Output only non-empty arguments

Shell Programming and Scripting
1

Hello, I am VERY new to shell scripting here, so please go easy. I have an assignment that requires creating a script using bash shell, outputting all command line arguments that are not empty ones such as " ", and showing total number of arguments. I know how to show the total with $# and all arguments with $*, but is there a way to not output empty arguments? I am trying if statements and each time, the arguments appear as white spaces instead of not showing at all.

2

Can you please post an example how you call your script with the parameters and how your script itself looks?
When doing so, use

```text
 and 
```

around the code, data or logs to separate it from descriptive text for better readability and preserving indention etc.

3 
#!/bin/sh 
if [ "$1" != "" ] then 
    echo "Argument 1: $1" 
     fi 
echo "arguments: [$*]" 
echo "Total number of command line arguments: $#"

Output:

"" 2 3 4
Argument 1: 
arguments: [ 2 3 4]
Total number of command line arguments: 4

I was just testing here with argument $1 and using random numbers as test arguments. There is the white space there before the 2.

4 
#!/bin/sh

VAR=$(echo $*| sed 's/  / /g')
echo "arguments: [$VAR]"
echo "Total number of command line arguments: $#"

exit 0
1 Like
5

Do you really need to pass a space as an argument? That just does not sound right IMO. There are better ways to handle when arguments are not passed.

6

That seems to do the trick, zaxxon. Thank you. To frank - yes, it was required as part of the exercise.

7

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