regsub inside braces only 1

[mark2atsafe]

I have a string, for example...

{attr 1} {a b c} attr2 abcd attr3 {a b c}

I need to replace all spaces with special characters (say !!!) but ONLY those spaces inside braces.

So in the above I need...

{attr!!!1} {a!!!b!!!c} attr2 abcd attr3 {a!!!b!!!c}

Any suggestions? At the moment I'm using...

while {[regexp (\{)+(.)+( ) $_string] !=0} {
regsub -all (\{)+(\[^ \])+( ) $_string {&!!!} _string2
}

...the problem being that on the second pass the regexp matches with "{attr!!!1} " giving "{attr!!!1}!!!{a!!!....."

What I need is a function that says regsub (but only up to the next } char)! Is there such a component in regsub? And is it possible to do all this inside a single command, without using the loop?

Any help would be greatly appreciated, since I've been banging my head on the wall over this one for ages.

Thanks!

Mark I

PS - What I'm actually looking for is an output of...

attr 1,a b c,attr2,abcd,attr3,a b c

...but if I can flag the correct space characters I can work out the rest OK. Of course, if there is an all in one solution, so much the better.

 

[feherke]

Hi

Try to make that [tt]while[/tt] abit less complicated :

while {[regsub -all {(\{[^{} ]*) } $_string {\1!!!} _string]} {}

Feherke.

http://rootshell.be/~feherke/

 

[mark2atsafe]

Hah! I knew there'd be a simpler solution. Now you can see why I'm not a programmer! Many thanks; this is an excellent solution.

Mark