ΠΠΎΠ·Π²ΠΎΠ»ΡΠ΅Ρ Π±ΠΈΡΡ ΡΠ»ΠΎΠΆΠ½ΡΠ΅ Π·Π°ΠΏΡΠΎΡΡ Π½Π° ΠΌΠ°Π»Π΅Π½ΡΠΊΠΈΠ΅ ΠΏΠΎΡΡΠ°ΠΏΠ½ΡΠ΅
WITH "SubQuery" as (SELECT * FROM "log"), "SubQuery2" as (SELECT * FROM "SubQuery" GROUP BY "id") SELECT * FROM "SubQuery2" LEFT JOIN "SubQuery" ON "SubQuery2"."id" = "SubQuery"."id"
Last updated 2 years ago