Shell Script pass Parameter inside a String

Question

how can I pass a parameter to a shell script that is a longer string with spaces and "-"? Here is what I currently have:

/usr/test.sh -g filesystem -q -p "-p \"/usr/\" -s \"some intern string\" -q"

The test.sh script recognizes two -p, two -q and some other things, that are not right. But the value behind the first -p should be one large string: -p "/usr/" -s "some intern string" -q

What is my mistake?

Thanks for any help,
greetings,
Chriss

This really smells like an XY problem. What are you doing with "${flags[parameter]}" ? Once you concatenate a string with spaces into another string with spaces, there's no reliable way to extract the original string. — glenn jackman
– glenn jackman, Commented Oct 22, 2015 at 17:58
Also, you're adding extra quotes into the values here: wasteParameter+=("\"$arg\"") — glenn jackman
– glenn jackman, Commented Oct 22, 2015 at 17:59
@glennjackman I am afraid, but the script does not run to this point. It interrupts at the point if [ $? != 0 ] ; then with the error, that it found some not allowed parameter. So there must be a problem with getopt I guess. — toddeTV
– toddeTV, Commented Oct 23, 2015 at 9:04

that other guy · Accepted Answer · 2015-10-22 17:40:19Z

1

One way of seeing what is being passed is running printf "<<%s>>\n" yourcommand, which will print each individual argument on a line surrounded by <<..>> for easy disambiguation:

$ printf "<<%s>>\n" /usr/test.sh -g filesystem -q -p "-p \"/usr/\" -s \"some intern string\" -q"
<</usr/test.sh>>
<<-g>>
<<filesystem>>
<<-q>>
<<-p>>
<<-p "/usr/" -s "some intern string" -q>>

As you can see, the argument to -p is indeed the string -p "/usr/" -s "some intern string" -q.

If test.sh doesn't work, it's a bug in test.sh, not in this invocation.

answered Oct 22, 2015 at 17:40

that other guy

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3 Comments

toddeTV Over a year ago

I added the important part out of my test.sh file and I really hope, you can explain my what I did wrong here. and sorry for stealing your time :(

that other guy Over a year ago

@christopher2007 That's a separate question, but what you pasted from test.sh is fine and if you put printf "<<%s>>\n" "${flags[parameter]}" at the end you get the expected value of <<-p "/usr/" -s "some intern string" -q>>

toddeTV Over a year ago

I am afraid, but the script does not run to this point. It interrupts at the point if [ $? != 0 ] ; then with the error, that it found some not allowed parameter. But as you mentioned, with printf "<<%s>>\n" /usr/test.sh -g filesystem -q -p "-p \"/usr/\" -s \"some intern string\" -q" the correct parts are displayed. Is there something wrong with my getopt?

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