passing arguments to jq filter

Question

Here is my config.json:

{
    "env": "dev",
    "dev": {
        "projects" : {
            "prj1": {
                "dependencies": {},
                "description": ""
            }
        }
    }
}

Here are my bash commands:

PRJNAME='prj1'

echo $PRJNAME

jq --arg v "$PRJNAME" '.dev.projects."$v"' config.json 
jq '.dev.projects.prj1' config.json

The output:

prj1
null
{
  "dependencies": {},
  "description": ""
}

So $PRJNAME is prj1, but the first invocation only outputs null.

Can someone help me?

Have you tried removing the `"` around `$v`? Why aren't you using the shell to fill in the variable like `jq ".dev.projects.$PRJNAME" config.json`? — Rambo Ramon, Jan 12 '16 at 14:46
@RamboRamon, using the shell to fill in the variable is error-prone -- think about if it contains characters like quotes that need to be escaped. `jq` is guaranteed to generate syntactically valid output. — Charles Duffy, Jan 12 '16 at 17:45

score 97 · Accepted Answer · answered Jan 12 '16 at 15:22

97

The jq program .dev.projects."$v" in your example will literally try to find a key named "$v". Try the following instead:

jq --arg v "$PRJNAME" '.dev.projects[$v]' config.json

answered Jan 12 '16 at 15:22

What if i want to pass 2 or more params ? Can you show me the example ? – lisi4ok Jan 12 '16 at 15:32
17

You can pass more than one argument by using the `--arg` instruction several times, as in: `jq --arg foo 1 --arg bar 2 -n '[$foo, $bar]'` – Jan 12 '16 at 15:58
1

Is it possible to pass multiple components in the arg? eg `PRJNAME='.dev.projects.prj1' jq --arg v "$PRJNAME" '[$v]' config.json `. Certainly that code doesn't work - it converts the arg to a string surrounded by square brackets, so I'm guessing that it doesn't like the dot notation... – lane Jan 17 '18 at 10:38
2

remove the '' from [$v] – datacruncher Mar 15 '19 at 15:55

score 34 · Answer 2 · edited Aug 10 '20 at 21:30

34

You can use --argjson too when you make your json.

--arg a v       # set variable $a to value <v>;
--argjson a v   # set variable $a to JSON value <v>;

edited Aug 10 '20 at 21:30

Gryu

2,102
2
16
29

answered Mar 03 '17 at 10:44

Sebastien DIAZ

351
3
2

14

You need to use `argjson` when passing a number or boolean which you don't want wrapped in quotes. – Tom Sep 14 '18 at 10:32

score 18 · Answer 3 · answered Mar 12 '18 at 10:27

18

As asked in a comment above there's a way to pass multiple argumets. Maybe there's a more elegant way, but it works.

If you are sure always all keys needed you can use this:

jq --arg key1 $k1 --arg key2 $k2 --arg key3 $k3 --arg key4 $k4 '.[$key1] | .[$key2] | .[$key3] | .[$key4] '

If the key isn't always used you could do it like this:

jq --arg key $k ' if key != "" then .[$key] else . end'

If key sometimes refers to an array:

jq --arg key $k ' if type == "array" then .[$key |tonumber] else .[$key] end'

of course you can combine these!

answered Mar 12 '18 at 10:27

LuWa

191
2
8

1

This works only for one var, when I try two vars it fails: OUTPUT=$(cat test.json | jq --arg env $ENV --arg app $APP '.[$env] .apps .[$app] .bar') – Broshi May 20 '19 at 10:32
.[$key |tonumber] was the trick for me, thanks! – Ignacio Tomas Crespo May 28 '21 at 11:18

score 5 · Answer 4 · answered Aug 12 '20 at 13:29

5

you can do this:


    key="dev.projects.prj1"
    filter=".$key"
    cat config.json | jq $filter

answered Aug 12 '20 at 13:29

Z.San

51
1
1

Or `jq $(echo ".$some_var")` for a one-liner :) thank you @Z.San for such a great the idea! – Nícolas Schirmer Mar 10 '23 at 22:53

score 0 · Answer 5 · answered Dec 19 '22 at 11:45

My example bash command to replace an array in a json file. Everything is in variables:

a=/opt/terminal/conf/config.json ; \
b='["alchohol"]' ; \
c=terminal.canceledInspections.resultSendReasons ; \
cat $a | jq .$c ; \
cat $a | jq --argjson b $b --arg c $c 'getpath($c / ".") = $b' | sponge $a ; \
cat $a | jq .$c

passing arguments to jq filter

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