JSON JQ if without else

Question

I use the following JQ command to filter out the JSON. My requirement is to filter out the JSON message if the expected node is present. Or else, do nothing. Hence, I use if, elif, ....

sed -n "s/.*Service - //p" $1/response.log* |
  jq "if (.requests | length) != 0 then .requests |= map(select(.id == \"123\"))
      elif (.result | length ) != 0  then .result |= map(select(.id== \"123\")) 
      else " "  end" > ~/result.log

Looks like else is mandatory here. I dont want to do anything inside the else block. Is there anyway, I can ignore else or just print some whitespce inside else.

In the above case, it prints double quotes " " in the result file.

Answer 1

You may want to use the idiom:

if CONDITION then WHATEVER else empty end

empty is a filter that outputs nothing at all -- not even null, which is after all something (namely a JSON value). It's a bit like a black hole, only blacker -- it will consume whatever it's offered, but unlike a black hole, it does not even emit Hawking radiation.

In your case, you have an "elif" so using "else empty" is probably what you want, but for reference, the above is exactly equivalent to:

select(CONDITION) | WHATEVER

P.S. My guess is that whatever the goal of the sed command, it could be done more reliably as part of the jq program, perhaps using walk/1.

UPDATE

After the release of jq 1.6, a change was made so that "if without else" has the semantics of "if _ then _ else . end", that is:

if P then Q end === if P then Q else . end