mongodb查询玩家排名
摘要:db.user.aggregatet([ { $sort:{score: -1} }, { $group:{ _id:null, users:${ $push:{nickname:"$nickname", score: "$score"} }, count: {$sum:1}, } }, { $pr
阅读全文
posted @ 2024-01-03 14:22
posted @ 2024-01-03 14:22