DT
Drizzle Team•11mo ago
XandoR

Difference between query modes

Heyo everyone ! 👋 Is there anyway I can get same result than result1 query using result2 syntax with db.select() ?
const result1 = await db.query.notifications.findMany({
where: eq(notifications.guild_id, interaction.guildId),
with: {
group: true,
},
});

const result2 = await db
.select()
.from(notifications)
.limit(1)
.where(eq(notifications.guild_id, interaction.guildId))
.innerJoin(groups, eq(notifications.group_id, groups._id))
.execute();
const result1 = await db.query.notifications.findMany({
where: eq(notifications.guild_id, interaction.guildId),
with: {
group: true,
},
});

const result2 = await db
.select()
.from(notifications)
.limit(1)
.where(eq(notifications.guild_id, interaction.guildId))
.innerJoin(groups, eq(notifications.group_id, groups._id))
.execute();
Or should I keep using db.query if I want to retrieve notifications with group inside of it ?
1 Reply
Angelelz
Angelelz•11mo ago
You could do it, but you'll need to use json array aggregation sql functions You can take a look at the query that result1 produces to see what functions are used fair warning, it will be complicated if you're not familiar with sql
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