Hello,

I have :

str=foo.txt
echo ${str%%.*}

which works.

But if I want to replace str by `basename $0` in echo..., I get "bad
substitution"

Is there a way to do it in 1 line ?
(I tried with eval() by whitout success...)

Thanks in advance.

On Wed, 25 Jul 2007 09:58:16 -0700, patrick wrote:

> Hello,
>
> I have :
>
> str=foo.txt
> echo ${str%%.*}
>
> which works.
>
> But if I want to replace str by `basename $0` in echo..., I get "bad
> substitution"
>
> Is there a way to do it in 1 line ?
> (I tried with eval() by whitout success...)
>
> Thanks in advance.

Not that I can see. That is the way things work.

You can use `basename $0` to get the NAME of a variable to expand, in
which case $0 had better not have a "." in its final component, as these
are not valid in variable names (unless you have ksh, and compound
variables).

patrick wrote:
> Hello,
>
> I have :
>
> str=foo.txt
> echo ${str%%.*}
>
> which works.
>
> But if I want to replace str by `basename $0` in echo..., I get "bad
> substitution"
>
> Is there a way to do it in 1 line ?
> (I tried with eval() by whitout success...)
>
> Thanks in advance.
>

If your intention is to increase speed, you can maybe do

str=${0##*/}; echo ${str%%.*}


--
Michael Tosch @ hp : com

On Jul 25, 12:32 pm, Michael Tosch
<eed...@NO.eed.SPAM.ericsson.PLS.se> wrote:
> patrick wrote:
> > Hello,
>
> > I have :
>
> > str=foo.txt
> > echo ${str%%.*}
>
> > which works.
>
> > But if I want to replace str by `basename $0` in echo..., I get "bad
> > substitution"
>
> > Is there a way to do it in 1 line ?
> > (I tried with eval() by whitout success...)
>
> > Thanks in advance.
>
> If your intention is to increase speed, you can maybe do
>
> str=${0##*/}; echo ${str%%.*}
>
> --
> Michael Tosch @ hp : com

Silly question, What is in his $0 var?
"bad substitution" is an odd evil message, hinting that his $0 has
some thing funny in it maybe ?
Just a thought.... JB

On Wed, 25 Jul 2007 12:53:17 -0700, johngnub wrote:

> On Jul 25, 12:32 pm, Michael Tosch
> <eed...@NO.eed.SPAM.ericsson.PLS.se> wrote:
>> patrick wrote:
>> > Hello,
>>
>> > I have :
>>
>> > str=foo.txt
>> > echo ${str%%.*}
>>
>> > which works.
>>
>> > But if I want to replace str by `basename $0` in echo..., I get "bad
>> > substitution"
>>
>> > Is there a way to do it in 1 line ?
>> > (I tried with eval() by whitout success...)
>>
>> > Thanks in advance.
>>
>> If your intention is to increase speed, you can maybe do
>>
>> str=${0##*/}; echo ${str%%.*}
>>
>> --
>> Michael Tosch @ hp : com
>
> Silly question, What is in his $0 var? "bad substitution" is an odd evil
> message, hinting that his $0 has some thing funny in it maybe ?
> Just a thought.... JB

Well, the example he gives is "foo.txt". The fact that he is trying to
strip off ".*" as a glob pattern implies that it probably has a "." in
it, which makes it an illegal name for a variable (unless he has compound
variables, which is pretty unlikely).

On 25 juil, 21:32, Michael Tosch <eed...@NO.eed.SPAM.ericsson.PLS.se>
wrote:
> patrick wrote:
> > Hello,
>
> > I have :
>
> > str=foo.txt
> > echo ${str%%.*}
>
> > which works.
>
> > But if I want to replace str by `basename $0` in echo..., I get "bad
> > substitution"
>
> > Is there a way to do it in 1 line ?
> > (I tried with eval() by whitout success...)
>
> > Thanks in advance.
>
> If your intention is to increase speed, you can maybe do
>
> str=${0##*/}; echo ${str%%.*}


Thanks for your answer, but the goal is just to do :

str=`basename $0`
echo ${str%%.*}

in 1 instruction instead of 2.
(if my script name is "foo.ksh", I get "foo")

On 26 Jul., 10:24, patrick <patrick.beltra...@caramail.com> wrote:
> On 25 juil, 21:32, Michael Tosch <eed...@NO.eed.SPAM.ericsson.PLS.se>
> wrote:
>
>
>
> > patrick wrote:
> > > Hello,
>
> > > I have :
>
> > > str=foo.txt
> > > echo ${str%%.*}
>
> > > which works.
>
> > > But if I want to replace str by `basename $0` in echo..., I get "bad
> > > substitution"
>
> > > Is there a way to do it in 1 line ?
> > > (I tried with eval() by whitout success...)
>
> > > Thanks in advance.
>
> > If your intention is to increase speed, you can maybe do
>
> > str=${0##*/}; echo ${str%%.*}
>
> Thanks for your answer, but the goal is just to do :

Goal? To achieve what? If you want to make a simple problem overly
complicated and less readable or less portable or less performant
you can have a look whether your shell supports ${var/.../...}
replacements and backreferences, otherwise I'd suggest to use the
standard solution that Michael proposed upthread

str=${0##*/}; echo ${str%%.*}

You can of course make use of external programs if the "one statement"
requirement is some homework task; have a look at the program expr(1)
as in

expr $var : regexp

where regexp contains the subexpression to be matched in var; what is
in regexp enclosed in \( and \) will be printed.
(http://www.scit.wlv.ac.uk/cgi-bin/mansec?1+expr)

Janis

>
> str=`basename $0`
> echo ${str%%.*}
>
> in 1 instruction instead of 2.
> (if my script name is "foo.ksh", I get "foo")

On 26 Jul., 10:24, patrick <patrick.beltra...@caramail.com> wrote:
> On 25 juil, 21:32, Michael Tosch <eed...@NO.eed.SPAM.ericsson.PLS.se>
> wrote:
>
>
>
> > patrick wrote:
> > > Hello,
>
> > > I have :
>
> > > str=foo.txt
> > > echo ${str%%.*}
>
> > > which works.
>
> > > But if I want to replace str by `basename $0` in echo..., I get "bad
> > > substitution"
>
> > > Is there a way to do it in 1 line ?
> > > (I tried with eval() by whitout success...)
>
> > > Thanks in advance.
>
> > If your intention is to increase speed, you can maybe do
>
> > str=${0##*/}; echo ${str%%.*}
>
> Thanks for your answer, but the goal is just to do :

Goal? To achieve what? If you want to make a simple problem overly
complicated and less readable or less portable or less performant
you can have a look whether your shell supports ${var/.../...}
replacements and backreferences, otherwise I'd suggest to use the
standard solution that Michael proposed upthread

str=${0##*/}; echo ${str%%.*}

You can of course make use of external programs if the "one statement"
requirement is some homework task; have a look at the program expr(1)
as in

expr $var : regexp

where regexp contains the subexpression to be matched in var; what is
in regexp enclosed in \( and \) will be printed.
(http://www.scit.wlv.ac.uk/cgi-bin/mansec?1+expr)

Janis

>
> str=`basename $0`
> echo ${str%%.*}
>
> in 1 instruction instead of 2.
> (if my script name is "foo.ksh", I get "foo")