Why does not work a sentece kept in a variable when is delivered in other script?

[Manuales]

Hi ..
i have a program which has the following variable:

VALOR="/usr/bin/ftp_-v_-n_-i_<_$logftp_>_${logftp}_out"

then, this value is sent to other script named script2:

script2 $VALOR

script2 contains many things between them:
....
...
echo $2 | sed 's/_/ /g'
...
...

it guess $2 is an instruction apart of be a command ... but it does not work as command !!!

it does not make the ftp indicated in that variable ... why?

can you help me??

Thanks ...

[Chan 007 · Accepted solution]

Hi,

when you do echo $2 |sed 's/_//g'

it just removes the "_"

So your command will fail,

i think you need to do sed s/_/ /g

Chan

[Sandman!]

Hi Manuales,

You need to pipe the output of sed to the shell in order for the command to be executed i.e.

replace...
echo $2 | sed 's/_/ /g'
with...
echo $2 | sed 's/_/ /g' | sh

Only then the ftp will be executed.

[James R. Ferguson]

Hi Manuales:

You don't need to perform the contortions of embedding underscore characters that you change to spaces. Consider:

# logftp="xxxxxx"
# VALOR="/usr/bin/ftp -v -n -i < ${logftp} > ${logftp}_out"
# ./showme DOTHISFORME "${VALOR}"

If ./showme looks like this below, then the second argument received is your ${VALOR} string and the first argument will be "DOTHISFORME":

# cat ./showme

#!/usr/bin/sh
echo $1 $2
exit 0

In 'showme' you could execute ${VALOR} by writing:

#!/usr/bin/sh
echo "I was told $1 $2"
$2
exit 0

Regards!

...JRF...

[Manuales]

Bad news ...
do not work either "${VALOR}" or "| sh" at the end of the sentence ...

i do not know why!!!??
some time i read about comand exe or exec or something like that, do you know it?
I'm using /uar/bin/sh shell
Please help !!

Thanks .

[Manuales]

Help ..

[James R. Ferguson]

Hi (again) Manuales:

OK, since you are using redirection we need to modify the wrapper I provided by adding an 'eval' statement (see the 'sh-posix(1)' manpages):

Consider again:

# logftp="xxxxxx"
# VALOR="/usr/bin/ftp -v -n -i < ${logftp} > ${logftp}_out"
# ./launch DOING "${VALOR}"

...where:

# cat ./launch

#!/usr/bin/sh
echo "$1 : $2"
eval $2 # <--- NOTE
exit 0

Regards!

...JRF...

[Antonio Cardoso_1]

Manuales,

if what you want to handle in your second script is
script2 param1 "/usr/bin/ftp -v -n -i < $logftp > ${logftp}_out"

i would write script1 with
VALOR="/usr/bin/ftp -v -n -i < ${logftp} > ${logftp}_out"
script2 param1 "${VALOR}"

and then in script2:
# retrieve first param
param1=$1
# retrieve rest of command line.
shift
param2=$*

hope this helps.
antonio.

[Sandman!]

Manuales,

What is stored inside "logftp" variable. Not sure if the hostname you're trying to ftp to is contained in $logftp. Anyway post the output of the following:

# echo $VALOR
# echo $logftp

thanks!

[Philippe Vervoort]

Manuales,

in VALOR, change the underscores in blanks

Then the line in script2 should read:
echo $*|sh

If you like to keep the underscores, you have to write it like ...<_${logftp}_>_${logftp}_out
because _$ seems to be something special for the shell and it 'eats' your variable.
In script2, then you should use:
echo $*|sed 's/_/ /g'|sh

Greetings,
Philippe