parameter expansion, substitution

parameter expansion, substitution

[S. Cowles]

i am not getting expected results in the following code snippet:
 	b="abc" ; b=${b/#abc%/d} ; echo ${b}
i expect the value of b to be changed to "d"; however, it remains "abc".

the following snippet does give the expected result:
 	b="abc" ; b=${b/#abc/d} ; echo ${b}

for this session, the following shell options are set: interactive monitor 
shinstdin zle.  the version of the shell being used is 4.3.17-dev-0, 
patchlevel 1.5604.

in my use case, i need to specify a pattern match at the end of the 
parameter.  have i misunderstood the syntax of the '%' end-of-string 
operator?  or is this problem something else?

thanks for any help.

Re: parameter expansion, substitution

[Mikael Magnusson]

On 9 March 2012 23:03, S. Cowles <scowles@ckhb.org> wrote:
> i am not getting expected results in the following code snippet:
>        b="abc" ; b=${b/#abc%/d} ; echo ${b}
> i expect the value of b to be changed to "d"; however, it remains "abc".
>
> the following snippet does give the expected result:
>        b="abc" ; b=${b/#abc/d} ; echo ${b}
>
> for this session, the following shell options are set: interactive monitor
> shinstdin zle.  the version of the shell being used is 4.3.17-dev-0,
> patchlevel 1.5604.
>
> in my use case, i need to specify a pattern match at the end of the
> parameter.  have i misunderstood the syntax of the '%' end-of-string
> operator?  or is this problem something else?
>
> thanks for any help.

It's possibly a bit counterintuitive, but both # and % go at the start
of the string, so you want
b=${b/#%abc/d}

-- 
Mikael Magnusson

Re: parameter expansion, substitution

[S. Cowles]

many thanks to you both, Bart and Mikael.

Re: parameter expansion, substitution

[Bart Schaefer]

On Mar 9,  2:03pm, S. Cowles wrote:
} Subject: parameter expansion, substitution
}
} i am not getting expected results in the following code snippet:
}  	b="abc" ; b=${b/#abc%/d} ; echo ${b}
} i expect the value of b to be changed to "d"; however, it remains "abc".

You've misunderstood.  You want:

  	b="abc" ; b=${b/#%abc/d} ; echo ${b}

The # or % or #% must appear at the beginning of the pattern, they are
not positional like ^ and $ in a regular expression.