String concat that keeps quotes - UNIX for Dummies Questions & Answers

duke · #1 (**permalink**) 07-22-2008

Hi All,

I hope you can help. i am concatenating String variables using the following method.

Code:

command="$command$x"

i have created a script which takes a set of args passed to the script using the $*

Code:

#example script
args=$*
count=0
for x in $args 
do
    count=`expr $count + 1`
    if [ "$count" -gt "3" ] 
        then
            command="$command$x"
    fi
done

echo $command

this seems to work if i write

Code:

example this is a test to see if it works

it should output
test to see if it works

Now i am finally coming to my question
if i write

Code:

example this is a test to "see if" it works

it still prints out
test to see if it works
it always seems to remove the quotes from the string. how can i keep the quotation in the output

Many thanks for your help

era · #2 (**permalink**) 07-22-2008

The quotation marks are not passed in literally. Compare:

Code:

vnix$ echo foo
foo
vnix$ echo "foo"
foo
vnix$ echo '"foo"'
"foo"

If you change the loop so that it prints every item it processes, you should find that "see if" is handled as a single argument.
See further http://www.grymoire.com/Unix/Quote.html

As an aside, note that "$@" should be preferred over $*

duke · #3 (**permalink**) 07-22-2008

Hi era,

I have noticed that
echo $@
or
echo $*

seem to remove the quotes from the arguments in the scripts. Is there a away to keep this

i.e example script
example this is an "example" test
echo $* or $@
echo's
this is an "example" test

rather than

this is an example test

zaxxon · #4 (**permalink**) 07-23-2008

If you want to keep the quotes you can escape them with a slash when writing the parameter like

Code:

./mach.ksh \"example\" test
$@: "example" test
$*: "example" test

era · #5 (**permalink**) 07-23-2008

Read the grymoire.com quoting tutorial from the link above. Quote characters are inherently special; you can't get literal quotes through in the shell without additional quoting or other escaping mechanisms.

duke · #6 (**permalink**) 07-23-2008

thanks era, i have been through the link you provide UNIX Shell Quote
this make perfect sense, unfortunately I am in big trouble i am using an application which allows me access to the arguments i cant make changes to the original commands that are passed,
so i cant wrap them round / or '

i can just use these arguments to create scripts which i use.

The problems is some of the commands that can be passed have quotations which seem to be disappear

so i assume there is no walk around this?

Many thanks

joeyg · #7 (**permalink**) 07-23-2008

Can you do something like

Code:

tr '"' '~'
your stuff, then
tr '~' '"'

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