String Concatenation in bash with space and putting quote around

问题

I got a very good response in my last question. The idea was to process N number of inputs from command line, save first 9 to variables and make a string with 10 to onward.

I found this to be the easiest solution.

var1="$1"
var2="$2"
var3="$3"
var4="public"
var5="$5"
var6="''"
var7="$7"
var8="$8"
var9="$9"
var10="$(shift 9; IFS=""; echo "$*")"

echo snmptrap $var1 $var2 $var3  $var4 $var5 $var6 $var7 $var8 $var9 "$var10"

snmptrap $var1 $var2 $var3  $var4 $var5 $var6 $var7 $var8 $var9 "$var10"

The output looks like this...

./snmptas  -v 2c -c "" 9.48.85.57 "" 1.3.6.1.4.1.2.6.201.3 s s  ABC DDEF EFFF

snmptrap -v 2c -c public 9.48.85.57 '' 1.3.6.1.4.1.2.6.201.3 s s ABCDDEFEFFF

But I wanted $var10 in this form

"ABC DDEF EFFF".

This needs to changed. It's taking the spaces off.

var10="$(shift 9; IFS=""; echo "$*")"

How can I make var10 = "ABC DDEF EFFF"?

Thanks

回答1:

A subshell with echo is not really appropriate there. Better this way:

shift 9
var10="$*"

If you want quotes around that as part of the value, then change the last line to:

var10=\""$*"\"

Unrelated to that, I'm wondering if your treatment for var6 does what you intended. I suspect that this will be closer to what you really intended:

snmptrap $var1 $var2 $var3  $var4 $var5 '' $var7 $var8 $var9 "$var10"

Or this:

var6=
snmptrap $var1 $var2 $var3  $var4 $var5 "$var6" $var7 $var8 $var9 "$var10"

But if my guess is wrong and your original treatment is working as intended then never mind, ignore these remarks.

回答2:

To get the variable var10 to contain quotes, you need something like this:

shift 9
printf -v "var10" '"%s"' "$*"

And, if you want to quote all your variables, such that missing ones have a place, use this:

#!/bin/bash
set -- -v 2c -c "" 9.48.85.57 "" 1.3.6.1.4.1.2.6.201.3 s s  ABC DDEF EFFF

for ((i=1;i<=9;i++)); do
    printf -v "var$i" '"%s"' "${!i}"
done

shift 9
printf -v "var10" '"%s"' "$*"

echo snmptrap $var{1..10}

To get the script to read the arguments passed instead of the ones written in the code, just comment out the set -- line.

The output will be:

$ ./snmptrap 
snmptrap "-v" "2c" "-c" "" "9.48.85.57" "" "1.3.6.1.4.1.2.6.201.3" "s" "s" "ABC DDEF EFFF"

That each parameter has quotes means no problem to use them as input to the script.

来源：https://stackoverflow.com/questions/37428689/string-concatenation-in-bash-with-space-and-putting-quote-around

标签

string

bash

whitespace

quote