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Bash shell adding extra single quotes

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# 15  
Old 07-13-2012
Or if you want the message to be optional:

Code:
read MESSAGE
[ -z "$MESSAGE" ] && cvs commit ./q.s
[ -z "$MESSAGE" ] || cvs commit -m "${MESSAGE}" ./q.s

# 16  
Old 07-13-2012
I have tried to get this to the bare essentials. What I want to do is pass a string from a.sh to b.sh and echo that to the screen.

Code:
#!/bin/bash -x
#a.sh
echo $1
./b.sh $1

#!/bin/bash -x
#b.sh
echo $1

$a.sh "test multi"
+ echo test multi
test multi
+ ./b.sh test multi
+ echo test
test

Why can't I get this to work???????? I have gone back through this thread and the only one I can get to work is the "eval" which is too susceptible to attack. Sorry, but I am just getting frustrated with myself.
Moderator's Comments:
Mod Comment code tags for code.
Last edited by Corona688; 07-13-2012 at 01:55 PM..
# 17  
Old 07-13-2012
If you don't put strings in quotes, the shell splits on them on spaces. The quotes are not considered part of the string -- they define whether splitting happens or not inside it.

Substitution happens inside double quotes, but not single ones. So use double quotes here.

"$1"

Note that -x might not show the quotes anyway... This is because -- I repeat -- the quotes are not part of the string. They're delimiters. They define where one string stops and ends, but aren't included in the data itself.

I also note that if you'd used my code as given, it would have worked; it included quotes where necessary...
# 18  
Old 07-13-2012
I removed the -x

Code:
#!/bin/bash
#a.sh
echo $1
./b.sh "$1"

Code:
#!/bin/bash
#b.sh
echo $1
STMT="cvs commit -m "$1" ./q.sh"
$STMT

How does that differ from your code as given?

This is my output trying to run either script.

Code:
$a.sh "This is"
This is
This is
cvs commit: nothing known about `is'
cvs [commit aborted]: correct above errors first!

Code:
$b.sh "This is"
This is
cvs commit: nothing known about `is'
cvs [commit aborted]: correct above errors first!
$


When I use this code:
Code:
#!/bin/bash
#b.sh
echo $1
STMT="cvs commit -m "$1" ./q.sh"
echo $STMT

the output is
Code:
cvs commit -m This is ./q.sh

That statement will not execute. This will
Code:
cvs commit -m "This is" ./q.sh

This is the cvs syntax
Code:
cvs [cvs-options] commit [command-options] [filename]

so if I don't have the double quotes this command
Code:
cvs commit -m This is ./q.s

h
will attempt to commit filename "is" with message "This". So I do need the double quotes, right? They ARE part of the string I am passing to the cvs statement.

---------- Post updated at 03:27 PM ---------- Previous update was at 03:25 PM ----------

I do not mean to sound testy or petty, I am just very frustrated with myself right now. Please keep trying for me and my sanity sake. Smilie Thanks
Last edited by Scrutinizer; 07-13-2012 at 05:29 PM.. Reason: code tags
# 19  
Old 07-13-2012
Try this (using Corona688's suggestion in #14):
Code:
#!/bin/bash -x
echo "$1"
cvs commit -m "$1" ./q.s

Code:
./test.sh "This is"

# 20  
Old 07-13-2012
Tried something new
Code:
#!/bin/bash -x
#b.sh
echo $1
STMT="cvs commit -m "\"$1\"" ./q.sh"
echo $STMT
STMT2="cvs commit -m "$1" ./q.sh"
echo $STMT2
STMT3="cvs commit -m \"$1\" ./q.sh"
echo $STMT3
STMT4="cvs commit -m \""$1"\" ./q.sh"
echo $STMT4

L$b.sh "First Second Third"
+ echo First Second Third
First Second Third
+ STMT='cvs commit -m "First Second Third" ./q.sh'
+ echo cvs commit -m '"First' Second 'Third"' ./q.sh
cvs commit -m "First Second Third" ./q.sh
+ STMT2='cvs commit -m First Second Third ./q.sh'
+ echo cvs commit -m First Second Third ./q.sh
cvs commit -m First Second Third ./q.sh
+ STMT3='cvs commit -m "First Second Third" ./q.sh'
+ echo cvs commit -m '"First' Second 'Third"' ./q.sh
cvs commit -m "First Second Third" ./q.sh
+ STMT4='cvs commit -m "First Second Third" ./q.sh'
+ echo cvs commit -m '"First' Second 'Third"' ./q.sh
cvs commit -m "First Second Third" ./q.sh

Still not right......
Last edited by Scott; 07-13-2012 at 06:27 PM.. Reason: Code tags
# 21  
Old 07-13-2012
Quote:
Originally Posted by hpodhrad
[..]
Code:
#!/bin/bash
#b.sh
echo $1
STMT="cvs commit -m "$1" ./q.sh"
$STMT

How does that differ from your code as given?

[..]
You are still using the indirect message of executing a string and for that you need eval, so this will not work...

If you want to "hide" the cvs command you could use a function as I suggested earlier. I'll give you an example:

Code:
#!/bin/bash -x

commit(){
  cvs commit -m "$1" ./q.s
}

echo "$1"
commit "$1"

Last edited by Scrutinizer; 07-13-2012 at 05:50 PM..
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