[ACCEPTED]-concatenate inputs in bash script-unix
function concatenate_args
{
string=""
for a in "$@" # Loop over arguments
do
if [[ "${a:0:1}" != "-" ]] # Ignore flags (first character is -)
then
if [[ "$string" != "" ]]
then
string+="_" # Delimeter
fi
string+="$a"
fi
done
echo "$string"
}
# Usage:
args="$(concatenate_args "$@")"
0
This is an ugly but simple solution:
echo $* | sed -e "s/ /_/g;s/[^_]*_//"
0
You can also use formatted strings to concatenate 1 args.
# assuming flag is first arg and optional
flag=$1
[[ $1 = ${1#-} ]] && unset $flag || shift
concat=$(printf '%s_' ${@})
echo ${concat%_} # to remove the trailing _
nJoy!
Here's a piece of code that I'm actually 6 proud of (it is very shell-style I think)
#!/bin/sh
firsttime=yes
for i in "$@"
do
test "$firsttime" && set -- && unset firsttime
test "${i%%-*}" && set -- "$@" "$i"
done
IFS=_ ; echo "$*"
I've 5 interpreted your question so as to remove 4 all arguments beginning with -
If you only want 3 to remove the beginning sequence of arguments 2 beginnnig with -:
#!/bin/sh
while ! test "${1%%-*}"
do
shift
done
IFS=_ ; echo "$*"
If you simply want to remove 1 the first argument:
#!/bin/sh
shift
IFS=_ ; printf %s\\n "$*"
flag="$1"
shift
oldIFS="$IFS"
IFS="_"
the_rest="$*"
IFS="$oldIFS"
In this context, "$*" is exactly what you're 6 looking for, it seems. It is seldom the 5 correct choice, but here's a case where 4 it really is the correct choice.
Alternatively, simply 3 loop and concatenate:
flag="$1"
shift
the_rest=""
pad=""
for arg in "$@"
do
the_rest="${the_rest}${pad}${arg}"
pad="_"
done
The $pad variable ensures 2 that you don't end up with a stray underscore 1 at the start of $the_rest.
#!/bin/bash
paramCat () {
for s in "$@"
do
case $s in
-*)
;;
*)
echo -n _${s}
;;
esac
done
}
catted="$(paramCat "$@")"
echo ${catted/_/}
0
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