CODEWITHSUNDEEP
bashjq
mrichter
mrichter

Reputation: 103

Bash/JQ - parse error: Expected separator between values at line 1, column 63

I'm having what I suspect is a quoting error in a bash script.

SECRET_VALUE_CMD="curl -s -L -H \"X-Vault-Token: $VAULT_TOKEN\" -X GET \"https://$VAULT_ADDR/v1/secret/$secret_path\""
SECRET_VALUE_RESPONSE=$(eval $SECRET_VALUE_CMD)
SECRET_VALUE=$(echo "$SECRET_VALUE_RESPONSE" | jq --raw-output '.data.value')

When I execute this in my script, I get the following to stderr:

parse error: Expected separator between values at line 1, column 63

and $SECRET_VALUE is blank.

An example of $SECRET_VALUE_RESPONSE is: 

{"request_id":"XXXX-YYYY..,"lease_id":"","renewable":false,"lease_duration":nnnnnn,"data":{"value":"secret-value"},"wrap_info":null,"warnings":null,"auth":null}

I've tried adding escaped quotes around the parameters to eval and echo, but can't seem to find a working combination. Any help would be greatly appreciated!

Upvotes: 2

Views: 3485

Answers (1)

janos
janos

Reputation: 124646

Don't use eval. You could create a function to execute curl, for example:

get_secret_value() {
    curl -s -L -H "X-Vault-Token: $VAULT_TOKEN" -X GET "https://$VAULT_ADDR/v1/secret/$secret_path"
}

secret_value=$(get_secret_value | jq --raw-output '.data.value')

Upvotes: 3

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