Command building & string escaping in Bash

Question

I am trying to build a Bash script that will take arguments from the command line as well as hard-coded arguments and pass them to a function, which in turn will construct a call to curl and perform an HTTP POST. Unfortunately, when I try passing strings with spaces in them, the script mangles the strings and the eventual call to curl fails to pass the arguments in the right order:

#!/bin/bash

API_URL="http://httpbin.org/"

docurl() {
    arg1=$1
    shift
    arg2=$1
    shift

    args=()
    while [ "$1" ]
    do
        arg_name="$1"
        shift
        arg_value="$1"
        shift

        args+=(-d $arg_name="$arg_value")
    done

    curl_result=$( curl -qSfs "$API_URL/post" -X POST -d arg1=$arg1 -d arg2=$arg2 ${args[@]} ) 2>/dev/null
    echo "$curl_result"
}

docurl foo1 foo2 title "$1" body "$2"

The script invocation would be something like this:

test2.sh "Hello" "Body of the message"

The output of the script as it stands is this:

{
  "form": {
    "body": "Body",
    "arg1": "foo1",
    "arg2": "foo2",
    "title": "Hello"
  },
  "headers": {
    "Host": "httpbin.org",
    "Connection": "close",
    "Content-Length": "41",
    "User-Agent": "curl/7.22.0 (x86_64-pc-linux-gnu) libcurl/7.22.0 OpenSSL/1.0.1 zlib/1.2.3.4 libidn/1.23 librtmp/2.3",
    "Content-Type": "application/x-www-form-urlencoded",
    "Accept": "*/*"
  },
  "files": {},
  "data": "",
  "url": "http://httpbin.org/post",
  "args": {},
  "origin": "xxx.xxx.xxx.xxx",
  "json": null
}

As you can see, in the form element, the fields "body" and "title" have been truncated. Can anybody let me know what on what I'm doing wrong here?

mirabilos · Accepted Answer · 2013-12-27 09:16:48Z

5

Use a bit more quotes!

    args+=(-d "$arg_name"="$arg_value")

And:

curl_result=$( curl -qSfs "$API_URL/post" -X POST -d arg1="$arg1" -d arg2="$arg2" "${args[@]}" ) 2>/dev/null

answered Dec 27, 2013 at 9:16

mirabilos

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1 Comment

mklement0 Over a year ago

Nice; it may make it clearer what happens to write args+=(-d "$arg_name=$arg_value") (double quotes around $arg_name=$arg_value as a whole) - the add-to-array assignment effectively adds 2 elements.

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