Convert arguments of a bash script to a JSON array for a web service

Question

I have a web service that requires a list of arguments in a JSON format. For instance, the JSON it could handle is like this:

{
  "args": ["basharg1", "bash arg 2 with spaces", "lastargs"]
}

How can I easily transform the arguments to a bash script to build a string that holds this kind of JSON? That string would then be used by curl to pas it to the web service.

Greg Edwards · Accepted Answer · 2014-09-16 15:25:29Z

3

On a Mac, a shorter version that uses glenn_jackman's approach is:

ARGS=""; for arg; do ARGS=$(printf '%s"%s"' "$ARGS", "$arg"); done;
ARGS=${ARGS#,}
JSON="{ \"args\": [$ARGS] }"
echo "$JSON"

answered Sep 16, 2014 at 15:25

Greg Edwards

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5 Comments

Alexander Mills Over a year ago

I think you can probably use single quotes around the JSON string, and then you won't have to escape the double quotes; have not confirmed that though.

Jean-Philippe Pellet Over a year ago

@AlexanderMills JSON strings are delimited by double quotes.

Alexander Mills Over a year ago

Yeah, I tried the single quote thing and it didn't work, so now we know :)

Alexander Mills Over a year ago

I am going to try this though '"{"foo":"bar"}"' (single quotes around double quotes)

leetbacoon Feb 4 at 21:41

doesn't automaticly escape special characters JSON needs escaped, like \x7F

Jean-Philippe Pellet · Accepted Answer · 2012-04-13 10:08:48Z

2

Here's how I ended up doing it. Feel free to suggest improvements.

#!/bin/bash

ARGS="\"$1\""
shift

while (( "$#" )); do
    ARGS="$ARGS, \"$1\""
    shift
done

ARGUMENTS_JSON="{ \"args\": [$ARGS] }"

echo "$ARGUMENTS_JSON"

answered Apr 13, 2012 at 10:08

Jean-Philippe Pellet

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2 Comments

glenn jackman Over a year ago

You could also write: ARGS=""; for arg; do printf -v ARGS '%s,"%s"' "$ARGS", "$arg"; done; ARGS=${ARGS#,}

Alexander Mills Over a year ago

how does this work? I want to put all the arguments to the script $1, $2, $3, etc, into a JSON array.

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