bash - Add all arguments except first to a string -

trying parse arguments 1 string code giving me errors can't find i:

test: line 3: =: command not found test: line 7: [i: command not found test: line 7: [i: command not found test: line 7: [i: command not found test: line 7: [i: command not found test: line 7: [i: command not found 

code bellow

#!/bin/sh  $msg=""  in $@ if [i -gt 1];     msg="$msg $i" fi done 

edit: thx help, got work. final solution if anyones interested:

#!/bin/sh  args="" in "${@:2}"     args="$args $i" done 

your specific error messages showing because:

• assigning variable not done $ character in $msg="", instead should using msg=""; and

• [ command, 1 should separated other words white space, shell doesn't think you're trying execute mythical [i command.

however, have couple of other problems. first value of i needs obtained $i, not i. using i on own give error along lines of:

-bash: [: i: integer expression expected 

because i not numeric value.

secondly, neither i nor $i going index can compare 1, $i -gt 1 expression not work. word $i expand value of argument, not index.

---

however, if want process first element of argument list, bash has c-like constructs make life lot easier:

for ((i = 2; <= $#; i++)) ;   # 2 argc inclusive.     echo processing ${!i} done 

running arguments hello name pax, result in:

processing processing name processing processing pax 

for constructing string containing arguments, use like:

msg="$2"                           # second arg (or blank if none). ((i = 3; <= $#; i++)) ;   # append other args.     msg="$msg ${!i}" done 

which give (for same arguments above):

[my name pax] 

although, in case, there's simpler approach doesn't involve (explicit) loops @ all:

msg="${@:2}"