iterate json into bash variable

Question

If I have a json string in a bash script, how could I convert that into an array?

{ "name": "foo", "id": "123" } { "name": "bar", "id": "456" }

what I want is to be prompted w/ the name, and that tells me what ID (into a variable) that I need to use.

something like

pick your poison:
1) foo
2) bar
#?

If I select 1, then id 123 would go into variable X, else if I select 2 then id 456 would go into variable X

Charles Duffy · Accepted Answer · 2015-03-17 18:51:35Z

5

One of the better approaches is to read the data in question into an associative array -- here, using jq for parsing.

declare -A data=( )
while IFS= read -r id name; do
  data[$id]=$name
done < <(jq -r '@text "\(.id) \(.name)"' <<<"$json_string")

Given this, you can then handle your menu:

echo "Pick your poison:"
for id in "${!data[@]}"; do
  name=${data[$id]}
  printf '%d) %s\n' "$id" "$name"
done
read -p '#?' selection
echo "User selected ${data[$selection]}"

answered Mar 17, 2015 at 18:51

Charles Duffy

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4 Comments

veilig Over a year ago

this is probably exactly what I need. but I keep getting a syntax error. "unexpected token <" done < <(jq -r '@text "\(.id) \(.name)"' <<<"$json_string")

glenn jackman Over a year ago

Make sure you're using bash, not sh

Charles Duffy Over a year ago

@veilig, you tagged your question bash, so I gave you an answer that works with bash. If you need an answer that works with /bin/sh, that's a (sligthly) different question; adjust the tags and I'll change the answer accordingly.

veilig Over a year ago

my bad - was my mistake.

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