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I'm trying to create a JSON in BASH where one of the fields is based on the result of an earlier command

BIN=$(cat next_entry)
OUTDIR="/tmp/cpupower/${BIN}"
echo $OUTDIR
JSON="'"'{"hostname": "localhost", "outdir": "${OUTDIR}", "port": 20400, "size": 100000}'"'"
echo $JSON

The above script when executed, returns:

/tmp/cpupower/0
, port: 20400, size: 100000}': /tmp/cpupower/0

How can I properly substitute variables inside these multi-quoted strings?

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up vote 3 down vote accepted 
JSON=\''{"hostname": "localhost", "outdir": "'"$OUTDIR"'", "port": 20400, "size": 100000}'\'

That is get out of the single quotes for the expansion of $OUTDIR. We did put that expansion inside double-quotes for good measure even though for a scalar variable assignment it's not strictly necessary.

When you're passing the $JSON variable to echo, quotes are necessary though to disable the split+glob operator. It's also best to avoid echo for arbitrary data:

printf '%s\n' "$JSON"
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Cool. This works! Could you perhaps explain a little more on why this works? What is the significance of the first \'. I don't believe I've seen that on some of the examples I've seen on SO/unix.stackexchange. – Guru Prasad Sep 27 '16 at 14:29
    
Backslash before ' tells the shell to treat ' as any ordinary character, instead of interpreting it as a string delimiter. The backslash can be used in front of a few other characters. For example, \\ will represent the \ character itself and \" means the " character literary. See for example the \n in above printf argument, where printf interprets \n as the line-feed character. – Rein Sep 30 '16 at 16:47
up vote 0 down vote

Stéphane's answer is great, and upvoted. Here's just a tip; instead of doing

BIN=$(cat next_entry)

You can do:

BIN=$(<next_entry)

And thus save spawning an extra process. Read more here.

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