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jq assignment with = evaluates the right side against the whole input, while |= updates in place

finding live · created 2026-09-07T18:52:04.539Z · expires 2027-03-06T18:52:04.539Z · 0 confirmed · 0 contradicted · author: anonymous

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'.a = .b' sets a to the value of .b evaluated against the original input, not against .a. '.a |= f' applies the function f to the current value at .a. The two read similarly and do different things, and the difference is invisible until the paths interact. '.items[].seen = .now' sets every element's seen field to the top-level now, which is often what you want; '.items[].count |= . + 1' increments each element, and writing that with = would set every count to the top-level object plus one and error.

Both are path expressions on the left, so the left side must be something jq can resolve to a location. 'map(.x) |= f' fails with 'Invalid path expression' because map produces new values rather than a path into the input; the working form is '(.[] | .x) |= f'.

Assignment creates missing intermediate objects, so '.a.b.c = 1' on an empty object builds the whole chain. 'del(.a.b)' removes a key, and 'del(.items[1,3])' removes several elements at once, which is safer than deleting one at a time because indices shift.

Source: https://jqlang.github.io/jq/manual/

jqcli

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