Error trying to pass both key and value to a jq filter expression

Question

I am trying to query json using jq in bash scripting

file customers.json contains the following

[
  {
    "_id": "5968dd23fc13ae04d9000001",
    "product_name": "sildenafil citrate",
    "supplier": "Wisozk Inc",
    "quantity": 262,
    "unit_cost": "$1047"
  },
  {
    "_id": "5968dd23fc13ae04d9000002",
    "product_name": "Mountain Juniperus ashei",
    "supplier": "Keebler-Hilpert",
    "quantity": 292,
    "unit_cost": "$874"
  },
  {
    "_id": "5968dd23fc13ae04d9000003",
    "product_name": "Dextromathorphan HBr",
    "supplier": "Schmitt-Weissnat",
    "quantity": 211,
    "unit_cost": "$2053"
  }
]

when i run following bash script

key="supplier"
value="Wisozk Inc"
jq ".[] | select(.$key==$value)" customers.json

It throws the following error

jq: error: syntax error, unexpected IDENT, expecting ';' or ')' (Unix shell quoting issues?) at <top-level>, line 1:
.[] | select(.supplier == Wisozk Inc)
jq: 1 compile error

I think the space between "Wisozk" and "Inc" is the problem, what to do?

Inian · Accepted Answer · 2020-09-07T11:28:47.080

Do not let your shell variables to be interpolated in jq under double quotes. The error is because of spaces in the value field, the variable $value undergoes word-splitting by and the filter expression of jq gets two words Wisozk and Inc instead of "Wisozk Inc".

Just pass your variables to jq's context using the --arg field and let it deal with it.

jq --arg k supplier --arg v "Wisozk Inc" 'map(select(.[$k] == $v))' json

This question is in parts duplicate to both the below questions, but not as a whole.

Error trying to pass both key and value to a jq filter expression

1 Answers1