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I have written a script to retrieve certain value from file.json . It works if I provide the value to jq select , but the variable doesn't seem to work (or I don't know how to use it).

#!/bin/sh
#this works ***
projectID=$(cat file.json | jq -r '.resource[] | select(.username=="myemail@hotmail.com") | .id')
echo "$projectID"
EMAILID=myemail@hotmail.com
#this does not work *** no value is printed
projectID=$(cat file.json | jq -r '.resource[] | select(.username=="$EMAILID") | .id')
echo "$projectID"
                A related issue: passing bash variable to jq filter has the slightly different syntax jq -r --arg var "$var" '.[$var]' stackoverflow.com/questions/34745451/…
– enharmonic
                Feb 19, 2020 at 18:11
                EMAILIDis between single quotes and does not get expanded. Write it as '"$EMAILID"' to take it out of the single quotes.
– user1934428
                Mar 24 at 8:53

Consider also passing in the shell variable (EMAILID) as a jq variable (here also EMAILID, for the sake of illustration):

   projectID=$(jq -r --arg EMAILID "$EMAILID" '
        .resource[]
        | select(.username==$EMAILID) 
        | .id' file.json)

Postscript

For the record, another possibility would be to use jq's env function for accessing environment variables. For example, consider this sequence of bash commands:

EMAILID=foo@bar.com  # not exported
EMAILID="$EMAILID" jq -n 'env.EMAILID'

The output is a JSON string:

"foo@bar.com"

shell arrays

Unfortunately, shell arrays are a different kettle of fish. Here are two SO resources regarding the ingestion of such arrays:

JQ - create JSON array using bash array with space

Convert bash array to json array and insert to file using jq

This is the only 100%-safe answer; it lets jq properly create the filter using the value, rather than using bash to create a string that jq interprets as a filter. (Consider what happens if the value of EMAILID contains a ).) – chepner Oct 16, 2016 at 2:35 My use case, it works! function n2 { termux-contact-list |jq -r --arg v1 "$1" '.[] | select(.name==$v1)|.number' } Calling: n2 Name1 – Timo Aug 19, 2018 at 7:12 fyi, jq's env function is changed between jq 1.4 and 1.5+, so references to env.EMAILID (in this answer's example) do not work in jq 1.4, thus recommending using the --arg EMAILID "$EMAILID" construct if you need to use jq 1.4 . Hope this helps; just took me a day+ to figure this out for myself ;) – m0j0hn Oct 4, 2018 at 20:15 In my case I had to use parenthesis around the variable name, and since it was inside a select I had to escape it too, like this jq --arg var "${SHELL_VAR}" '.UserPools[] | select(.Name == "\($var)-some-string")' – Federico May 23, 2019 at 2:24 Arguments passed using --arg will be passed as strings. If you want to pass data of a different JSON type, use --argjson, which will parse the string as JSON, e.g., --argjson myObj '{"a": [1, 2]}' – BallpointBen Feb 17, 2020 at 16:59

Little unrelated but I will still put it here, For other practical purposes shell variables can be used as -

value=10
jq  '."key" = "'"$value"'"' file.json

Posting it here as it might help others. In string it might be necessary to pass the quotes to jq. To do the following with jq:

.items[] | select(.name=="string")

in bash you could do

EMAILID=$1
projectID=$(cat file.json | jq -r '.resource[] | select(.username=='\"$EMAILID\"') | .id')

essentially escaping the quotes and passing it on to jq

It's a quote issue, you need :

projectID=$(
  cat file.json | jq -r ".resource[] | select(.username==\"$EMAILID\") | .id"

If you put single quotes to delimit the main string, the shell takes $EMAILID literally.

"Double quote" every literal that contains spaces/metacharacters and every expansion: "$var", "$(command "$var")", "${array[@]}", "a & b". Use 'single quotes' for code or literal $'s: 'Costs $5 US', ssh host 'echo "$HOSTNAME"'. See
http://mywiki.wooledge.org/Quotes
http://mywiki.wooledge.org/Arguments
http://wiki.bash-hackers.org/syntax/words

In my case error as well, similar if not using quotes at all: termux-contact-list |jq -r '.[] | select(.name=='$1')|.number'. Result: jq Compile error – Timo Aug 19, 2018 at 7:04 I'm not sure why this comment has been voted up so many times; yes, double quotes allows for variable expansion, but jq doesn't like single quotes: jq: error: syntax error, unexpected INVALID_CHARACTER (Unix shell quoting issues?) – jdelman May 29, 2019 at 19:01 If your input is a regular json, then its values would be double (not single) quoted. What works: Using double quotes for entire arg of jq and escaping (with `\`) any double quotes within, like @asid answered. – GuSuku Nov 14, 2019 at 17:35 It does NOT work. If you put single quotes to delimit the main string, the shell takes $EMAILID literally. Assuming you set $EMAILID="myemail@hotmail.com", the command cat file.json | jq -r ".resource[] | select(.username=='$EMAILID') | .id" actually runs as cat file.json | jq -r ".resource[] | select(.username==myemail@hotmail.com) | .id" and incur a compile error for jq. – axiqia Nov 26, 2020 at 12:35 This got me on the way to solving my issue. My final working string UPLOADED_ARNS=$(aws devicefarm list-uploads --arn ${PROJECT_ARN} | \ jq ".uploads[] | {arn: .arn, name: .name} | select (.name | contains(\"$FILE_EXTENSION\"))" | \ jq --raw-output '.arn') – Josh Graham Sep 30, 2021 at 18:19 Hey DukeLion, it would be nice to precise the min required version to have this working. Your solution didn't work on jq version 1.6. – tkrishtop Sep 15, 2022 at 9:48 #this works *** projectID=$(cat file.json | jq -r '.resource[] | select(.username=="myemail@hotmail.com") | .id') echo "$projectID" EMAILID=myemail@hotmail.com # Use --arg to pass the variable to jq. This should work: projectID=$(cat file.json | jq --arg EMAILID $EMAILID -r '.resource[] | select(.username=="$EMAILID") | .id') echo "$projectID"

See here, which is where I found this solution: https://github.com/stedolan/jq/issues/626

I had to replace select(.username=="$EMAILID") with select(.username==$EMAILID) to make it work. (I removed the quotes). I'm using jq version 1.6. – Erik Sjölund Jan 17 at 16:43

I know is a bit later to reply, sorry. But that works for me.

export K8S_public_load_balancer_url="$(kubectl get services -n ${TENANT}-production -o wide | grep "ingress-nginx-internal$" | awk '{print $4}')"

And now I am able to fetch and pass the content of the variable to jq

export TF_VAR_public_load_balancer_url="$(aws elbv2 describe-load-balancers --region eu-west-1 | jq -r '.LoadBalancers[] | select (.DNSName == "'$K8S_public_load_balancer_url'") | .LoadBalancerArn')"

In my case I needed to use double quote and quote to access the variable value.

Cheers.

I also faced same issue of variable substitution with jq. I found that --arg is the option which must be used with square bracket [] otherwise it won't work.. I am giving you sample example below:

RUNNER_TOKEN=$(aws secretsmanager get-secret-value --secret-id $SECRET_ID | jq '.SecretString|fromjson' | jq --arg kt $SECRET_KEY -r '.[$kt]' | tr -d '"')

In case where we want to append some string to the variable value and we are using the escaped double quotes, for example appending .crt to a variable CERT_TYPE; the following should work:

$ CERT_TYPE=client.reader
$ cat certs.json | jq -r ".\"${CERT_TYPE}\".crt" #### This will *not* work #####
$ cat certs.json | jq -r ".\"${CERT_TYPE}.crt\""
        

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