Kysely•5mo ago

Is there way to select every field from table using `select` instead of `selectAll`?

I have very wide table and writing every field by hand is time consuming. selectAll does the job, but I also need to populate some relations and build json_aggs. Is there way to either select every column from table (and then build jsonObject) or extend selectAll with some extra fields?

db
  .selectFrom('document')
  .select((eb)=> [
      'dokument.*',
      // some json_agg for example
  ])
  .executeTakeFirstOrThrow()

db
  .selectFrom('document')
  .select((eb)=> [
      'dokument.*',
      // some json_agg for example
  ])
  .executeTakeFirstOrThrow()

Using raw sql it could be done with

SELECT d*, json_obj(w*) as "warehouse"
FROM document d
LEFT JOIN warehouse w ON w.id = d.warehouse_id

SELECT d*, json_obj(w*) as "warehouse"
FROM document d
LEFT JOIN warehouse w ON w.id = d.warehouse_id

please dont waste your time on building this query, just a little hint will be super helpfull! 😉

Solution:

.selectAll('document')
.select(eb => ...)

.selectAll('document')
.select(eb => ...)

...

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1 Reply

Solution

koskimas•5mo ago

.selectAll('document')
.select(eb => ...)

.selectAll('document')
.select(eb => ...)

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Is there way to select every field from table using `select` instead of `selectAll`?