, so it's good you kept asking. :b:

---------- Post updated at 02:53 PM ---------- Previous update was at 02:47 PM ----------

Just FYI, there is one additional case you probably already know about, but here goes anyway. You can use the \ "backslash" to "escape" the dollar sign, and prevent the shell from interpreting $var as a variable, as follows.

$ echo "\$x"
$x
10

Important to note that this substitution happens right now. If you change the value of one of the variables used later, the other string won't alter with it.